Explanation:
Okay, let's see. I need to find the value of x between 0 and 180 degrees such that tan 53° tan 81° tan x° equals tan 53° + tan 81° + tan x°. Hmm, that seems like a trigonometric equation. I remember there are some identities involving tangent functions, maybe the tangent addition formula or something related to product-to-sum formulas? Or maybe there's a way to rearrange the equation to make it easier?

Let me write down the equation again to keep track:

tan 53° * tan 81° * tan x° = tan 53° + tan 81° + tan x°

First, maybe I can denote tan 53° as A and tan 81° as B to simplify the equation. So then the equation becomes:

A * B * tan x = A + B + tan x

Let me rearrange this equation to solve for tan x. Let me bring all terms to one side:

A * B * tan x - tan x - A - B = 0

Factor tan x from the first two terms:

tan x (A * B - 1) - (A + B) = 0

Then:

tan x (A * B - 1) = A + B

So:

tan x = (A + B) / (A * B - 1)

Hmm, that's interesting. Let me write that down:

tan x = (tan 53° + tan 81°) / (tan 53° * tan 81° - 1)

Wait a second, this looks similar to the tangent addition formula. The formula for tan(A + B) is (tan A + tan B) / (1 - tan A tan B). But here, in the denominator, it's tan A tan B - 1, which is negative of (1 - tan A tan B). So maybe there's a connection here.

Let me recall:

tan(A + B) = (tan A + tan B) / (1 - tan A tan B)

Comparing that to our expression for tan x:

tan x = [tan A + tan B] / [tan A tan B - 1] = - [tan A + tan B] / [1 - tan A tan B] = - tan(A + B)

So that means tan x = - tan(A + B). Therefore, tan x = - tan(53° + 81°). Let me compute 53 + 81: that's 134 degrees. So tan x = - tan 134°.

But tan(-θ) = -tan θ, so tan x = tan(-134°). However, tangent has a period of 180°, so tan x = tan(-134° + 180°*k), where k is an integer. So possible solutions for x are -134° + 180°k. But since x is between 0° and 180°, let's find k such that x falls into this interval.

Let me compute for k=1: -134 + 180 = 46°. For k=0: -134°, which is negative, not in the interval. For k=2: -134 + 360 = 226°, which is more than 180°, so not in the interval. Therefore, the only solution in 0° < x < 180° is 46°. But wait, let me check that.

Wait, tan x = -tan 134°, which is tan(-134°). But tan(-134°) = tan(180° - 134° - 180°)? Wait, no. Let me think again. Since tangent has period 180°, tan θ = tan(θ + 180°*k). So tan(-134°) is equal to tan(-134° + 180°) which is tan(46°). Because -134 + 180 is 46. So yes, tan x = tan 46°. Therefore, x = 46° + 180°*k. But since x is between 0 and 180, the only possible solution is 46°. Is that correct?

But let me verify this. Let me compute tan 53° * tan 81° * tan 46° and check if it equals tan 53° + tan 81° + tan 46°. Maybe I should calculate these values numerically to confirm.

First, let me recall approximate values:

tan 53°: I know that tan 60° is √3 ≈ 1.732, tan 45° is 1, so 53° is close to 53.13°, which is a known value in some contexts. Wait, in a 3-4-5 triangle, the angles are approximately 37°, 53°, and 90°, so tan 53° ≈ 4/3 ≈ 1.333... But let me check with calculator.

Wait, if I can't use a calculator, maybe I can recall that tan(37°) ≈ 3/4 = 0.75, tan(53°) ≈ 4/3 ≈ 1.333, tan(60°) ≈ 1.732, tan(81°) is close to tan(90° - 9°) which is cot 9°, and cot 9° is about 6.3138. Let me check tan 81° ≈ tan(90° - 9°) = cot 9°, which is approximately 6.3138. Let me confirm that 81° is 81 degrees.

But if I approximate tan 53° ≈ 1.333, tan 81° ≈ 6.3138, tan 46°: tan(45° + 1°) ≈ 1 + tan 1° / (1 - tan 1°) but that might not be helpful. Alternatively, tan 45° = 1, tan 46° is slightly more than 1. Let me approximate tan 46° ≈ 1.0355 (if I recall that tan 45° is 1, tan 46° ≈ 1.0355). But maybe I can use exact values?

Alternatively, maybe there's a better way. Let me check if 53° + 81° + 46° equals 180°? 53 + 81 + 46 = 53 + 81 is 134, plus 46 is 180. Oh! That's interesting. So 53°, 81°, and 46° add up to 180°, which is a straight line. So maybe there's some relationship here.

Wait, if three angles add up to 180°, then they are angles of a triangle? Not necessarily, unless they are the internal angles. But maybe there's some identity here. Let me think. If x + 53° + 81° = 180°, then x = 46°, which is the value we found earlier. So maybe there's a trigonometric identity that connects this?

Alternatively, let's consider that in a triangle, if angles A, B, C are such that A + B + C = 180°, then tan A + tan B + tan C = tan A tan B tan C. Wait, that's an identity! Yes, in a triangle, since A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C. So if x, 53°, and 81° add up to 180°, then the equation given in the problem would hold. Let me check this.

Suppose that x + 53° + 81° = 180°, so x = 180° - 53° - 81° = 180° - 134° = 46°, as before. Then, according to the identity tan A + tan B + tan C = tan A tan B tan C, then tan 53° + tan 81° + tan 46° = tan 53° tan 81° tan 46°, which is exactly the equation given in the problem. Therefore, x must be 46°, which is the answer.

But why does this identity hold for angles summing to 180°? Let me recall. If A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C. Let me verify this identity.

Starting with A + B + C = π. So, A + B = π - C. Taking tangent of both sides:

tan(A + B) = tan(π - C) = -tan C

But tan(A + B) = (tan A + tan B) / (1 - tan A tan B) = -tan C

Therefore:

(tan A + tan B) / (1 - tan A tan B) = -tan C

Multiply both sides by (1 - tan A tan B):

tan A + tan B = -tan C (1 - tan A tan B)

tan A + tan B = -tan C + tan A tan B tan C

Bring all terms to left-hand side:

tan A + tan B + tan C - tan A tan B tan C = 0

Therefore:

tan A + tan B + tan C = tan A tan B tan C

Which is the identity. So, yes, if three angles add up to π (180°), then the sum of their tangents equals the product of their tangents. So in our problem, we have tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°, which is exactly the identity above if x°, 53°, and 81° add up to 180°, so x = 180° - 53° - 81° = 46°, so x = 46°. Therefore, the answer is 46°.

But let me check with approximate values to make sure that there isn't any mistake in my reasoning.

First, compute tan 53°, tan 81°, tan 46°.

Using approximate values:

tan 53° ≈ 1.3270 (exact value can be found with calculator, but since I don't have one, but maybe I can use known values). However, if I recall that tan(53°) ≈ 1.333, tan(46°) ≈ 1.0355, tan(81°) ≈ 6.3138.

Calculating left-hand side: tan53 * tan81 * tan46 ≈ 1.333 * 6.3138 * 1.0355.

First, 1.333 * 6.3138 ≈ let's calculate 1.333 * 6 = 8, 1.333 * 0.3138 ≈ 0.418, so total approx 8.418. Then 8.418 * 1.0355 ≈ approx 8.418 + 8.418 * 0.0355 ≈ 8.418 + 0.298 ≈ 8.716.

Now the right-hand side: tan53 + tan81 + tan46 ≈ 1.333 + 6.3138 + 1.0355 ≈ 1.333 + 6.3138 is approx 7.6468 + 1.0355 ≈ 8.6823.

Hmm, the left side approx 8.716, right side approx 8.6823. There's a slight discrepancy here. Maybe my approximated values for tan 53°, tan 81°, tan 46° are not precise enough?

Alternatively, maybe my assumption that x=46° is correct, and the minor difference is due to approximated tan values?

Alternatively, maybe I need to calculate more accurately.

Alternatively, perhaps there's another way.

Alternatively, let me check if x = 46° satisfies the original equation.

Alternatively, since we have an identity that if A + B + C = 180°, then tan A + tan B + tan C = tan A tan B tan C, so if x = 46°, then 53 + 81 + 46 = 180, so it should satisfy the equation. Therefore, unless there's an error in the identity, it should be correct.

Wait, perhaps my approximate calculation was wrong. Let me try to compute tan 53°, tan 81°, tan 46° more accurately.

Let me recall that tan(53°) is approximately 1.3270448216, tan(81°) is approximately 6.3137515147, tan(46°) is approximately 1.0355303138.

Calculating left-hand side: 1.3270448216 * 6.3137515147 * 1.0355303138.

First multiply 1.3270448216 * 6.3137515147:

Let me compute that:

Approximately:

First, 1.327 * 6.3137 ≈ Let's compute 1 * 6.3137 = 6.3137, 0.327 * 6.3137 ≈ Let me compute 0.3 * 6.3137 = 1.8941, 0.027 * 6.3137 ≈ 0.1705, so total ≈ 1.8941 + 0.1705 ≈ 2.0646. Therefore total ≈ 6.3137 + 2.0646 ≈ 8.3783.

Now multiply this by tan 46° ≈ 1.03553:

So 8.3783 * 1.03553 ≈ Let me calculate 8.3783 * 1 = 8.3783, plus 8.3783 * 0.03553 ≈ approx.

First, 8.3783 * 0.03 = 0.2513, 8.3783 * 0.00553 ≈ approx 0.0463. Total ≈ 0.2513 + 0.0463 ≈ 0.2976. Therefore total ≈ 8.3783 + 0.2976 ≈ 8.6759.

Now the right-hand side: tan 53 + tan 81 + tan 46 ≈ 1.32704 + 6.31375 + 1.03553 ≈ 1.32704 + 6.31375 = 7.64079 + 1.03553 ≈ 8.67632.

So the left-hand side ≈ 8.6759 and the right-hand side ≈ 8.6763. The difference is about 0.0004, which is due to rounding errors in the approximate values of the tangents. Therefore, this confirms that x = 46° is indeed the correct solution.

Therefore, the answer is 46 degrees.

But let me make sure there are no other solutions. The original equation was tan x = -tan(134°). Since tan is periodic with period 180°, so tan x = tan(-134° + 180°k). The general solution is x = -134° + 180°k. Within 0° < x < 180°, we need to find k such that x is in this interval.

Let me solve for k:

For k=1: -134 + 180 = 46°, which is between 0 and 180.

For k=0: -134°, which is negative, not in the interval.

For k=2: -134 + 360 = 226°, which is more than 180°, so outside the interval.

Therefore, only x=46° is valid. So there are no other solutions. Hence, x=46° is the unique solution in the given interval.

Therefore, the answer is 46°, so \boxed{46}.

**Final Answer**
\boxed{46}

We are given the equation:

$$
\tan 53^\circ \tan 81^\circ \tan x^\circ = \tan 53^\circ + \tan 81^\circ + \tan x^\circ
$$

and are asked to find the value of $ x $ such that $ 0 < x < 180 $.

---

### Step 1: Introduce Variables

Let’s denote:

- $ A = \tan 53^\circ $
- $ B = \tan 81^\circ $
- $ T = \tan x^\circ $

Then the given equation becomes:

$$
A \cdot B \cdot T = A + B + T
$$

---

### Step 2: Rearrange the Equation

Subtract both sides to form:

$$
A \cdot B \cdot T - A - B - T = 0
$$

Bring all terms involving $ T $ to one side:

$$
T(A \cdot B - 1) = A + B
$$

Now solve for $ T $:

$$
\tan x^\circ = \frac{\tan 53^\circ + \tan 81^\circ}{\tan 53^\circ \cdot \tan 81^\circ - 1}
$$

---

### Step 3: Use Trigonometric Identity

Recall the tangent addition formula:

$$
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}
$$

This is similar in form to our expression for $ \tan x^\circ $, but with a sign change in the denominator. Hence:

$$
\tan x^\circ = -\tan(53^\circ + 81^\circ) = -\tan(134^\circ)
$$

Recall that $ \tan(-\theta) = -\tan \theta $, so:

$$
\tan x^\circ = \tan(-134^\circ)
$$

Since $ \tan(\theta) = \tan(\theta + 180^\circ) $, we can write:

$$
\tan x^\circ = \tan(46^\circ)
$$

This is because $ -134^\circ + 180^\circ = 46^\circ $.

So, we get:

$$
x = 46^\circ
$$

---

### Step 4: Confirm That $ x = 46^\circ $ Is a Valid Solution

Let’s check if this satisfies the identity:

In a triangle, if the three angles add up to $ 180^\circ $, it holds that:

$$
\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C
$$

Check:

$$
53^\circ + 81^\circ + 46^\circ = 180^\circ
$$

So the identity applies, and:

$$
\tan 53^\circ + \tan 81^\circ + \tan 46^\circ = \tan 53^\circ \cdot \tan 81^\circ \cdot \tan 46^\circ
$$

Thus, the original equation is satisfied.

---

### Step 5: Uniqueness of the Solution

The general solution for $ \tan x^\circ = \tan(-134^\circ) $ is:

$$
x = -134^\circ + 180^\circ k \quad \text{for integer } k
$$

We seek $ x \in (0^\circ, 180^\circ) $. Trying $ k = 1 $:

$$
x = -134^\circ + 180^\circ = 46^\circ
$$

Trying $ k = 0 $:

$$
x = -134^\circ \quad \text{(not in range)}
$$

Trying $ k = 2 $:

$$
x = -134^\circ + 360^\circ = 226^\circ \quad \text{(not in range)}
$$

Thus, the only value for $ x $ in the desired interval is $ \boxed{46} $.

---

### Final Answer

$$
\boxed{46}
$$

[Assistant]
Explanation:
Okay, let's see. I need to find the value of x between 0 and 180 degrees such that tan 53° tan 81° tan x° equals tan 53° + tan 81° + tan x°. Hmm, that seems like a trigonometric equation. I remember there are some identities involving tangent functions, maybe the tangent addition formula or something related to product-to-sum formulas? Or maybe there's a way to rearrange the equation to make it easier?

Let me write down the equation again to keep track:

tan 53° * tan 81° * tan x° = tan 53° + tan 81° + tan x°

First, maybe I can denote tan 53° as A and tan 81° as B to simplify the equation. So then the equation becomes:

A * B * tan x = A + B + tan x

Let me rearrange this equation to solve for tan x. Let me bring all terms to one side:

A * B * tan x - tan x - A - B = 0

Factor tan x from the first two terms:

tan x (A * B - 1) - (A + B) = 0

Then:

tan x (A * B - 1) = A + B

So:

tan x = (A + B) / (A * B - 1)

Hmm, that's interesting. Let me write that down:

tan x = (tan 53° + tan 81°) / (tan 53° * tan 81° - 1)

Wait a second, this looks similar to the tangent addition formula. The formula for tan(A + B) is (tan A + tan B) / (1 - tan A tan B). But here, in the denominator, it's tan A tan B - 1, which is negative of (1 - tan A tan B). So maybe there's a connection here.

Let me recall:

tan(A + B) = (tan A + tan B) / (1 - tan A tan B)

Comparing that to our expression for tan x:

tan x = [tan A + tan B] / [tan A tan B - 1] = - [tan A + tan B] / [1 - tan A tan B] = - tan(A + B)

So that means tan x = - tan(A + B). Therefore, tan x = - tan(53° + 81°). Let me compute 53 + 81: that's 134 degrees. So tan x = - tan 134°.

But tan(-θ) = -tan θ, so tan x = tan(-134°). However, tangent has a period of 180°, so tan x = tan(-134° + 180°*k), where k is an integer. So possible solutions for x are -134° + 180°k. But since x is between 0° and 180°, let's find k such that x falls into this interval.

Let me compute for k=1: -134 + 180 = 46°. For k=0: -134°, which is negative, not in the interval. For k=2: -134 + 360 = 226°, which is more than 180°, so not in the interval. Therefore, the only solution in 0° < x < 180° is 46°. But wait, let me check that.

Wait, tan x = -tan 134°, which is tan(-134°). But tan(-134°) = tan(180° - 134° - 180°)? Wait, no. Let me think again. Since tangent has period 180°, tan θ = tan(θ + 180°*k). So tan(-134°) is equal to tan(-134° + 180°) which is tan(46°). Because -134 + 180 is 46. So yes, tan x = tan 46°. Therefore, x = 46° + 180°*k. But since x is between 0 and 180, the only possible solution is 46°. Is that correct?

But let me verify this. Let me compute tan 53° * tan 81° * tan 46° and check if it equals tan 53° + tan 81° + tan 46°. Maybe I should calculate these values numerically to confirm.

First, let me recall approximate values:

tan 53°: I know that tan 60° is √3 ≈ 1.732, tan 45° is 1, so 53° is close to 53.13°, which is a known value in some contexts. Wait, in a 3-4-5 triangle, the angles are approximately 37°, 53°, and 90°, so tan 53° ≈ 4/3 ≈ 1.333... But let me check with calculator.

Wait, if I can't use a calculator, maybe I can recall that tan(37°) ≈ 3/4 = 0.75, tan(53°) ≈ 4/3 ≈ 1.333, tan(60°) ≈ 1.732, tan(81°) is close to tan(90° - 9°) which is cot 9°, and cot 9° is about 6.3138. Let me check tan 81° ≈ tan(90° - 9°) = cot 9°, which is approximately 6.3138. Let me confirm that 81° is 81 degrees.

But if I approximate tan 53° ≈ 1.333, tan 81° ≈ 6.3138, tan 46°: tan(45° + 1°) ≈ 1 + tan 1° / (1 - tan 1°) but that might not be helpful. Alternatively, tan 45° = 1, tan 46° is slightly more than 1. Let me approximate tan 46° ≈ 1.0355 (if I recall that tan 45° is 1, tan 46° ≈ 1.0355). But maybe I can use exact values?

Alternatively, maybe there's a better way. Let me check if 53° + 81° + 46° equals 180°? 53 + 81 + 46 = 53 + 81 is 134, plus 46 is 180. Oh! That's interesting. So 53°, 81°, and 46° add up to 180°, which is a straight line. So maybe there's some relationship here.

Wait, if three angles add up to 180°, then they are angles of a triangle? Not necessarily, unless they are the internal angles. But maybe there's some identity here. Let me think. If x + 53° + 81° = 180°, then x = 46°, which is the value we found earlier. So maybe there's a trigonometric identity that connects this?

Alternatively, let's consider that in a triangle, if angles A, B, C are such that A + B + C = 180°, then tan A + tan B + tan C = tan A tan B tan C. Wait, that's an identity! Yes, in a triangle, since A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C. So if x, 53°, and 81° add up to 180°, then the equation given in the problem would hold. Let me check this.

Suppose that x + 53° + 81° = 180°, so x = 180° - 53° - 81° = 180° - 134° = 46°, as before. Then, according to the identity tan A + tan B + tan C = tan A tan B tan C, then tan 53° + tan 81° + tan 46° = tan 53° tan 81° tan 46°, which is exactly the equation given in the problem. Therefore, x must be 46°, which is the answer.

But why does this identity hold for angles summing to 180°? Let me recall. If A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C. Let me verify this identity.

Starting with A + B + C = π. So, A + B = π - C. Taking tangent of both sides:

tan(A + B) = tan(π - C) = -tan C

But tan(A + B) = (tan A + tan B) / (1 - tan A tan B) = -tan C

Therefore:

(tan A + tan B) / (1 - tan A tan B) = -tan C

Multiply both sides by (1 - tan A tan B):

tan A + tan B = -tan C (1 - tan A tan B)

tan A + tan B = -tan C + tan A tan B tan C

Bring all terms to left-hand side:

tan A + tan B + tan C - tan A tan B tan C = 0

Therefore:

tan A + tan B + tan C = tan A tan B tan C

Which is the identity. So, yes, if three angles add up to π (180°), then the sum of their tangents equals the product of their tangents. So in our problem, we have tan 53° tan 81° tan x° = tan 53° + tan 81° + tan x°, which is exactly the identity above if x°, 53°, and 81° add up to 180°, so x = 180° - 53° - 81° = 46°, so x = 46°. Therefore, the answer is 46°.

But let me check with approximate values to make sure that there isn't any mistake in my reasoning.

First, compute tan 53°, tan 81°, tan 46°.

Using approximate values:

tan 53° ≈ 1.3270 (exact value can be found with calculator, but since I don't have one, but maybe I can use known values). However, if I recall that tan(53°) ≈ 1.333, tan(46°) ≈ 1.0355, tan(81°) ≈ 6.3138.

Calculating left-hand side: tan53 * tan81 * tan46 ≈ 1.333 * 6.3138 * 1.0355.

First, 1.333 * 6.3138 ≈ let's calculate 1.333 * 6 = 8, 1.333 * 0.3138 ≈ 0.418, so total approx 8.418. Then 8.418 * 1.0355 ≈ approx 8.418 + 8.418 * 0.0355 ≈ 8.418 + 0.298 ≈ 8.716.

Now the right-hand side: tan53 + tan81 + tan46 ≈ 1.333 + 6.3138 + 1.0355 ≈ 1.333 + 6.3138 is approx 7.6468 + 1.0355 ≈ 8.6823.

Hmm, the left side approx 8.716, right side approx 8.6823. There's a slight discrepancy here. Maybe my approximated values for tan 53°, tan 81°, tan 46° are not precise enough?

Alternatively, maybe my assumption that x=46° is correct, and the minor difference is due to approximated tan values?

Alternatively, maybe I need to calculate more accurately.

Alternatively, perhaps there's another way.

Alternatively, let me check if x = 46° satisfies the original equation.

Alternatively, since we have an identity that if A + B + C = 180°, then tan A + tan B + tan C = tan A tan B tan C, so if x = 46°, then 53 + 81 + 46 = 180, so it should satisfy the equation. Therefore, unless there's an error in the identity, it should be correct.

Wait, perhaps my approximate calculation was wrong. Let me try to compute tan 53°, tan 81°, tan 46° more accurately.

Let me recall that tan(53°) is approximately 1.3270448216, tan(81°) is approximately 6.3137515147, tan(46°) is approximately 1.0355303138.

Calculating left-hand side: 1.3270448216 * 6.3137515147 * 1.0355303138.

First multiply 1.3270448216 * 6.3137515147:

Let me compute that:

Approximately:

First, 1.327 * 6.3137 ≈ Let's compute 1 * 6.3137 = 6.3137, 0.327 * 6.3137 ≈ Let me compute 0.3 * 6.3137 = 1.8941, 0.027 * 6.3137 ≈ 0.1705, so total ≈ 1.8941 + 0.1705 ≈ 2.0646. Therefore total ≈ 6.3137 + 2.0646 ≈ 8.3783.

Now multiply this by tan 46° ≈ 1.03553:

So 8.3783 * 1.03553 ≈ Let me calculate 8.3783 * 1 = 8.3783, plus 8.3783 * 0.03553 ≈ approx.

First, 8.3783 * 0.03 = 0.2513, 8.3783 * 0.00553 ≈ approx 0.0463. Total ≈ 0.2513 + 0.0463 ≈ 0.2976. Therefore total ≈ 8.3783 + 0.2976 ≈ 8.6759.

Now the right-hand side: tan 53 + tan 81 + tan 46 ≈ 1.32704 + 6.31375 + 1.03553 ≈ 1.32704 + 6.31375 = 7.64079 + 1.03553 ≈ 8.67632.

So the left-hand side ≈ 8.6759 and the right-hand side ≈ 8.6763. The difference is about 0.0004, which is due to rounding errors in the approximate values of the tangents. Therefore, this confirms that x = 46° is indeed the correct solution.

Therefore, the answer is 46 degrees.

But let me make sure there are no other solutions. The original equation was tan x = -tan(134°). Since tan is periodic with period 180°, so tan x = tan(-134° + 180°k). The general solution is x = -134° + 180°k. Within 0° < x < 180°, we need to find k such that x is in this interval.

Let me solve for k:

For k=1: -134 + 180 = 46°, which is between 0 and 180.

For k=0: -134°, which is negative, not in the interval.

For k=2: -134 + 360 = 226°, which is more than 180°, so outside the interval.

Therefore, only x=46° is valid. So there are no other solutions. Hence, x=46° is the unique solution in the given interval.

Therefore, the answer is 46°, so \boxed{46}.

**Final Answer**
\boxed{46}

We are given the equation:

$$
\tan 53^\circ \tan 81^\circ \tan x^\circ = \tan 53^\circ + \tan 81^\circ + \tan x^\circ
$$

and are asked to find the value of $ x $ such that $ 0 < x < 180 $.

---

### Step 1: Introduce Variables

Let’s denote:

- $ A = \tan 53^\circ $
- $ B = \tan 81^\circ $
- $ T = \tan x^\circ $

Then the given equation becomes:

$$
A \cdot B \cdot T = A + B + T
$$

---

### Step 2: Rearrange the Equation

Subtract both sides to form:

$$
A \cdot B \cdot T - A - B - T = 0
$$

Bring all terms involving $ T $ to one side:

$$
T(A \cdot B - 1) = A + B
$$

Now solve for $ T $:

$$
\tan x^\circ = \frac{\tan 53^\circ + \tan 81^\circ}{\tan 53^\circ \cdot \tan 81^\circ - 1}
$$

---

### Step 3: Use Trigonometric Identity

Recall the tangent addition formula:

$$
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}
$$

This is similar in form to our expression for $ \tan x^\circ $, but with a sign change in the denominator. Hence:

$$
\tan x^\circ = -\tan(53^\circ + 81^\circ) = -\tan(134^\circ)
$$

Recall that $ \tan(-\theta) = -\tan \theta $, so:

$$
\tan x^\circ = \tan(-134^\circ)
$$

Since $ \tan(\theta) = \tan(\theta + 180^\circ) $, we can write:

$$
\tan x^\circ = \tan(46^\circ)
$$

This is because $ -134^\circ + 180^\circ = 46^\circ $.

So, we get:

$$
x = 46^\circ
$$

---

### Step 4: Confirm That $ x = 46^\circ $ Is a Valid Solution

Let’s check if this satisfies the identity:

In a triangle, if the three angles add up to $ 180^\circ $, it holds that:

$$
\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C
$$

Check:

$$
53^\circ + 81^\circ + 46^\circ = 180^\circ
$$

So the identity applies, and:

$$
\tan 53^\circ + \tan 81^\circ + \tan 46^\circ = \tan 53^\circ \cdot \tan 81^\circ \cdot \tan 46^\circ
$$

Thus, the original equation is satisfied.

---

### Step 5: Uniqueness of the Solution

The general solution for $ \tan x^\circ = \tan(-134^\circ) $ is:

$$
x = -134^\circ + 180^\circ k \quad \text{for integer } k
$$

We seek $ x \in (0^\circ, 180^\circ) $. Trying $ k = 1 $:

$$
x = -134^\circ + 180^\circ = 46^\circ
$$

Trying $ k = 0 $:

$$
x = -134^\circ \quad \text{(not in range)}
$$

Trying $ k = 2 $:

$$
x = -134^\circ + 360^\circ = 226^\circ \quad \text{(not in range)}
$$

Thus, the only value for $ x $ in the desired interval is $ \boxed{46} $.

---

### Final Answer

$$
\boxed{46}
$$

Solution:
Isolating $\tan x^\circ,$ we find
\begin{align*}
\tan x &= \frac{\tan 53^\circ + \tan 81^\circ}{\tan 53^\circ \tan 81^\circ - 1} \\
&= -\frac{\tan 53^\circ + \tan 81^\circ}{1 - \tan 53^\circ \tan 81^\circ}.
\end{align*}From the angle addition formula, this is equal to
\[-\tan (53^\circ + 81^\circ) = -\tan 134^\circ = \tan 46^\circ.\]Therefore, $x = \boxed{46}.$


Explanation:
Okay, so I have this problem here: Let z be a complex number such that z^5 = 1 and z ≠ 1. I need to compute the expression z + 1/z + z² + 1/z². Hmm, let me think about how to approach this.

First, I remember that if z^5 = 1, then z is a 5th root of unity. Since z ≠ 1, it's one of the primitive 5th roots of unity. There are four such roots, right? They are e^(2πik/5) for k = 1, 2, 3, 4. But maybe I don't need to get into Euler's formula yet. Let me see.

The expression to compute is z + 1/z + z² + 1/z². Maybe I can combine terms or find some relationships using the fact that z^5 = 1. Let me recall that if z^5 = 1, then z^5 - 1 = 0, which factors as (z - 1)(z^4 + z³ + z² + z + 1) = 0. Since z ≠ 1, then z^4 + z³ + z² + z + 1 = 0. That seems useful.

So, maybe I can find an expression for z + 1/z + z² + 1/z² using this equation. Let me note that 1/z is equal to z^4, because z^5 = 1 implies that z^5 = 1, so multiplying both sides by z^{-1} gives z^4 = z^{-1}. Therefore, 1/z = z^4. Similarly, 1/z² = z³. So substituting these into the original expression:

z + z^4 + z² + z³.

Therefore, the expression simplifies to z + z² + z³ + z^4. But from the equation above, z^4 + z³ + z² + z + 1 = 0. Therefore, z + z² + z³ + z^4 = -1. So the value of the original expression is -1? Wait, that seems straightforward, but let me check again.

Original expression: z + 1/z + z² + 1/z². If 1/z = z^4 and 1/z² = z³, then replacing those gives z + z^4 + z² + z³. Which is exactly z + z² + z³ + z^4. And from the equation z^4 + z³ + z² + z + 1 = 0, so subtracting 1 from both sides gives z^4 + z³ + z² + z = -1. Therefore, the original expression is -1. Is that correct?

Wait, but maybe I should verify this with a concrete example. Let me pick a specific 5th root of unity and compute the expression numerically to check.

Let me take z = e^(2πi/5). Let me compute z + 1/z + z² + 1/z².

First, compute z. e^(2πi/5) is a complex number on the unit circle. Similarly, 1/z is e^(-2πi/5). Similarly, z² is e^(4πi/5), and 1/z² is e^(-4πi/5). Therefore, the expression becomes e^(2πi/5) + e^(-2πi/5) + e^(4πi/5) + e^(-4πi/5).

But e^(θ) + e^(-θ) = 2cosθ. So this is 2cos(2π/5) + 2cos(4π/5). Let me compute these values numerically.

cos(2π/5) is approximately cos(72°) ≈ 0.3090. Then 2cos(72°) ≈ 0.618.

cos(4π/5) is cos(144°) ≈ -0.8090. Then 2cos(144°) ≈ -1.618.

Adding them together: 0.618 - 1.618 = -1.0. So indeed, the numerical value is -1. That matches the previous result. So the answer is -1?

But wait, maybe I should check another root? Let me take z = e^(4πi/5). Then 1/z is e^(-4πi/5), z² is e^(8πi/5) which is e^(-2πi/5) because 8π/5 - 2π = 8π/5 - 10π/5 = -2π/5. Similarly, 1/z² is e^(-8πi/5) = e^(2πi/5). Therefore, the expression is e^(4πi/5) + e^(-4πi/5) + e^(-2πi/5) + e^(2πi/5). Which again is 2cos(4π/5) + 2cos(2π/5) = same as before, which is -1. So same result.

Therefore, it seems that regardless of which root we take (other than 1), the expression evaluates to -1. Therefore, the answer should be -1. But let me think again if there is another way to get this.

Alternatively, since z^5 = 1, so z ≠ 0, so maybe we can consider writing the original expression as:

z + z^4 + z² + z³. Which is same as (z + z² + z³ + z^4). And since z^4 + z³ + z² + z + 1 = 0, then z + z² + z³ + z^4 = -1. Therefore, the expression is -1. Therefore, answer is -1. So that seems straightforward.

Wait, but maybe I need to check if there's a mistake in assuming that 1/z = z^4? Let me verify that. Since z^5 = 1, so z^5 - 1 = 0. Therefore, z^5 = 1, so multiplying both sides by z^{-1}, we get z^4 = z^{-1}. Therefore, yes, 1/z = z^4. Similarly, 1/z² = z^{-2} = z^{5 - 2} = z^3. Therefore, correct.

Therefore, substituting, the original expression becomes z + z^4 + z² + z³. Which is the sum of all roots except 1, which equals -1. So that's correct. Therefore, the answer is \boxed{-1}.

But let me check if there is another approach. Maybe using minimal polynomials or something else. Let me think.

Alternatively, since z is a root of z^4 + z³ + z² + z + 1 = 0. Maybe we can compute the expression z + z² + z³ + z^4, which is equal to -1, as above. Alternatively, maybe we can group terms.

But perhaps we can also think about z + 1/z. Let me note that z + 1/z = z + z^4. Similarly, z² + 1/z² = z² + z³. Therefore, the original expression is (z + z^4) + (z² + z³). Which is the same as the sum of all roots except 1, which is -1. Therefore, same result.

Alternatively, maybe if I let S = z + z² + z³ + z^4, then S = -1. Therefore, the answer is -1.

Alternatively, suppose I want to compute z + 1/z + z² + 1/z². Let me compute z + 1/z first. Let me denote w = z + 1/z. Similarly, let me compute w² or something. But maybe that's more complicated. Let me try.

So w = z + 1/z = z + z^4. Then, compute w squared: w² = (z + z^4)^2 = z² + 2z^5 + z^8. But z^5 = 1, so z^5 = 1. So z^8 = z^(5 + 3) = z^3. Therefore, w² = z² + 2(1) + z³ = z² + z³ + 2. Similarly, maybe we can compute something else. However, we also have the original expression as w + (z² + 1/z²). Let me denote v = z² + 1/z² = z² + z³. Therefore, the original expression is w + v. But we have w² = z² + z³ + 2, so z² + z³ = w² - 2. Therefore, original expression is w + (w² - 2) = w² + w - 2. But I need to find w. But maybe this is getting more complicated. Alternatively, since we already know that z + z² + z³ + z^4 = -1, which is w + v = -1, and if we have that w² = v + 2, then substituting v = -1 - w, so w² = (-1 - w) + 2 => w² = 1 - w => w² + w - 1 = 0. Solving quadratic equation: w = [-1 ± sqrt(1 + 4)] / 2 = [-1 ± sqrt(5)] / 2. But since w = z + z^4, which is 2cos(2π/5) ≈ 0.618, which is positive. Therefore, the positive root would be [-1 + sqrt(5)] / 2 ≈ (-1 + 2.236)/2 ≈ 0.618, which matches. However, then original expression is w + v = -1. So maybe this is redundant, but it shows that even if we take this approach, we still get that the original expression is -1. Therefore, regardless of approach, the answer is -1.

Alternatively, if I think of the original expression as z + z^4 + z² + z³, which is the sum of all roots except 1, which is known to be -1. Therefore, the answer is -1.

Therefore, I think I can be confident that the answer is \boxed{-1}.

**Final Answer**
\boxed{-1}

To evaluate the expression
$$
z + \frac{1}{z} + z^2 + \frac{1}{z^2}
$$
given that $ z^5 = 1 $ and $ z \neq 1 $, we begin by recognizing that $ z $ is a **primitive 5th root of unity**. That is, $ z \neq 1 $, but $ z^5 = 1 $, which implies that $ z $ lies on the unit circle in the complex plane and satisfies the polynomial equation:

$$
z^5 - 1 = 0
$$

This factors as:

$$
(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0
$$

Since $ z \neq 1 $, the roots of the equation are the roots of the quartic polynomial:

$$
z^4 + z^3 + z^2 + z + 1 = 0
$$

This gives us a key identity:

$$
z^4 + z^3 + z^2 + z = -1
$$

---

Now, the expression in the problem is:

$$
z + \frac{1}{z} + z^2 + \frac{1}{z^2}
$$

We can simplify the terms $ \frac{1}{z} $ and $ \frac{1}{z^2} $. Since $ z^5 = 1 $, it follows that $ \frac{1}{z} = z^{-1} = z^{5 - 1} = z^4 $, and $ \frac{1}{z^2} = z^{-2} = z^{5 - 2} = z^3 $. Substituting these into the expression gives:

$$
z + z^4 + z^2 + z^3
$$

Which is exactly:

$$
z + z^2 + z^3 + z^4
$$

This sum is the sum of all roots of the equation $ z^4 + z^3 + z^2 + z + 1 = 0 $, except for $ z = 1 $. According to the identity we derived earlier, this sum is:

$$
z + z^2 + z^3 + z^4 = -1
$$

---

Therefore, the value of the original expression is:

$$
\boxed{-1}
$$

[Assistant]
Explanation:
Okay, so I have this problem here: Let z be a complex number such that z^5 = 1 and z ≠ 1. I need to compute the expression z + 1/z + z² + 1/z². Hmm, let me think about how to approach this.

First, I remember that if z^5 = 1, then z is a 5th root of unity. Since z ≠ 1, it's one of the primitive 5th roots of unity. There are four such roots, right? They are e^(2πik/5) for k = 1, 2, 3, 4. But maybe I don't need to get into Euler's formula yet. Let me see.

The expression to compute is z + 1/z + z² + 1/z². Maybe I can combine terms or find some relationships using the fact that z^5 = 1. Let me recall that if z^5 = 1, then z^5 - 1 = 0, which factors as (z - 1)(z^4 + z³ + z² + z + 1) = 0. Since z ≠ 1, then z^4 + z³ + z² + z + 1 = 0. That seems useful.

So, maybe I can find an expression for z + 1/z + z² + 1/z² using this equation. Let me note that 1/z is equal to z^4, because z^5 = 1 implies that z^5 = 1, so multiplying both sides by z^{-1} gives z^4 = z^{-1}. Therefore, 1/z = z^4. Similarly, 1/z² = z³. So substituting these into the original expression:

z + z^4 + z² + z³.

Therefore, the expression simplifies to z + z² + z³ + z^4. But from the equation above, z^4 + z³ + z² + z + 1 = 0. Therefore, z + z² + z³ + z^4 = -1. So the value of the original expression is -1? Wait, that seems straightforward, but let me check again.

Original expression: z + 1/z + z² + 1/z². If 1/z = z^4 and 1/z² = z³, then replacing those gives z + z^4 + z² + z³. Which is exactly z + z² + z³ + z^4. And from the equation z^4 + z³ + z² + z + 1 = 0, so subtracting 1 from both sides gives z^4 + z³ + z² + z = -1. Therefore, the original expression is -1. Is that correct?

Wait, but maybe I should verify this with a concrete example. Let me pick a specific 5th root of unity and compute the expression numerically to check.

Let me take z = e^(2πi/5). Let me compute z + 1/z + z² + 1/z².

First, compute z. e^(2πi/5) is a complex number on the unit circle. Similarly, 1/z is e^(-2πi/5). Similarly, z² is e^(4πi/5), and 1/z² is e^(-4πi/5). Therefore, the expression becomes e^(2πi/5) + e^(-2πi/5) + e^(4πi/5) + e^(-4πi/5).

But e^(θ) + e^(-θ) = 2cosθ. So this is 2cos(2π/5) + 2cos(4π/5). Let me compute these values numerically.

cos(2π/5) is approximately cos(72°) ≈ 0.3090. Then 2cos(72°) ≈ 0.618.

cos(4π/5) is cos(144°) ≈ -0.8090. Then 2cos(144°) ≈ -1.618.

Adding them together: 0.618 - 1.618 = -1.0. So indeed, the numerical value is -1. That matches the previous result. So the answer is -1?

But wait, maybe I should check another root? Let me take z = e^(4πi/5). Then 1/z is e^(-4πi/5), z² is e^(8πi/5) which is e^(-2πi/5) because 8π/5 - 2π = 8π/5 - 10π/5 = -2π/5. Similarly, 1/z² is e^(-8πi/5) = e^(2πi/5). Therefore, the expression is e^(4πi/5) + e^(-4πi/5) + e^(-2πi/5) + e^(2πi/5). Which again is 2cos(4π/5) + 2cos(2π/5) = same as before, which is -1. So same result.

Therefore, it seems that regardless of which root we take (other than 1), the expression evaluates to -1. Therefore, the answer should be -1. But let me think again if there is another way to get this.

Alternatively, since z^5 = 1, so z ≠ 0, so maybe we can consider writing the original expression as:

z + z^4 + z² + z³. Which is same as (z + z² + z³ + z^4). And since z^4 + z³ + z² + z + 1 = 0, then z + z² + z³ + z^4 = -1. Therefore, the expression is -1. Therefore, answer is -1. So that seems straightforward.

Wait, but maybe I need to check if there's a mistake in assuming that 1/z = z^4? Let me verify that. Since z^5 = 1, so z^5 - 1 = 0. Therefore, z^5 = 1, so multiplying both sides by z^{-1}, we get z^4 = z^{-1}. Therefore, yes, 1/z = z^4. Similarly, 1/z² = z^{-2} = z^{5 - 2} = z^3. Therefore, correct.

Therefore, substituting, the original expression becomes z + z^4 + z² + z³. Which is the sum of all roots except 1, which equals -1. So that's correct. Therefore, the answer is \boxed{-1}.

But let me check if there is another approach. Maybe using minimal polynomials or something else. Let me think.

Alternatively, since z is a root of z^4 + z³ + z² + z + 1 = 0. Maybe we can compute the expression z + z² + z³ + z^4, which is equal to -1, as above. Alternatively, maybe we can group terms.

But perhaps we can also think about z + 1/z. Let me note that z + 1/z = z + z^4. Similarly, z² + 1/z² = z² + z³. Therefore, the original expression is (z + z^4) + (z² + z³). Which is the same as the sum of all roots except 1, which is -1. Therefore, same result.

Alternatively, maybe if I let S = z + z² + z³ + z^4, then S = -1. Therefore, the answer is -1.

Alternatively, suppose I want to compute z + 1/z + z² + 1/z². Let me compute z + 1/z first. Let me denote w = z + 1/z. Similarly, let me compute w² or something. But maybe that's more complicated. Let me try.

So w = z + 1/z = z + z^4. Then, compute w squared: w² = (z + z^4)^2 = z² + 2z^5 + z^8. But z^5 = 1, so z^5 = 1. So z^8 = z^(5 + 3) = z^3. Therefore, w² = z² + 2(1) + z³ = z² + z³ + 2. Similarly, maybe we can compute something else. However, we also have the original expression as w + (z² + 1/z²). Let me denote v = z² + 1/z² = z² + z³. Therefore, the original expression is w + v. But we have w² = z² + z³ + 2, so z² + z³ = w² - 2. Therefore, original expression is w + (w² - 2) = w² + w - 2. But I need to find w. But maybe this is getting more complicated. Alternatively, since we already know that z + z² + z³ + z^4 = -1, which is w + v = -1, and if we have that w² = v + 2, then substituting v = -1 - w, so w² = (-1 - w) + 2 => w² = 1 - w => w² + w - 1 = 0. Solving quadratic equation: w = [-1 ± sqrt(1 + 4)] / 2 = [-1 ± sqrt(5)] / 2. But since w = z + z^4, which is 2cos(2π/5) ≈ 0.618, which is positive. Therefore, the positive root would be [-1 + sqrt(5)] / 2 ≈ (-1 + 2.236)/2 ≈ 0.618, which matches. However, then original expression is w + v = -1. So maybe this is redundant, but it shows that even if we take this approach, we still get that the original expression is -1. Therefore, regardless of approach, the answer is -1.

Alternatively, if I think of the original expression as z + z^4 + z² + z³, which is the sum of all roots except 1, which is known to be -1. Therefore, the answer is -1.

Therefore, I think I can be confident that the answer is \boxed{-1}.

**Final Answer**
\boxed{-1}

To evaluate the expression
$$
z + \frac{1}{z} + z^2 + \frac{1}{z^2}
$$
given that $ z^5 = 1 $ and $ z \neq 1 $, we begin by recognizing that $ z $ is a **primitive 5th root of unity**. That is, $ z \neq 1 $, but $ z^5 = 1 $, which implies that $ z $ lies on the unit circle in the complex plane and satisfies the polynomial equation:

$$
z^5 - 1 = 0
$$

This factors as:

$$
(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0
$$

Since $ z \neq 1 $, the roots of the equation are the roots of the quartic polynomial:

$$
z^4 + z^3 + z^2 + z + 1 = 0
$$

This gives us a key identity:

$$
z^4 + z^3 + z^2 + z = -1
$$

---

Now, the expression in the problem is:

$$
z + \frac{1}{z} + z^2 + \frac{1}{z^2}
$$

We can simplify the terms $ \frac{1}{z} $ and $ \frac{1}{z^2} $. Since $ z^5 = 1 $, it follows that $ \frac{1}{z} = z^{-1} = z^{5 - 1} = z^4 $, and $ \frac{1}{z^2} = z^{-2} = z^{5 - 2} = z^3 $. Substituting these into the expression gives:

$$
z + z^4 + z^2 + z^3
$$

Which is exactly:

$$
z + z^2 + z^3 + z^4
$$

This sum is the sum of all roots of the equation $ z^4 + z^3 + z^2 + z + 1 = 0 $, except for $ z = 1 $. According to the identity we derived earlier, this sum is:

$$
z + z^2 + z^3 + z^4 = -1
$$

---

Therefore, the value of the original expression is:

$$
\boxed{-1}
$$

Solution:
Since $z^5 = 1,$ $z^5 - 1 = 0,$ which factors as
\[(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0.\]Since $z \neq 1,$ $z^4 + z^3 + z^2 + z + 1 = 0.$

Then
\[z + \frac{1}{z} + z^2 + \frac{1}{z^2} = \frac{z^3 + z + z^4 + 1}{z^2} = \frac{-z^2}{z^2} = \boxed{-1}.\]


Explanation:
Okay, so I need to convert the binary number 10101001110₂ to octal. Hmm, let me remember how to do this. I think there's a method where you can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent. Let me try to recall the exact steps.

First, I know that each digit in an octal number represents three binary digits because 8 is 2³. So, if I can split the binary number into groups of three bits, each group can be converted to an octal digit. But wait, the binary number given is 10101001110. Let me count the digits to see how many there are. Let me write it down: 1 0 1 0 1 0 0 1 1 1 0. Let me count: 1 (1), 0 (2), 1 (3), 0 (4), 1 (5), 0 (6), 0 (7), 1 (8), 1 (9), 1 (10), 0 (11). So there are 11 digits. Now, since we need to group them into sets of three, starting from the right, but if the number of digits isn't a multiple of three, we need to add leading zeros to make it so. Let me check: 11 divided by 3 is 3 with a remainder of 2. So, there are two digits on the left that won't form a complete group. Therefore, I need to add one leading zero to make it 12 digits, which is divisible by three. Wait, 11 digits: adding one zero on the left would make it 12 digits, which is 4 groups of three.

Let me write that down. Original binary number: 1 0 1 0 1 0 0 1 1 1 0. Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0. Now, let me split into groups of three starting from the right. Wait, but when grouping for octal, I think it's from the right. Let me confirm that. Yes, because the rightmost group is the least significant bits. So, starting from the right, we split into groups of three. Let me try that.

Original binary number is 10101001110. Let me write it with spaces to separate the groups. Let me count from the right:

Starting from the right, the last three digits are 110. Then moving left, the next three digits are 010 (wait, let me check):

Wait, let me write the binary number again: 1 0 1 0 1 0 0 1 1 1 0. Let me index the positions from right to left, starting at position 0:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digits:    1  0 1 0 1 0 0 1 1 1 0

Wait, maybe I need to reverse it. Let me write the binary number as a string: 10101001110. Let's reverse it to make grouping from the right easier. So reversed: 01110010101. Let me count the digits: 11 digits. Now, grouping into sets of three from the left (since reversed, the original right is now left). Let me group them:

First group (leftmost): 011

Second group: 100

Third group: 101

Fourth group: 01 (wait, that's only two digits). Wait, no, total digits are 11, reversed. Let me check again. Original number is 11 digits. When reversed, it's still 11 digits. Let me split into groups of three starting from the left (since reversed, which corresponds to original right):

Group 1: 011

Group 2: 100

Group 3: 101

Group 4: 01 (remaining digits). Wait, but there are 11 digits. Let me split into groups:

Group 1: first three digits: 011

Group 2: next three digits: 100

Group 3: next three digits: 101

Group 4: remaining two digits: 01. Wait, but we need to add leading zeros to make this a group of three. Wait, since we are grouping from the left (reversed), so perhaps I need to pad with leading zeros? Wait, maybe I confused.

Alternatively, maybe I should not reverse it. Let me try again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0. Let me separate into groups of three starting from the right. Since there are 11 digits, the rightmost group is three digits, then moving left. Let me count:

Starting from the right:

Positions 0-2: last three digits: positions 0 (rightmost) is 0, position 1 is 1, position 2 is 1. Wait, the original number is 1 0 1 0 1 0 0 1 1 1 0. Let me index them from right (position 0) to left (position 10). Let me write them with positions:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

So, starting from position 0 (rightmost):

Group 1 (positions 0-2): digits at positions 0,1,2: 0,1,1 → which is 011

Group 2 (positions 3-5): digits at positions 3,4,5: 1,0,0 → 100

Group 3 (positions 6-8): digits at positions 6,7,8: 0,1,0 → 010

Group 4 (positions 9-10): digits at positions 9,10: 0,1 → but we need three digits. Wait, positions 9-10 are only two digits. Therefore, we need to add a leading zero to make it three digits. So group 4: 0 0 1? Wait, positions 9 is 0 and position 10 is 1. Wait, hold on. Let me check:

Position 9: digit is 0 (from the original list: position 9 is 0)

Position 10: digit is 1 (from the original list: position 10 is 1)

So, positions 9 and 10 are digits 0 and 1. To make a group of three, we need to add a leading zero. So group 4 would be 0 0 1? Wait, no. Wait, groups are from right to left. Wait, group 1 is positions 0-2, group 2 positions 3-5, group 3 positions 6-8, group 4 positions 9-11? But there is no position 11. So we need to add a leading zero to the leftmost group. So the leftmost group (positions 9-10) has only two digits, so add a zero in front. Therefore, group 4 becomes 0 0 1? Wait, position 10 is the leftmost digit (original number is 1 0 1 0 1 0 0 1 1 1 0). Wait, maybe I should think of it differently.

Alternatively, perhaps it's easier to add leading zeros to the original binary number so that the total number of digits is a multiple of three. The original number has 11 digits. 11 divided by 3 is 3 with remainder 2. Therefore, we need to add 1 leading zero to make it 12 digits. Let me do that. Adding one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0. Now, let's split into groups of three starting from the right:

Let me write the number as 010101001110. Now, splitting into groups of three starting from the right:

From the right: the last three digits are 110, then the next three digits are 001, then the next three digits are 010, and the remaining three digits are 10? Wait, no. Wait, let me split:

The number is: 0 1 0 1 0 1 0 0 1 1 1 0. Let me separate into groups:

Starting from the right:

Group 1: last three digits: 1 1 0 → 110

Group 2: previous three digits: 0 0 1 → 001

Group 3: previous three digits: 0 1 0 → 010

Group 4: remaining digits: 0 1 → wait, there are two digits left. Wait, original number after adding one leading zero is 12 digits:

Positions (left to right): 0 1 0 1 0 1 0 0 1 1 1 0

Wait, splitting into groups from right:

Starting from the right, every three digits:

Group 1: positions 9-11: 1 1 0

Group 2: positions 6-8: 0 0 1

Group 3: positions 3-5: 0 1 0

Group 4: positions 0-2: 0 1 0

Wait, maybe I need to write the number as:

Let me write the 12-digit binary number as:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the right:

Starting from the right (which is the last three digits):

Group 1: 1 1 0 → rightmost three digits

Group 2: next three digits to the left: 0 0 1

Group 3: next three digits: 0 1 0

Group 4: remaining digits: 0 1 0

Wait, but let me check:

Original number after adding leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three:

From the right:

First three: positions 9-11: digits at positions 9,10,11? Wait, maybe indexing is confusing. Let me write the number with spaces:

Let me write the entire binary number with spaces between each group:

Original number: 0 1 0 1 0 1 0 0 1 1 1 0

If I split into groups of three from the right:

Start from the end:

Group 1: 1 1 0 → the last three digits

Group 2: previous three digits: 0 0 1

Group 3: previous three digits: 0 1 0

Group 4: remaining digits: 0 1 0

Wait, but how does that split? Let me count:

Total digits: 12. Split into four groups of three:

Group 1: digits 10-12? Wait, perhaps it's clearer to split as follows:

Let me write the binary number as follows, adding spaces between groups:

Starting from the right:

First group (rightmost): 110

Second group: 001

Third group: 010

Fourth group: 010

Wait, but how does that split? Wait, let's take the binary number:

Original (after adding leading zero): 0 1 0 1 0 1 0 0 1 1 1 0

Let me write it with indices from 0 (leftmost) to 11 (rightmost):

Position: 0 1 2 3 4 5 6 7 8 9 10 11

Digits:   0 1 0 1 0 1 0 0 1 1 1 0

Wait, actually, if I added a leading zero, the original number was 10101001110, which is 11 digits. Adding a leading zero makes it 12 digits: 010101001110. So positions 0-11 (left to right):

Position 0: 0

1:1

2:0

3:1

4:0

5:1

6:0

7:0

8:1

9:1

10:1

11:0

Wait, so from position 0 to 11. Now, splitting into groups of three starting from the right (position 11). So the rightmost group is positions 9-11? Wait, no. Wait, when grouping from the right, each group is three bits, so starting at the rightmost bit (position 11), the first group is positions 9, 10, 11? Wait, no. Let me clarify:

The rightmost group is the last three digits, which would be positions 9, 10, 11? Wait, no. If the entire number is 12 bits, positions 0 to 11, with 0 being the leftmost and 11 being the rightmost. Wait, that seems counterintuitive. Normally, position 0 is the rightmost, but if we are considering the number as written, with position 0 being the leftmost, then grouping from the right would be different.

This is getting confusing. Maybe it's better to write the binary number with spaces between groups. Let me try again.

Original binary number is 010101001110 (after adding a leading zero). Let me separate it into groups of three starting from the right:

Take the number: 010101001110

Split into groups of three:

Starting from the right, we have:

- The last three digits: 110

- The previous three digits: 001

- The previous three digits: 010

- The remaining digits: 010

Wait, let me check:

Number: 0 1 0 1 0 1 0 0 1 1 1 0

Grouping from the right:

Start at the end:

First group: last three digits: 1 1 0

Second group: previous three: 0 0 1

Third group: previous three: 0 1 0

Fourth group: remaining digits: 0 1 0

Wait, but then the fourth group is 010. Wait, but how does that split? Let me write it as:

010 101 001 110? No, that would be splitting from the left. Wait, perhaps confusion arises from direction. Maybe I should write the binary number and then split it as follows:

Original (with leading zero): 0 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three, starting from the right:

From the right, the first three bits are the last three digits: positions 10, 11? Wait, no. Wait, if the number is 12 bits long, positions 0 to 11 (left to right), then the rightmost bit is position 11. To group from the right, the first group is bits 9, 10, 11? Wait, no, bits 9, 10, 11 would be the last three bits if the number is written as positions 0-11.

Wait, this is getting too confusing. Maybe I should convert the binary number to decimal first, then to octal? Although that might take longer, but maybe it's easier for verification.

Alternatively, perhaps I can use the fact that each octal digit corresponds to three binary digits, so I can convert each group of three bits to octal. However, since the number isn't a multiple of three bits, we need to pad with leading zeros.

Let me try that approach again. Original binary number: 10101001110. Let's count the number of bits: 11 bits. As mentioned, we need to add one leading zero to make it 12 bits. So adding a zero at the beginning: 010101001110. Now, split into groups of three starting from the left?

Wait, but octal conversion groups from the right. Wait, confusion again. Let me check online... Wait, no, I can't. But I should recall.

Actually, when converting binary to octal, it's standard to group the binary digits into sets of three, starting from the right. If there are not enough digits to make a group of three on the left, you add leading zeros. So for example, if you have 11 bits, you add one leading zero, making it 12 bits, then split into four groups of three bits starting from the right. Wait, but if you add leading zeros, the groups would be on the left. Wait, maybe it's better to split the binary number into groups of three bits, starting from the right. Let me take the original number without adding a zero first and see:

Original binary number: 10101001110. Let me write it with spaces separating groups of three starting from the right:

Starting from the right, the last three bits are 110. Then moving left, the next three bits are 001. Wait, but let me see:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups from the right:

Group1: 1 1 0 → rightmost three bits: positions 8,9,10? Wait, no. Let me index the bits from right to left as positions 0 to 10.

Original number:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

So positions 0-10. Now, grouping into three:

Group 0 (rightmost): positions 0-2: 0,1,1 → 011

Group 1: positions 3-5: 1,0,0 → 100

Group 2: positions 6-8: 0,1,0 → 010

Group 3: positions 9-11: but there is no position 11. So there's one bit left: position 9: 0. Therefore, we need to pad with two leading zeros to make group 3: 000? Wait, no. Wait, group 3 would consist of position 9 and the bits to its left. Wait, if we have only one bit left, which is position 9 (digit 0), then we need to add two zeros to the left of it to make a group of three. Therefore, group 3 would be 000? Wait, but the original digit at position 9 is 0. So group 3 is 0 (from position 9) plus two zeros? Wait, no. Wait, how does grouping work?

If there's one bit left (position 9: 0), to make a group of three, we add two zeros to the left of this bit? So group 3 is 000? Wait, but that would be incorrect because the original bit is 0. Wait, no. Let me think. When grouping from the right, if there is a partial group on the left (i.e., the leftmost group has less than three bits), we pad with leading zeros. So in this case, group 3 would be the remaining bits: position 9 (0), and since we need three bits, we add two leading zeros. So group 3 is 0 0 0? But the original bit is 0, so maybe it's 000. Wait, but where does the 0 come from? The position 9 is 0. So if we need three bits, and there's only one bit, we add two zeros in front. So group 3 is 000, but the original bit was 0, but the other bits are zero? Wait, this is confusing.

Alternatively, maybe I should take the original number, add leading zeros to make the total number of bits a multiple of three, then split into groups.

Original number: 10101001110 (11 bits). Add one leading zero to make it 12 bits: 010101001110. Now, split into four groups of three bits from the left:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, is that correct? Let me check:

The number is 010101001110. Split into four groups of three from the left:

First three bits: 010

Next three: 101

Next three: 001

Next three: 110

Yes. Then each group can be converted to octal.

Now, converting each group to octal:

Group 1: 010₂ = 2₈

Group 2: 101₂ = 5₈

Group 3: 001₂ = 1₈

Group 4: 110₂ = 6₈

Therefore, combining them: 2 5 1 6 → 2516₈?

Wait, but the groups were split from the left, but octal conversion is usually done from the right. Wait, but if we added leading zeros to make it a multiple of three, then grouping from the left is okay?

Wait, let me verify with another method. Let me convert the binary number to decimal first and then to octal. Maybe that will help me check.

Original binary number: 10101001110.

Let me convert this to decimal. Let's remember that each bit represents a power of 2, starting from the right (position 0). Let me write down the positions:

Binary number: 1 0 1 0 1 0 0 1 1 1 0

Positions (from right): Let me assign positions from 0 to 10 (right to left):

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

Wait, no, if we consider the rightmost digit as position 0, then:

Original number is 10101001110. Let me write it with positions from right to left:

Starting from the right:

Position 0: 0 (rightmost)

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1 (leftmost)

Wait, let me confirm:

Original binary number: positions from right (0) to left (10):

Digit positions:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, that seems complex. Let me write the binary number as:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

So, leftmost digit is position 10 (value 2^10), then next is position 9 (2^9), down to position 0 (2^0). Wait, if I consider the leftmost digit as the highest power.

So, the value is:

1*(2^10) + 0*(2^9) + 1*(2^8) + 0*(2^7) + 1*(2^6) + 0*(2^5) + 0*(2^4) + 1*(2^3) + 1*(2^2) + 1*(2^1) + 0*(2^0)

Calculating each term:

1*(1024) + 0*(512) + 1*(256) + 0*(128) + 1*(64) + 0*(32) + 0*(16) + 1*(8) + 1*(4) + 1*(2) + 0*(1)

Adding them up:

1024 + 0 + 256 + 0 + 64 + 0 + 0 + 8 + 4 + 2 + 0 =

Let me compute step by step:

Start with 1024.

+256 = 1280

+64 = 1344

+8 = 1352

+4 = 1356

+2 = 1358

So decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

1358 divided by 8: 1358 / 8 = 169.75 → 169 with a remainder of 6 (since 169*8=1352, 1358-1352=6)

169 divided by 8: 169 /8=21.125 → 21 with remainder 1 (21*8=168, 169-168=1)

21 divided by 8: 2 with remainder 5 (2*8=16, 21-16=5)

2 divided by 8: 0 with remainder 2

So the remainders, from last to first, are 2, 5, 1, 6. Therefore, the octal number is 2516₈. Which matches the earlier result when grouping with leading zeros.

Therefore, the binary number 10101001110₂ is equal to 2516 in octal. So my initial grouping was correct even though I added the leading zero and grouped from the left. Wait, but why?

Wait, when adding the leading zero to make the total bits a multiple of three, grouping from the left or right should be equivalent in terms of the octal digits? Wait, because adding leading zeros doesn't change the value. So, if I add a leading zero, then split into groups from the left, converting each group, that should be equivalent to splitting from the right. Therefore, the result is 2516₈.

Therefore, the answer is 2516 octal. But let me check once more with the grouping method.

Original binary number after adding leading zero: 010101001110. Split into groups of three from the left:

Group1: 010 (which is 2)

Group2: 101 (which is 5)

Group3: 001 (which is 1)

Group4: 110 (which is 6)

So, the octal number is 2 5 1 6 → 2516. Which is same as above. Therefore, correct.

Alternatively, if I had grouped from the right without adding leading zeros, but with padding, I might have gotten the same.

But to make sure, let me try that method again. Original binary number is 10101001110 (11 bits). To group from the right, need to pad with one leading zero, making it 12 bits: 010101001110. Then grouping from the right:

Split into groups of three:

Starting from the right:

Group1: 110 (rightmost three bits)

Group2: 001

Group3: 010

Group4: 010 (leftmost three bits)

Wait, but if I split into groups from the right, the first group is the rightmost three bits, then next three, etc. Let me write the 12-bit number: 0 1 0 1 0 1 0 0 1 1 1 0.

Split into groups from right:

Group1: bits 9-11 (positions 9,10,11? Wait, with positions from left to right, which is confusing. Let me think of it as bits:

Number is 0 1 0 1 0 1 0 0 1 1 1 0.

Let me write it as:

[0][1][0][1][0][1][0][0][1][1][1][0]

Split from the right:

Starting at the end, which is the 12th bit (rightmost):

Group1: bits 10-12? Wait, no. Let me split the number into groups of three from the right, so:

First group: last three bits: bits 10, 11, 12? No. Wait, the entire number has 12 bits. Let me index them from right (position 0) to left (position 11):

Position: 11 10 9 8 7 6 5 4 3 2 1 0

Bits:     0  1 0 1 0 1 0 0 1 1 1 0

Wait, this is getting too tangled. Alternatively, take the binary number as a string: 010101001110.

Split into groups of three from the right:

Start from the end:

- The last three digits: 110 → which is the last three bits.

- Then the previous three digits: 001 → next three bits.

- Then the previous three digits: 010 → next three bits.

- Then the remaining digits: 010 → which is the first three bits.

Therefore, groups are 010, 010, 001, 110 → but when grouping from the right, the order would be reversed. Wait, no. Wait, the groups from the right are:

Group1: 110 (rightmost)

Group2: 001

Group3: 010

Group4: 010 (leftmost)

Therefore, converting each group:

Group1: 110₂ = 6₈

Group2: 001₂ = 1₈

Group3: 010₂ = 2₈

Group4: 010₂ = 2₈

Therefore, combining them in order from left to right (since group4 is the leftmost group): 2 2 1 6 → 2216₈? But this contradicts the previous result of 2516. Wait, that can't be. There must be an error here.

Wait, this inconsistency shows that my grouping method was flawed. Previously, when I added leading zero and split from the left, I got 2516, but grouping from the right without proper padding gave me 2216, which is different. Clearly, I must have messed up the grouping.

Wait, let me use the decimal conversion result of 1358, which is 2516 in octal. Therefore, my second grouping from the right must have been wrong. Let me try again.

Original binary number after adding leading zero: 010101001110. Split into groups of three from the right:

The number is 0 1 0 1 0 1 0 0 1 1 1 0.

Let me write this as:

From left to right: 0 1 0 1 0 1 0 0 1 1 1 0

To split into groups of three from the right:

Start from the end (rightmost three bits): 1 1 0 → which is the last three bits. Then the next three bits to the left are 0 0 1. Then the next three bits are 0 1 0. Then the remaining three bits are 0 1 0. Wait, but that seems same as before. So converting these groups:

110 → 6, 001 → 1, 010 → 2, 010 → 2. But the octal number would be 2 2 1 6? Wait, but if the groups are ordered from right to left, then when converted, they should be written in the same order as the groups. However, when you group from the right, the first group is the least significant digit. So octal digits are ordered from left to right, with the leftmost group being the most significant. Wait, so if I have groups from right to left as [group1, group2, group3, group4], then when written as octal number, it's group4 group3 group2 group1. Wait, so let me clarify:

When you split from the right, group1 is the least significant octal digit, group2 is the next, etc. Therefore, the octal number is group4 group3 group2 group1.

In our case, groups from right:

group1: 110 → 6

group2: 001 → 1

group3: 010 → 2

group4: 010 → 2

So octal number is group4 group3 group2 group1: 2 2 1 6 → 2216? But decimal conversion says it should be 2516. There's a contradiction here. What's wrong?

Ah! I think I messed up the grouping. Let me check again the binary number after adding the leading zero: 010101001110. Let me write it as a string: "010101001110". Let me split it into groups of three starting from the right:

The string is: 0 1 0 1 0 1 0 0 1 1 1 0

Let me count the bits from the right. Starting from the end (which is the last digit, '0'), moving left:

Group1: last three digits: 1 1 0 → "110"

Group2: previous three digits: 0 0 1 → "001"

Group3: previous three digits: 0 1 0 → "010"

Group4: previous three digits: 0 1 0 → "010"

Wait, but that's the same as before. Then the octal digits would be group4, group3, group2, group1: "010", "010", "001", "110" → which is 2, 2, 1, 6 → 2216? But decimal conversion says 2516. Hence, inconsistency. Therefore, my grouping is wrong. There's an error here. What's the problem?

Wait, let me check my decimal conversion again. Original binary number: 10101001110. Let me verify the decimal value again.

Breaking down the binary number:

From left to right: 1 0 1 0 1 0 0 1 1 1 0.

Let me compute each bit's value:

Position 10: 1 * 2^10 = 1024

Position 9: 0 * 2^9 = 0

Position 8: 1 * 2^8 = 256

Position 7: 0 * 2^7 = 0

Position 6: 1 * 2^6 = 64

Position 5: 0 * 2^5 = 0

Position 4: 0 * 2^4 = 0

Position 3: 1 * 2^3 = 8

Position 2: 1 * 2^2 = 4

Position 1: 1 * 2^1 = 2

Position 0: 0 * 2^0 = 0

Adding them up: 1024 + 256 + 64 + 8 + 4 + 2 = 1024 + 256 = 1280; 1280 + 64 = 1344; 1344 + 8 = 1352; 1352 + 4 = 1356; 1356 + 2 = 1358. Yes, that is correct. Decimal value is 1358. Converting 1358 to octal: 1358 divided by 8 is 169.75, remainder 6. 169 divided by 8 is 21.125, remainder 1. 21 divided by 8 is 2.625, remainder 5. 2 divided by 8 is 0.25, remainder 2. So octal is 2 5 1 6, which is 2516₈. So this is correct.

But when I grouped the binary number as 010101001110 into groups from the right, I had groups 110, 001, 010, 010 which gave me octal digits 6, 1, 2, 2 → 2216. But why discrepancy?

Ah! I think I made a mistake in grouping the binary number. Let me check the binary number again. After adding the leading zero, the binary number is 010101001110. Let me write it as:

Positions from left to right:

0 1 0 1 0 1 0 0 1 1 1 0

Wait, if I split this into groups of three from the right, the groups should be:

Starting from the right:

- Last three bits: positions 9-11? Wait, if the entire number is 12 bits, then positions 0-11 from left to right. Wait, no, confusion again. Let me take the string "010101001110" and split it into groups of three from the right:

- Starting from the end, take three characters: '110' → group1

Then move left three characters: '001' → group2

Then move left three: '010' → group3

Then move left three: '010' → group4

Wait, but the entire string is 12 characters. So:

Group1: 110 (positions 9-11)

Group2: 001 (positions 6-8)

Group3: 010 (positions 3-5)

Group4: 010 (positions 0-2)

Wait, but if the string is "0 1 0 1 0 1 0 0 1 1 1 0", then positions from left to right are 0-11. So group4 (positions 0-2) is '010', group3 (positions 3-5) is '101', group2 (positions 6-8) is '001', group1 (positions 9-11) is '110'. Wait, this is different from before!

Wait, if I split the string into groups of three from the right, it's important to split the string into groups of three, starting from the end. Let me write the string as follows:

"0" "1" "0" "1" "0" "1" "0" "0" "1" "1" "1" "0"

Now, starting from the end:

Group1: last three characters: "1" "1" "0" → "110"

Group2: previous three: "0" "1" "1" → "011" ? Wait, no. Wait, after taking the last three, the previous three is the next three to the left.

Wait, the string is: positions 0-11 (left to right):

0 1 0 1 0 1 0 0 1 1 1 0

Wait, if I take the last three characters: positions 9,10,11: digits at positions 9:1, 10:1, 11:0 → "110"

Then previous three: positions 6,7,8: digits at 6:0, 7:0, 8:1 → "001"

Then previous three: positions 3,4,5: digits at 3:1, 4:0, 5:1 → "101"

Then previous three: positions 0,1,2: digits at 0:0, 1:1, 2:0 → "010"

Therefore, the groups are, from right to left: "110", "001", "101", "010". Therefore, converting these groups:

Group1 (rightmost): 110₂ = 6

Group2: 001₂ = 1

Group3: 101₂ = 5

Group4 (leftmost): 010₂ = 2

Therefore, the octal number is group4 group3 group2 group1: 2 5 1 6 → 2516₈, which matches the decimal conversion. So earlier mistake was in my grouping when I thought group3 was "010" instead of "101". Therefore, the correct groups are "010", "101", "001", "110" when grouped from the left, but when grouped from the right, the order is "110", "001", "101", "010", which correspond to octal digits 6, 1, 5, 2, so writing them in order from group4 to group1 (left to right): 2 5 1 6 → 2516. Yes! Therefore, the correct grouping is 2516.

This shows that it's vital to group correctly, either from the right (with proper identification of each group) or from the left after padding. So the binary number converted to octal is 2516.

Thus, the answer is \boxed{2516}.

**Final Answer**
The binary number $10101001110_2$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ into its octal equivalent, we can use the fact that **each octal digit corresponds to three binary digits**. Therefore, the strategy is to **group the binary digits into sets of three, starting from the right**, and convert each group to its octal equivalent.

---

### Step 1: Count the Number of Digits

The binary number is:

$$
1\ 0\ 1\ 0\ 1\ 0\ 0\ 1\ 1\ 1\ 0 \quad \text{(11 digits)}
$$

Since 11 is not divisible by 3, we **add one leading zero** to make the total number of bits equal to 12, which is divisible by 3.

The adjusted binary number becomes:

$$
0\ 1\ 0\ 1\ 0\ 1\ 0\ 0\ 1\ 1\ 1\ 0
$$

---

### Step 2: Group into Sets of Three (From Right to Left)

Break this into groups of 3 bits, starting from the right:

$$
010\ 101\ 001\ 110
$$

Note: Grouping from the right ensures that the least significant bits are grouped correctly.

---

### Step 3: Convert Each Binary Group to Octal

Now, convert each group from binary to octal:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the Octal Digits

Putting the octal digits together:

$$
2\ 5\ 1\ 6
$$

---

### Final Answer

$$
\boxed{2516}
$$

[Assistant]
Explanation:
Okay, so I need to convert the binary number 10101001110₂ to octal. Hmm, let me remember how to do this. I think there's a method where you can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent. Let me try to recall the exact steps.

First, I know that each digit in an octal number represents three binary digits because 8 is 2³. So, if I can split the binary number into groups of three bits, each group can be converted to an octal digit. But wait, the binary number given is 10101001110. Let me count the digits to see how many there are. Let me write it down: 1 0 1 0 1 0 0 1 1 1 0. Let me count: 1 (1), 0 (2), 1 (3), 0 (4), 1 (5), 0 (6), 0 (7), 1 (8), 1 (9), 1 (10), 0 (11). So there are 11 digits. Now, since we need to group them into sets of three, starting from the right, but if the number of digits isn't a multiple of three, we need to add leading zeros to make it so. Let me check: 11 divided by 3 is 3 with a remainder of 2. So, there are two digits on the left that won't form a complete group. Therefore, I need to add one leading zero to make it 12 digits, which is divisible by three. Wait, 11 digits: adding one zero on the left would make it 12 digits, which is 4 groups of three.

Let me write that down. Original binary number: 1 0 1 0 1 0 0 1 1 1 0. Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0. Now, let me split into groups of three starting from the right. Wait, but when grouping for octal, I think it's from the right. Let me confirm that. Yes, because the rightmost group is the least significant bits. So, starting from the right, we split into groups of three. Let me try that.

Original binary number is 10101001110. Let me write it with spaces to separate the groups. Let me count from the right:

Starting from the right, the last three digits are 110. Then moving left, the next three digits are 010 (wait, let me check):

Wait, let me write the binary number again: 1 0 1 0 1 0 0 1 1 1 0. Let me index the positions from right to left, starting at position 0:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digits:    1  0 1 0 1 0 0 1 1 1 0

Wait, maybe I need to reverse it. Let me write the binary number as a string: 10101001110. Let's reverse it to make grouping from the right easier. So reversed: 01110010101. Let me count the digits: 11 digits. Now, grouping into sets of three from the left (since reversed, the original right is now left). Let me group them:

First group (leftmost): 011

Second group: 100

Third group: 101

Fourth group: 01 (wait, that's only two digits). Wait, no, total digits are 11, reversed. Let me check again. Original number is 11 digits. When reversed, it's still 11 digits. Let me split into groups of three starting from the left (since reversed, which corresponds to original right):

Group 1: 011

Group 2: 100

Group 3: 101

Group 4: 01 (remaining digits). Wait, but there are 11 digits. Let me split into groups:

Group 1: first three digits: 011

Group 2: next three digits: 100

Group 3: next three digits: 101

Group 4: remaining two digits: 01. Wait, but we need to add leading zeros to make this a group of three. Wait, since we are grouping from the left (reversed), so perhaps I need to pad with leading zeros? Wait, maybe I confused.

Alternatively, maybe I should not reverse it. Let me try again.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0. Let me separate into groups of three starting from the right. Since there are 11 digits, the rightmost group is three digits, then moving left. Let me count:

Starting from the right:

Positions 0-2: last three digits: positions 0 (rightmost) is 0, position 1 is 1, position 2 is 1. Wait, the original number is 1 0 1 0 1 0 0 1 1 1 0. Let me index them from right (position 0) to left (position 10). Let me write them with positions:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

So, starting from position 0 (rightmost):

Group 1 (positions 0-2): digits at positions 0,1,2: 0,1,1 → which is 011

Group 2 (positions 3-5): digits at positions 3,4,5: 1,0,0 → 100

Group 3 (positions 6-8): digits at positions 6,7,8: 0,1,0 → 010

Group 4 (positions 9-10): digits at positions 9,10: 0,1 → but we need three digits. Wait, positions 9-10 are only two digits. Therefore, we need to add a leading zero to make it three digits. So group 4: 0 0 1? Wait, positions 9 is 0 and position 10 is 1. Wait, hold on. Let me check:

Position 9: digit is 0 (from the original list: position 9 is 0)

Position 10: digit is 1 (from the original list: position 10 is 1)

So, positions 9 and 10 are digits 0 and 1. To make a group of three, we need to add a leading zero. So group 4 would be 0 0 1? Wait, no. Wait, groups are from right to left. Wait, group 1 is positions 0-2, group 2 positions 3-5, group 3 positions 6-8, group 4 positions 9-11? But there is no position 11. So we need to add a leading zero to the leftmost group. So the leftmost group (positions 9-10) has only two digits, so add a zero in front. Therefore, group 4 becomes 0 0 1? Wait, position 10 is the leftmost digit (original number is 1 0 1 0 1 0 0 1 1 1 0). Wait, maybe I should think of it differently.

Alternatively, perhaps it's easier to add leading zeros to the original binary number so that the total number of digits is a multiple of three. The original number has 11 digits. 11 divided by 3 is 3 with remainder 2. Therefore, we need to add 1 leading zero to make it 12 digits. Let me do that. Adding one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0. Now, let's split into groups of three starting from the right:

Let me write the number as 010101001110. Now, splitting into groups of three starting from the right:

From the right: the last three digits are 110, then the next three digits are 001, then the next three digits are 010, and the remaining three digits are 10? Wait, no. Wait, let me split:

The number is: 0 1 0 1 0 1 0 0 1 1 1 0. Let me separate into groups:

Starting from the right:

Group 1: last three digits: 1 1 0 → 110

Group 2: previous three digits: 0 0 1 → 001

Group 3: previous three digits: 0 1 0 → 010

Group 4: remaining digits: 0 1 → wait, there are two digits left. Wait, original number after adding one leading zero is 12 digits:

Positions (left to right): 0 1 0 1 0 1 0 0 1 1 1 0

Wait, splitting into groups from right:

Starting from the right, every three digits:

Group 1: positions 9-11: 1 1 0

Group 2: positions 6-8: 0 0 1

Group 3: positions 3-5: 0 1 0

Group 4: positions 0-2: 0 1 0

Wait, maybe I need to write the number as:

Let me write the 12-digit binary number as:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the right:

Starting from the right (which is the last three digits):

Group 1: 1 1 0 → rightmost three digits

Group 2: next three digits to the left: 0 0 1

Group 3: next three digits: 0 1 0

Group 4: remaining digits: 0 1 0

Wait, but let me check:

Original number after adding leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three:

From the right:

First three: positions 9-11: digits at positions 9,10,11? Wait, maybe indexing is confusing. Let me write the number with spaces:

Let me write the entire binary number with spaces between each group:

Original number: 0 1 0 1 0 1 0 0 1 1 1 0

If I split into groups of three from the right:

Start from the end:

Group 1: 1 1 0 → the last three digits

Group 2: previous three digits: 0 0 1

Group 3: previous three digits: 0 1 0

Group 4: remaining digits: 0 1 0

Wait, but how does that split? Let me count:

Total digits: 12. Split into four groups of three:

Group 1: digits 10-12? Wait, perhaps it's clearer to split as follows:

Let me write the binary number as follows, adding spaces between groups:

Starting from the right:

First group (rightmost): 110

Second group: 001

Third group: 010

Fourth group: 010

Wait, but how does that split? Wait, let's take the binary number:

Original (after adding leading zero): 0 1 0 1 0 1 0 0 1 1 1 0

Let me write it with indices from 0 (leftmost) to 11 (rightmost):

Position: 0 1 2 3 4 5 6 7 8 9 10 11

Digits:   0 1 0 1 0 1 0 0 1 1 1 0

Wait, actually, if I added a leading zero, the original number was 10101001110, which is 11 digits. Adding a leading zero makes it 12 digits: 010101001110. So positions 0-11 (left to right):

Position 0: 0

1:1

2:0

3:1

4:0

5:1

6:0

7:0

8:1

9:1

10:1

11:0

Wait, so from position 0 to 11. Now, splitting into groups of three starting from the right (position 11). So the rightmost group is positions 9-11? Wait, no. Wait, when grouping from the right, each group is three bits, so starting at the rightmost bit (position 11), the first group is positions 9, 10, 11? Wait, no. Let me clarify:

The rightmost group is the last three digits, which would be positions 9, 10, 11? Wait, no. If the entire number is 12 bits, positions 0 to 11, with 0 being the leftmost and 11 being the rightmost. Wait, that seems counterintuitive. Normally, position 0 is the rightmost, but if we are considering the number as written, with position 0 being the leftmost, then grouping from the right would be different.

This is getting confusing. Maybe it's better to write the binary number with spaces between groups. Let me try again.

Original binary number is 010101001110 (after adding a leading zero). Let me separate it into groups of three starting from the right:

Take the number: 010101001110

Split into groups of three:

Starting from the right, we have:

- The last three digits: 110

- The previous three digits: 001

- The previous three digits: 010

- The remaining digits: 010

Wait, let me check:

Number: 0 1 0 1 0 1 0 0 1 1 1 0

Grouping from the right:

Start at the end:

First group: last three digits: 1 1 0

Second group: previous three: 0 0 1

Third group: previous three: 0 1 0

Fourth group: remaining digits: 0 1 0

Wait, but then the fourth group is 010. Wait, but how does that split? Let me write it as:

010 101 001 110? No, that would be splitting from the left. Wait, perhaps confusion arises from direction. Maybe I should write the binary number and then split it as follows:

Original (with leading zero): 0 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three, starting from the right:

From the right, the first three bits are the last three digits: positions 10, 11? Wait, no. Wait, if the number is 12 bits long, positions 0 to 11 (left to right), then the rightmost bit is position 11. To group from the right, the first group is bits 9, 10, 11? Wait, no, bits 9, 10, 11 would be the last three bits if the number is written as positions 0-11.

Wait, this is getting too confusing. Maybe I should convert the binary number to decimal first, then to octal? Although that might take longer, but maybe it's easier for verification.

Alternatively, perhaps I can use the fact that each octal digit corresponds to three binary digits, so I can convert each group of three bits to octal. However, since the number isn't a multiple of three bits, we need to pad with leading zeros.

Let me try that approach again. Original binary number: 10101001110. Let's count the number of bits: 11 bits. As mentioned, we need to add one leading zero to make it 12 bits. So adding a zero at the beginning: 010101001110. Now, split into groups of three starting from the left?

Wait, but octal conversion groups from the right. Wait, confusion again. Let me check online... Wait, no, I can't. But I should recall.

Actually, when converting binary to octal, it's standard to group the binary digits into sets of three, starting from the right. If there are not enough digits to make a group of three on the left, you add leading zeros. So for example, if you have 11 bits, you add one leading zero, making it 12 bits, then split into four groups of three bits starting from the right. Wait, but if you add leading zeros, the groups would be on the left. Wait, maybe it's better to split the binary number into groups of three bits, starting from the right. Let me take the original number without adding a zero first and see:

Original binary number: 10101001110. Let me write it with spaces separating groups of three starting from the right:

Starting from the right, the last three bits are 110. Then moving left, the next three bits are 001. Wait, but let me see:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups from the right:

Group1: 1 1 0 → rightmost three bits: positions 8,9,10? Wait, no. Let me index the bits from right to left as positions 0 to 10.

Original number:

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

So positions 0-10. Now, grouping into three:

Group 0 (rightmost): positions 0-2: 0,1,1 → 011

Group 1: positions 3-5: 1,0,0 → 100

Group 2: positions 6-8: 0,1,0 → 010

Group 3: positions 9-11: but there is no position 11. So there's one bit left: position 9: 0. Therefore, we need to pad with two leading zeros to make group 3: 000? Wait, no. Wait, group 3 would consist of position 9 and the bits to its left. Wait, if we have only one bit left, which is position 9 (digit 0), then we need to add two zeros to the left of it to make a group of three. Therefore, group 3 would be 000? Wait, but the original digit at position 9 is 0. So group 3 is 0 (from position 9) plus two zeros? Wait, no. Wait, how does grouping work?

If there's one bit left (position 9: 0), to make a group of three, we add two zeros to the left of this bit? So group 3 is 000? Wait, but that would be incorrect because the original bit is 0. Wait, no. Let me think. When grouping from the right, if there is a partial group on the left (i.e., the leftmost group has less than three bits), we pad with leading zeros. So in this case, group 3 would be the remaining bits: position 9 (0), and since we need three bits, we add two leading zeros. So group 3 is 0 0 0? But the original bit is 0, so maybe it's 000. Wait, but where does the 0 come from? The position 9 is 0. So if we need three bits, and there's only one bit, we add two zeros in front. So group 3 is 000, but the original bit was 0, but the other bits are zero? Wait, this is confusing.

Alternatively, maybe I should take the original number, add leading zeros to make the total number of bits a multiple of three, then split into groups.

Original number: 10101001110 (11 bits). Add one leading zero to make it 12 bits: 010101001110. Now, split into four groups of three bits from the left:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Wait, is that correct? Let me check:

The number is 010101001110. Split into four groups of three from the left:

First three bits: 010

Next three: 101

Next three: 001

Next three: 110

Yes. Then each group can be converted to octal.

Now, converting each group to octal:

Group 1: 010₂ = 2₈

Group 2: 101₂ = 5₈

Group 3: 001₂ = 1₈

Group 4: 110₂ = 6₈

Therefore, combining them: 2 5 1 6 → 2516₈?

Wait, but the groups were split from the left, but octal conversion is usually done from the right. Wait, but if we added leading zeros to make it a multiple of three, then grouping from the left is okay?

Wait, let me verify with another method. Let me convert the binary number to decimal first and then to octal. Maybe that will help me check.

Original binary number: 10101001110.

Let me convert this to decimal. Let's remember that each bit represents a power of 2, starting from the right (position 0). Let me write down the positions:

Binary number: 1 0 1 0 1 0 0 1 1 1 0

Positions (from right): Let me assign positions from 0 to 10 (right to left):

Position: 10 9 8 7 6 5 4 3 2 1 0

Digit:    1  0 1 0 1 0 0 1 1 1 0

Wait, no, if we consider the rightmost digit as position 0, then:

Original number is 10101001110. Let me write it with positions from right to left:

Starting from the right:

Position 0: 0 (rightmost)

Position 1: 1

Position 2: 1

Position 3: 1

Position 4: 0

Position 5: 0

Position 6: 1

Position 7: 0

Position 8: 1

Position 9: 0

Position 10: 1 (leftmost)

Wait, let me confirm:

Original binary number: positions from right (0) to left (10):

Digit positions:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, that seems complex. Let me write the binary number as:

From left to right: 1 0 1 0 1 0 0 1 1 1 0

So, leftmost digit is position 10 (value 2^10), then next is position 9 (2^9), down to position 0 (2^0). Wait, if I consider the leftmost digit as the highest power.

So, the value is:

1*(2^10) + 0*(2^9) + 1*(2^8) + 0*(2^7) + 1*(2^6) + 0*(2^5) + 0*(2^4) + 1*(2^3) + 1*(2^2) + 1*(2^1) + 0*(2^0)

Calculating each term:

1*(1024) + 0*(512) + 1*(256) + 0*(128) + 1*(64) + 0*(32) + 0*(16) + 1*(8) + 1*(4) + 1*(2) + 0*(1)

Adding them up:

1024 + 0 + 256 + 0 + 64 + 0 + 0 + 8 + 4 + 2 + 0 =

Let me compute step by step:

Start with 1024.

+256 = 1280

+64 = 1344

+8 = 1352

+4 = 1356

+2 = 1358

So decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

1358 divided by 8: 1358 / 8 = 169.75 → 169 with a remainder of 6 (since 169*8=1352, 1358-1352=6)

169 divided by 8: 169 /8=21.125 → 21 with remainder 1 (21*8=168, 169-168=1)

21 divided by 8: 2 with remainder 5 (2*8=16, 21-16=5)

2 divided by 8: 0 with remainder 2

So the remainders, from last to first, are 2, 5, 1, 6. Therefore, the octal number is 2516₈. Which matches the earlier result when grouping with leading zeros.

Therefore, the binary number 10101001110₂ is equal to 2516 in octal. So my initial grouping was correct even though I added the leading zero and grouped from the left. Wait, but why?

Wait, when adding the leading zero to make the total bits a multiple of three, grouping from the left or right should be equivalent in terms of the octal digits? Wait, because adding leading zeros doesn't change the value. So, if I add a leading zero, then split into groups from the left, converting each group, that should be equivalent to splitting from the right. Therefore, the result is 2516₈.

Therefore, the answer is 2516 octal. But let me check once more with the grouping method.

Original binary number after adding leading zero: 010101001110. Split into groups of three from the left:

Group1: 010 (which is 2)

Group2: 101 (which is 5)

Group3: 001 (which is 1)

Group4: 110 (which is 6)

So, the octal number is 2 5 1 6 → 2516. Which is same as above. Therefore, correct.

Alternatively, if I had grouped from the right without adding leading zeros, but with padding, I might have gotten the same.

But to make sure, let me try that method again. Original binary number is 10101001110 (11 bits). To group from the right, need to pad with one leading zero, making it 12 bits: 010101001110. Then grouping from the right:

Split into groups of three:

Starting from the right:

Group1: 110 (rightmost three bits)

Group2: 001

Group3: 010

Group4: 010 (leftmost three bits)

Wait, but if I split into groups from the right, the first group is the rightmost three bits, then next three, etc. Let me write the 12-bit number: 0 1 0 1 0 1 0 0 1 1 1 0.

Split into groups from right:

Group1: bits 9-11 (positions 9,10,11? Wait, with positions from left to right, which is confusing. Let me think of it as bits:

Number is 0 1 0 1 0 1 0 0 1 1 1 0.

Let me write it as:

[0][1][0][1][0][1][0][0][1][1][1][0]

Split from the right:

Starting at the end, which is the 12th bit (rightmost):

Group1: bits 10-12? Wait, no. Let me split the number into groups of three from the right, so:

First group: last three bits: bits 10, 11, 12? No. Wait, the entire number has 12 bits. Let me index them from right (position 0) to left (position 11):

Position: 11 10 9 8 7 6 5 4 3 2 1 0

Bits:     0  1 0 1 0 1 0 0 1 1 1 0

Wait, this is getting too tangled. Alternatively, take the binary number as a string: 010101001110.

Split into groups of three from the right:

Start from the end:

- The last three digits: 110 → which is the last three bits.

- Then the previous three digits: 001 → next three bits.

- Then the previous three digits: 010 → next three bits.

- Then the remaining digits: 010 → which is the first three bits.

Therefore, groups are 010, 010, 001, 110 → but when grouping from the right, the order would be reversed. Wait, no. Wait, the groups from the right are:

Group1: 110 (rightmost)

Group2: 001

Group3: 010

Group4: 010 (leftmost)

Therefore, converting each group:

Group1: 110₂ = 6₈

Group2: 001₂ = 1₈

Group3: 010₂ = 2₈

Group4: 010₂ = 2₈

Therefore, combining them in order from left to right (since group4 is the leftmost group): 2 2 1 6 → 2216₈? But this contradicts the previous result of 2516. Wait, that can't be. There must be an error here.

Wait, this inconsistency shows that my grouping method was flawed. Previously, when I added leading zero and split from the left, I got 2516, but grouping from the right without proper padding gave me 2216, which is different. Clearly, I must have messed up the grouping.

Wait, let me use the decimal conversion result of 1358, which is 2516 in octal. Therefore, my second grouping from the right must have been wrong. Let me try again.

Original binary number after adding leading zero: 010101001110. Split into groups of three from the right:

The number is 0 1 0 1 0 1 0 0 1 1 1 0.

Let me write this as:

From left to right: 0 1 0 1 0 1 0 0 1 1 1 0

To split into groups of three from the right:

Start from the end (rightmost three bits): 1 1 0 → which is the last three bits. Then the next three bits to the left are 0 0 1. Then the next three bits are 0 1 0. Then the remaining three bits are 0 1 0. Wait, but that seems same as before. So converting these groups:

110 → 6, 001 → 1, 010 → 2, 010 → 2. But the octal number would be 2 2 1 6? Wait, but if the groups are ordered from right to left, then when converted, they should be written in the same order as the groups. However, when you group from the right, the first group is the least significant digit. So octal digits are ordered from left to right, with the leftmost group being the most significant. Wait, so if I have groups from right to left as [group1, group2, group3, group4], then when written as octal number, it's group4 group3 group2 group1. Wait, so let me clarify:

When you split from the right, group1 is the least significant octal digit, group2 is the next, etc. Therefore, the octal number is group4 group3 group2 group1.

In our case, groups from right:

group1: 110 → 6

group2: 001 → 1

group3: 010 → 2

group4: 010 → 2

So octal number is group4 group3 group2 group1: 2 2 1 6 → 2216? But decimal conversion says it should be 2516. There's a contradiction here. What's wrong?

Ah! I think I messed up the grouping. Let me check again the binary number after adding the leading zero: 010101001110. Let me write it as a string: "010101001110". Let me split it into groups of three starting from the right:

The string is: 0 1 0 1 0 1 0 0 1 1 1 0

Let me count the bits from the right. Starting from the end (which is the last digit, '0'), moving left:

Group1: last three digits: 1 1 0 → "110"

Group2: previous three digits: 0 0 1 → "001"

Group3: previous three digits: 0 1 0 → "010"

Group4: previous three digits: 0 1 0 → "010"

Wait, but that's the same as before. Then the octal digits would be group4, group3, group2, group1: "010", "010", "001", "110" → which is 2, 2, 1, 6 → 2216? But decimal conversion says 2516. Hence, inconsistency. Therefore, my grouping is wrong. There's an error here. What's the problem?

Wait, let me check my decimal conversion again. Original binary number: 10101001110. Let me verify the decimal value again.

Breaking down the binary number:

From left to right: 1 0 1 0 1 0 0 1 1 1 0.

Let me compute each bit's value:

Position 10: 1 * 2^10 = 1024

Position 9: 0 * 2^9 = 0

Position 8: 1 * 2^8 = 256

Position 7: 0 * 2^7 = 0

Position 6: 1 * 2^6 = 64

Position 5: 0 * 2^5 = 0

Position 4: 0 * 2^4 = 0

Position 3: 1 * 2^3 = 8

Position 2: 1 * 2^2 = 4

Position 1: 1 * 2^1 = 2

Position 0: 0 * 2^0 = 0

Adding them up: 1024 + 256 + 64 + 8 + 4 + 2 = 1024 + 256 = 1280; 1280 + 64 = 1344; 1344 + 8 = 1352; 1352 + 4 = 1356; 1356 + 2 = 1358. Yes, that is correct. Decimal value is 1358. Converting 1358 to octal: 1358 divided by 8 is 169.75, remainder 6. 169 divided by 8 is 21.125, remainder 1. 21 divided by 8 is 2.625, remainder 5. 2 divided by 8 is 0.25, remainder 2. So octal is 2 5 1 6, which is 2516₈. So this is correct.

But when I grouped the binary number as 010101001110 into groups from the right, I had groups 110, 001, 010, 010 which gave me octal digits 6, 1, 2, 2 → 2216. But why discrepancy?

Ah! I think I made a mistake in grouping the binary number. Let me check the binary number again. After adding the leading zero, the binary number is 010101001110. Let me write it as:

Positions from left to right:

0 1 0 1 0 1 0 0 1 1 1 0

Wait, if I split this into groups of three from the right, the groups should be:

Starting from the right:

- Last three bits: positions 9-11? Wait, if the entire number is 12 bits, then positions 0-11 from left to right. Wait, no, confusion again. Let me take the string "010101001110" and split it into groups of three from the right:

- Starting from the end, take three characters: '110' → group1

Then move left three characters: '001' → group2

Then move left three: '010' → group3

Then move left three: '010' → group4

Wait, but the entire string is 12 characters. So:

Group1: 110 (positions 9-11)

Group2: 001 (positions 6-8)

Group3: 010 (positions 3-5)

Group4: 010 (positions 0-2)

Wait, but if the string is "0 1 0 1 0 1 0 0 1 1 1 0", then positions from left to right are 0-11. So group4 (positions 0-2) is '010', group3 (positions 3-5) is '101', group2 (positions 6-8) is '001', group1 (positions 9-11) is '110'. Wait, this is different from before!

Wait, if I split the string into groups of three from the right, it's important to split the string into groups of three, starting from the end. Let me write the string as follows:

"0" "1" "0" "1" "0" "1" "0" "0" "1" "1" "1" "0"

Now, starting from the end:

Group1: last three characters: "1" "1" "0" → "110"

Group2: previous three: "0" "1" "1" → "011" ? Wait, no. Wait, after taking the last three, the previous three is the next three to the left.

Wait, the string is: positions 0-11 (left to right):

0 1 0 1 0 1 0 0 1 1 1 0

Wait, if I take the last three characters: positions 9,10,11: digits at positions 9:1, 10:1, 11:0 → "110"

Then previous three: positions 6,7,8: digits at 6:0, 7:0, 8:1 → "001"

Then previous three: positions 3,4,5: digits at 3:1, 4:0, 5:1 → "101"

Then previous three: positions 0,1,2: digits at 0:0, 1:1, 2:0 → "010"

Therefore, the groups are, from right to left: "110", "001", "101", "010". Therefore, converting these groups:

Group1 (rightmost): 110₂ = 6

Group2: 001₂ = 1

Group3: 101₂ = 5

Group4 (leftmost): 010₂ = 2

Therefore, the octal number is group4 group3 group2 group1: 2 5 1 6 → 2516₈, which matches the decimal conversion. So earlier mistake was in my grouping when I thought group3 was "010" instead of "101". Therefore, the correct groups are "010", "101", "001", "110" when grouped from the left, but when grouped from the right, the order is "110", "001", "101", "010", which correspond to octal digits 6, 1, 5, 2, so writing them in order from group4 to group1 (left to right): 2 5 1 6 → 2516. Yes! Therefore, the correct grouping is 2516.

This shows that it's vital to group correctly, either from the right (with proper identification of each group) or from the left after padding. So the binary number converted to octal is 2516.

Thus, the answer is \boxed{2516}.

**Final Answer**
The binary number $10101001110_2$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ into its octal equivalent, we can use the fact that **each octal digit corresponds to three binary digits**. Therefore, the strategy is to **group the binary digits into sets of three, starting from the right**, and convert each group to its octal equivalent.

---

### Step 1: Count the Number of Digits

The binary number is:

$$
1\ 0\ 1\ 0\ 1\ 0\ 0\ 1\ 1\ 1\ 0 \quad \text{(11 digits)}
$$

Since 11 is not divisible by 3, we **add one leading zero** to make the total number of bits equal to 12, which is divisible by 3.

The adjusted binary number becomes:

$$
0\ 1\ 0\ 1\ 0\ 1\ 0\ 0\ 1\ 1\ 1\ 0
$$

---

### Step 2: Group into Sets of Three (From Right to Left)

Break this into groups of 3 bits, starting from the right:

$$
010\ 101\ 001\ 110
$$

Note: Grouping from the right ensures that the least significant bits are grouped correctly.

---

### Step 3: Convert Each Binary Group to Octal

Now, convert each group from binary to octal:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the Octal Digits

Putting the octal digits together:

$$
2\ 5\ 1\ 6
$$

---

### Final Answer

$$
\boxed{2516}
$$

Solution: Since $2^3=8$, we may convert between base 2 and base 8 representations by replacing each block of three digits in base 2 with its equivalent in base 8. In this case, we begin by noticing that the last three digits are worth $110_2=6_8$. The next block of three digits is $001_2=1_8$. Continuing, we find that the next two digits (moving right-to-left) are $101_2=5_8$ and $010_2=2_8$. Altogether, we find that $10101001110_{2}=\boxed{2516_8}$.