Explanation:
Okay, so I need to find the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $. The problem gives me the equation:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \ldots $ are positive integers. My task is to compute $ b $.

First, I remember that continued fractions for cube roots might not be as straightforward as square roots, but maybe there's a systematic way to approach this. Let me start by recalling what a continued fraction is. A continued fraction is an expression of the form:

$$
x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dotsb}}}
$$

where each $ a_i $ is a positive integer. So, in this case, $ \sqrt[3]{2} $ is expressed as $ a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} $. Therefore, the first term $ a $ is the integer part of $ \sqrt[3]{2} $, and the rest is the continued fraction part.

Let me compute $ \sqrt[3]{2} $ numerically to get an idea. The cube root of 2 is approximately 1.25992105. Therefore, the integer part $ a $ should be 1. So, $ a = 1 $. Then, the remaining part of the continued fraction is $ \sqrt[3]{2} - 1 \approx 0.25992105 $.

So, the equation becomes:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Therefore, subtracting 1 from both sides:

$$
\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Taking reciprocal of both sides:

$$
\cfrac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

So, the left-hand side (LHS) here is $ \frac{1}{\sqrt[3]{2} - 1} $. Let me compute that value numerically.

First, $ \sqrt[3]{2} \approx 1.25992105 $, so $ \sqrt[3]{2} - 1 \approx 0.25992105 $, so reciprocal is approximately $ 1 / 0.25992105 \approx 3.8473221 $. So, that value is approximately 3.8473.

Therefore, the next term in the continued fraction is $ b $, which is the integer part of this value. Since 3.8473 is between 3 and 4, so $ b = 3 $. Wait, but let me confirm this more accurately.

But before I proceed, I need to be careful. Maybe there's a better way to compute $ b $ without relying solely on decimal approximations, especially because continued fractions can sometimes be tricky with irrational numbers.

Alternatively, perhaps I can manipulate the equation algebraically. Let me try that.

Starting with $ \frac{1}{\sqrt[3]{2} - 1} $. Let me rationalize this expression. To do that, I need to multiply numerator and denominator by something to eliminate the cube root in the denominator.

Let me recall that for expressions involving cube roots, multiplying by the conjugate can be a bit complex. The denominator is $ \sqrt[3]{2} - 1 $. Let me denote $ x = \sqrt[3]{2} $. Then, the denominator is $ x - 1 $, and the expression is $ \frac{1}{x - 1} $.

Since $ x^3 = 2 $, maybe I can express $ \frac{1}{x - 1} $ in terms of $ x $. Let me try to rationalize $ \frac{1}{x - 1} $.

Multiply numerator and denominator by $ x^2 + x + 1 $, because $ (x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1 $.

Oh! That's brilliant! Because $ x^3 - 1 = (x - 1)(x^2 + x + 1) = 1 $, so indeed, multiplying numerator and denominator by $ x^2 + x + 1 $ gives:

$$
\frac{1}{x - 1} \times \frac{x^2 + x + 1}{x^2 + x + 1} = \frac{x^2 + x + 1}{1} = x^2 + x + 1
$$

Therefore, $ \frac{1}{x - 1} = x^2 + x + 1 $. Since $ x = \sqrt[3]{2} $, substituting back:

$$
\frac{1}{\sqrt[3]{2} - 1} = (\sqrt[3]{2})^2 + \sqrt[3]{2} + 1
$$

So, this expression simplifies to $ (\sqrt[3]{2})^2 + \sqrt[3]{2} + 1 $, which is approximately $ 1.5874 + 1.2599 + 1 \approx 3.8473 $, which matches the decimal value I obtained earlier.

But now, how does this help me? The expression $ \frac{1}{\sqrt[3]{2} - 1} $ is equal to $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $. But how does this relate to the continued fraction?

Well, from earlier steps, we had:

$$
\frac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

Therefore, we can write:

$$
\sqrt[3]{4} + \sqrt[3]{2} + 1 = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

Since $ b $ is the integer part of $ \frac{1}{\sqrt[3]{2} - 1} $, which is approximately 3.8473, so $ b = 3 $. But maybe I can confirm this algebraically as well.

But let me check if $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $ is indeed equal to approximately 3.8473. Let me compute each term:

- $ \sqrt[3]{4} \approx 1.5874 $
- $ \sqrt[3]{2} \approx 1.2599 $
- So adding them up: 1.5874 + 1.2599 + 1 ≈ 1.5874 + 1.2599 = 2.8473 + 1 = 3.8473. Yep, that's correct.

Therefore, the value of $ b $ is 3? But wait, is there a way to confirm that the continued fraction expansion after 3 is indeed... Let me think.

Wait, so if $ \frac{1}{\sqrt[3]{2} - 1} = 3 + \frac{1}{c + \dotsb} $, then perhaps we can write:

Let me denote $ y = \frac{1}{\sqrt[3]{2} - 1} \approx 3.8473 $, so $ y = 3 + \frac{1}{c + \dotsb} $. Therefore, the fractional part is $ y - 3 \approx 0.8473 $, so $ \frac{1}{c + \dotsb} \approx 0.8473 $, so $ c + \dotsb \approx \frac{1}{0.8473} \approx 1.18 \ldots $. Therefore, $ c $ must be 1, because the integer part is 1. But I need to check this more accurately.

But maybe there's a better way to compute $ c $ by repeating the process. Let me try to formalize this.

Let me denote $ y_0 = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 3.8473 $.

So, $ b = \lfloor y_0 \rfloor = 3 $, so then the next term is $ \frac{1}{y_0 - b} = \frac{1}{y_0 - 3} $.

So let me compute $ y_1 = \frac{1}{y_0 - 3} $. Since $ y_0 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, then $ y_0 - 3 = \sqrt[3]{4} + \sqrt[3]{2} + 1 - 3 = \sqrt[3]{4} + \sqrt[3]{2} - 2 $.

So $ y_1 = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} - 2} $. Let me compute this expression.

Again, perhaps rationalizing or manipulating algebraically. Let me denote $ x = \sqrt[3]{2} $. Then $ \sqrt[3]{4} = x^2 $, and $ y_0 - 3 = x^2 + x - 2 $. So $ y_1 = \frac{1}{x^2 + x - 2} $.

Let me factor the denominator. Let me see: $ x^2 + x - 2 $. If we treat this as a quadratic in x, maybe we can factor it? Let me check.

Let me try to factor $ x^2 + x - 2 $. The discriminant is $ 1^2 - 4(1)(-2) = 1 + 8 = 9 $. So the roots are $ \frac{-1 \pm 3}{2} $. So, $ \frac{-1 + 3}{2} = 1 $, and $ \frac{-1 - 3}{2} = -2 $. Therefore, the quadratic factors as $ (x - 1)(x + 2) $. Wait:

$ (x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2 $. Yes, correct.

Therefore, $ y_1 = \frac{1}{(x - 1)(x + 2)} $, where $ x = \sqrt[3]{2} $. Interesting.

But how does this help me? Maybe I can perform partial fractions or find another way to simplify this. Alternatively, maybe express this in terms of cube roots. Let me see.

Alternatively, perhaps I can express $ \frac{1}{(x - 1)(x + 2)} $ as $ \frac{A}{x - 1} + \frac{B}{x + 2} $. Let me try partial fractions.

Let me set:

$$
\frac{1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}
$$

Multiply both sides by $ (x - 1)(x + 2) $:

$$
1 = A(x + 2) + B(x - 1)
$$

Expanding:

$$
1 = A x + 2A + B x - B = (A + B)x + (2A - B)
$$

Therefore, equating coefficients:

- Coefficient of x: $ A + B = 0 $
- Constant term: $ 2A - B = 1 $

From the first equation: $ B = -A $. Substitute into the second equation:

$ 2A - (-A) = 3A = 1 \implies A = 1/3 $, so $ B = -1/3 $.

Therefore,

$$
\frac{1}{(x - 1)(x + 2)} = \frac{1}{3(x - 1)} - \frac{1}{3(x + 2)}
$$

Therefore, $ y_1 = \frac{1}{3(x - 1)} - \frac{1}{3(x + 2)} $.

But we already know that $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore:

$$
\frac{1}{3(x - 1)} = \frac{x^2 + x + 1}{3}
$$

So substituting back into $ y_1 $:

$$
y_1 = \frac{x^2 + x + 1}{3} - \frac{1}{3(x + 2)}
$$

Hmm. Hmm. Maybe this isn't helpful. Alternatively, let me compute $ y_1 = \frac{1}{(x - 1)(x + 2)} $. Let me plug in the value of $ x \approx 1.2599 $:

So, $ x + 2 \approx 3.2599 $, and $ x - 1 \approx 0.2599 $. So, denominator is approximately 0.2599 * 3.2599 ≈ 0.2599 * 3.2599 ≈ let me compute:

First, approximate 0.2599 * 3.2599:

Well, 0.25 * 3.25 = 0.8125, but since it's 0.2599 * 3.2599, which is a bit more. Let me compute:

Let me approximate 3.2599 ≈ 3.26, 0.2599 ≈ 0.26. So 0.26 * 3.26 ≈ 0.26 * 3 + 0.26 * 0.26 ≈ 0.78 + 0.0676 ≈ 0.8476. So the denominator is approximately 0.8476, so $ y_1 = 1 / 0.8476 ≈ 1.18 $.

Therefore, $ y_1 ≈ 1.18 $. Therefore, the next term $ c $ is the integer part of $ y_1 $, which is 1. So $ c = 1 $, and then the next term is $ \frac{1}{y_1 - 1} \approx \frac{1}{0.18} ≈ 5.555... $. Therefore, the next term $ d = 5 $. But perhaps this is getting too deep into the continued fraction. However, since the problem only asks for $ b $, maybe I can stop here? But wait, the problem says that $ a, b, c, d $ are positive integers. So I need to confirm that $ b = 3 $, but maybe there's something wrong here? Let me think again.

Wait, but I had earlier that $ \frac{1}{\sqrt[3]{2} - 1} = 3 + \frac{1}{c + \dotsb} $. So if I have that $ \frac{1}{\sqrt[3]{2} - 1} $ equals $ y_0 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is exactly equal to 3.847... So the integer part is indeed 3, so $ b = 3 $. Therefore, perhaps the answer is 3? But maybe I need to check if there's a mistake here?

Alternatively, maybe there's a mistake in my assumption that $ \frac{1}{\sqrt[3]{2} - 1} = x^2 + x + 1 $. Let me check that again.

Let me compute $ (x - 1)(x^2 + x + 1) $. Expanding:

$ x*(x^2 + x + 1) - 1*(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1 $. Since $ x^3 = 2 $, so this equals $ 2 - 1 = 1 $. Therefore, yes, $ (x - 1)(x^2 + x + 1) = 1 $, so $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore, that's correct. Therefore, the previous calculation is correct.

Therefore, $ y_0 = x^2 + x + 1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 1.5874 + 1.2599 + 1 = approx 3.8473. So floor of that is 3, so $ b = 3 $.

But maybe I should check some references or known continued fractions for cube roots? Wait, but since I don't have references here, I need to verify this another way.

Alternatively, perhaps I can use the definition of continued fractions. The continued fraction expansion of a number is found by taking the integer part, then taking reciprocal of the fractional part, and repeating.

So starting with $ \sqrt[3]{2} \approx 1.2599 $, so $ a = 1 $. Then, the fractional part is $ \sqrt[3]{2} - 1 \approx 0.2599 $. Taking reciprocal: $ 1 / 0.2599 ≈ 3.847 $, so the next term $ b = 3 $. Then, take the fractional part of that, which is $ 0.847 $, reciprocal is approx 1.18, so next term is 1. Then fractional part is 0.18, reciprocal approx 5.55, next term 5, etc.

But this seems to confirm that $ b = 3 $. So perhaps the answer is 3?

But wait, let me check with another approach. Maybe using the continued fraction expansion algorithm.

Let me recall that the continued fraction algorithm for a real number $ \alpha $ is:

- Let $ a_0 = \lfloor \alpha \rfloor $
- Let $ \alpha_1 = 1 / (\alpha - a_0) $
- Then $ a_1 = \lfloor \alpha_1 \rfloor $, $ \alpha_2 = 1 / (\alpha_1 - a_1) $, etc.

Applying this to $ \alpha = \sqrt[3]{2} \approx 1.2599 $:

- $ a_0 = 1 $
- $ \alpha_1 = 1 / (\sqrt[3]{2} - 1) \approx 1 / 0.2599 ≈ 3.847 $
- $ a_1 = \lfloor 3.847 \rfloor = 3 $
- $ \alpha_2 = 1 / (3.847 - 3) ≈ 1 / 0.847 ≈ 1.18 $
- $ a_2 = \lfloor 1.18 \rfloor = 1 $
- $ \alpha_3 = 1 / (1.18 - 1) = 1 / 0.18 ≈ 5.555 $
- $ a_3 = 5 $
- And so on.

So, according to this, the continued fraction expansion of $ \sqrt[3]{2} $ starts as [1; 3, 1, 5, ...], so $ a = 1 $, $ b = 3 $, $ c = 1 $, $ d = 5 $, etc. Therefore, the value of $ b $ is 3.

Therefore, I think the answer is 3. However, just to be thorough, let me check if there's a possibility that my calculation of $ \alpha_1 $ is incorrect. Let me recast the problem using algebra.

Given that $ \sqrt[3]{2} = a + \frac{1}{b + \frac{1}{c + \dots}} $. We found $ a = 1 $, and then $ \frac{1}{\sqrt[3]{2} - 1} = b + \frac{1}{c + \dots} $.

Earlier, we found that $ \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $:

Since $ \sqrt[3]{4} = (\sqrt[3]{2})^2 $. Let me denote $ x = \sqrt[3]{2} $, so expression becomes $ x^2 + x + 1 $. Let me check if this is equal to $ \frac{1}{x - 1} $:

Earlier, we saw that $ (x - 1)(x^2 + x + 1) = 1 $, so yes, $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore, that's correct. Therefore, the value of $ \frac{1}{x - 1} $ is indeed $ x^2 + x + 1 $, which is approximately 3.847 as we saw. Therefore, the continued fraction expansion gives us $ b = 3 $, since the integer part is 3.

Therefore, unless there's a mistake in this entire reasoning, $ b = 3 $.

But to be 100% sure, let me check if someone else has computed the continued fraction of cube root of 2. Wait, since I can't access external resources, I need to rely on my own calculations. Let me try another approach.

Suppose we assume that the continued fraction for $ \sqrt[3]{2} $ is [1; 3, 1, 5, ...], as per previous steps, then let me see if the convergents approach the value.

First convergent is 1.

Second convergent is 1 + 1/3 = 4/3 ≈ 1.333...

Third convergent is 1 + 1/(3 + 1/1) = 1 + 1/(4) = 5/4 = 1.25

Fourth convergent is 1 + 1/(3 + 1/(1 + 1/5)) = 1 + 1/(3 + 1/(6/5)) = 1 + 1/(3 + 5/6) = 1 + 1/(23/6) = 1 + 6/23 ≈ 1 + 0.2609 ≈ 1.2609

Compare with $ \sqrt[3]{2} \approx 1.2599 $. So, the fourth convergent is approx 1.2609, which is slightly larger than the actual value. Then the next convergent would involve more terms. So, as we can see, the convergents approach the actual value. Therefore, the continued fraction expansion seems to be accurate.

Alternatively, maybe I can compute more terms. Let me try.

We had up to $ \alpha_3 ≈ 5.555 $, so $ a_3 = 5 $, then $ \alpha_4 = 1 / (5.555 - 5) = 1 / 0.555 ≈ 1.8 $, so $ a_4 = 1 $, and so on.

Therefore, the continued fraction is [1; 3, 1, 5, 1, ...]. So the sequence is 1; 3, 1, 5, 1, ...

Therefore, in this case, $ b = 3 $, so I think that the answer is 3.

Alternatively, perhaps there is a mistake in assuming that all steps are correct. Let me check once more the first step.

Original value: $ \sqrt[3]{2} \approx 1.2599 $. Then, subtract integer part 1: approx 0.2599. Take reciprocal: approx 3.847. So, the continued fraction is 1 + 1/(3 + ...). Therefore, yes, b = 3.

Therefore, unless there is some hidden trick or something I'm missing, I think the answer is 3.

But just to be thorough, let me try to see if there's an alternative expression or identity that might suggest otherwise. For example, could there be a continued fraction representation where b is different?

Alternatively, perhaps I made a mistake in the algebra earlier when rationalizing. Let me check that again.

We had $ \frac{1}{\sqrt[3]{2} - 1} $. Let me rationalize it again. Let me denote $ x = \sqrt[3]{2} $, then $ x^3 = 2 $. Multiply numerator and denominator by $ x^2 + x + 1 $:

Numerator becomes $ x^2 + x + 1 $. Denominator becomes $ (x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1 $. Therefore, yes, the expression is indeed $ x^2 + x + 1 $. Therefore, correct. Therefore, the value is $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is approximately 3.847.

Hence, the continued fraction term b is 3. Therefore, I think this is the correct answer.

Wait, but maybe I should check with another method. For example, suppose I assume that the continued fraction is [1; 3, 1, 5, ...], then compute the value of the continued fraction up to certain terms and check if it approximates $ \sqrt[3]{2} $.

So, first convergent: 1.

Second convergent: 1 + 1/3 = 4/3 ≈ 1.333...

Third convergent: 1 + 1/(3 + 1/1) = 1 + 1/4 = 5/4 = 1.25

Fourth convergent: 1 + 1/(3 + 1/(1 + 1/5)) = 1 + 1/(3 + 1/(6/5)) = 1 + 1/(3 + 5/6) = 1 + 1/(23/6) = 1 + 6/23 ≈ 1 + 0.26087 ≈ 1.26087

Fifth convergent: Let me compute it. The fifth term is 1, so we have:

Continued fraction up to [1; 3, 1, 5, 1]. So, the convergent is:

Starting from the end:

Let me denote the continued fraction as:

1 + 1/(3 + 1/(1 + 1/(5 + 1/1)))

Compute innermost first:

5 + 1/1 = 6

Then, 1 + 1/6 = 7/6

Then, 3 + 1/(7/6) = 3 + 6/7 = 27/7

Then, 1 + 1/(27/7) = 1 + 7/27 = 34/27 ≈ 1.259259...

Compare with $ \sqrt[3]{2} ≈ 1.259921 $. So, the fifth convergent is 34/27 ≈ 1.259259, which is slightly less than the actual value.

Then, the next term would be... Let me compute the next term. Let me go further.

Previously, after the fifth term (which was 1), the next term $ \alpha_4 $ was approx 1.8, so $ a_4 = 1 $, then $ \alpha_5 = 1 / (1.8 - 1) = 1 / 0.8 = 1.25 $. Therefore, the sixth term is 1, and so on.

So, sixth convergent would be:

Continued fraction [1; 3, 1, 5, 1, 1]. So, compute:

Starting from the end:

1 + 1/1 = 2

Then, 5 + 1/2 = 11/2

Then, 1 + 1/(11/2) = 1 + 2/11 = 13/11

Then, 3 + 1/(13/11) = 3 + 11/13 = 50/13 ≈ 3.846...

Then, 1 + 1/(50/13) = 1 + 13/50 = 63/50 = 1.26

Wait, that doesn't seem to align. Maybe I need to compute it step by step.

Alternatively, perhaps it's getting too complex, but the point is that with each convergent, we get closer to the actual value. The fifth convergent is approx 1.259259, and the actual value is approx 1.259921. The difference is about 0.00066, which is small.

If we take more terms, we would approach closer. Therefore, this further supports that the continued fraction is [1; 3, 1, 5, 1, ...], so $ b = 3 $.

Therefore, after all these verifications, I think that the value of $ b $ is indeed 3.

But just to ensure, let me think if there is a mathematical reason why $ b $ cannot be something else. For example, suppose I thought maybe $ b = 2 $, would that be possible?

If $ b = 2 $, then the continued fraction would be $ 1 + 1/(2 + ...) $, so the value would be $ 1 + 1/(something) $. Let me compute $ 1 + 1/(2 + ...) $. If the rest of the continued fraction is such that $ 2 + ... $ is about 3.847 / 1? Wait, no. Wait, the value after subtracting 1 is approx 0.2599, which is 1 / 3.847. If $ b = 2 $, then the next term would be 2 + ..., which would make the reciprocal $ 1/(2 + ...) = 0.2599 $, so $ 2 + ... = 1 / 0.2599 ≈ 3.847 $. Therefore, the continued fraction after $ b = 2 $ would be $ 2 + ... = 3.847 $, which would mean that the ... part is $ 3.847 - 2 = 1.847 $, and then taking reciprocal would be approx 0.541... So, then $ c $ would be 0? But $ c $ must be a positive integer. Therefore, this is impossible. Therefore, if we had $ b = 2 $, then the next term would require a fractional part that leads to $ c $ being zero or negative, which is not allowed. Therefore, $ b $ cannot be 2. Similarly, if $ b = 4 $, then the reciprocal would be 1 / (something). Let me check:

If $ b = 4 $, then the value after subtracting 1 would be approx 0.2599 = 1/(4 + ...). Therefore, 4 + ... = 1 / 0.2599 ≈ 3.847, so ... = 3.847 - 4 = -0.153, which is negative. But since continued fractions have positive integers, this is impossible. Therefore, $ b = 4 $ is invalid. Hence, the only possible value is $ b = 3 $.

Therefore, this further confirms that $ b = 3 $ is the correct answer.

Thus, after all these steps and checks, I can be confident that the value of $ b $ is 3.

**Final Answer**
\boxed{3}

To determine the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $, we begin by examining the structure of the provided identity:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \ldots $ are positive integers.

---

### Step 1: Determine the Integer Part

We start by approximating the cube root:

$$
\sqrt[3]{2} \approx 1.25992105
$$

Thus, the integer part $ a = \lfloor \sqrt[3]{2} \rfloor = 1 $, and the fractional part is:

$$
\sqrt[3]{2} - 1 \approx 0.25992105
$$

Rewriting the identity:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Subtracting 1 from both sides:

$$
\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Taking the reciprocal:

$$
\cfrac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

So, the task reduces to evaluating this reciprocal expression.

---

### Step 2: Algebraic Rationalization

Let $ x = \sqrt[3]{2} $. Then:

$$
\frac{1}{\sqrt[3]{2} - 1} = \frac{1}{x - 1}
$$

To rationalize, we multiply the numerator and denominator by the factor $ x^2 + x + 1 $, which is the conjugate expression for $ x^3 - 1 $:

$$
(x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1
$$

Thus:

$$
\frac{1}{x - 1} = x^2 + x + 1
$$

Now substitute $ x = \sqrt[3]{2} $:

$$
\frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

Numerically:

- $ \sqrt[3]{4} = (\sqrt[3]{2})^2 \approx 1.5874 $
- $ \sqrt[3]{2} \approx 1.259921 $
- Sum: $ 1.5874 + 1.259921 + 1 \approx 3.8473 $

Thus:

$$
\frac{1}{\sqrt[3]{2} - 1} \approx 3.8473
$$

---

### Step 3: Extract the First Continued Fraction Term

From:

$$
\frac{1}{\sqrt[3]{2} - 1} = b + \text{(continued fraction)}
$$

we find the integer part $ b = \lfloor 3.8473 \rfloor = 3 $

Hence, the value of $ b $ is:

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $. The problem gives me the equation:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \ldots $ are positive integers. My task is to compute $ b $.

First, I remember that continued fractions for cube roots might not be as straightforward as square roots, but maybe there's a systematic way to approach this. Let me start by recalling what a continued fraction is. A continued fraction is an expression of the form:

$$
x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dotsb}}}
$$

where each $ a_i $ is a positive integer. So, in this case, $ \sqrt[3]{2} $ is expressed as $ a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} $. Therefore, the first term $ a $ is the integer part of $ \sqrt[3]{2} $, and the rest is the continued fraction part.

Let me compute $ \sqrt[3]{2} $ numerically to get an idea. The cube root of 2 is approximately 1.25992105. Therefore, the integer part $ a $ should be 1. So, $ a = 1 $. Then, the remaining part of the continued fraction is $ \sqrt[3]{2} - 1 \approx 0.25992105 $.

So, the equation becomes:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Therefore, subtracting 1 from both sides:

$$
\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Taking reciprocal of both sides:

$$
\cfrac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

So, the left-hand side (LHS) here is $ \frac{1}{\sqrt[3]{2} - 1} $. Let me compute that value numerically.

First, $ \sqrt[3]{2} \approx 1.25992105 $, so $ \sqrt[3]{2} - 1 \approx 0.25992105 $, so reciprocal is approximately $ 1 / 0.25992105 \approx 3.8473221 $. So, that value is approximately 3.8473.

Therefore, the next term in the continued fraction is $ b $, which is the integer part of this value. Since 3.8473 is between 3 and 4, so $ b = 3 $. Wait, but let me confirm this more accurately.

But before I proceed, I need to be careful. Maybe there's a better way to compute $ b $ without relying solely on decimal approximations, especially because continued fractions can sometimes be tricky with irrational numbers.

Alternatively, perhaps I can manipulate the equation algebraically. Let me try that.

Starting with $ \frac{1}{\sqrt[3]{2} - 1} $. Let me rationalize this expression. To do that, I need to multiply numerator and denominator by something to eliminate the cube root in the denominator.

Let me recall that for expressions involving cube roots, multiplying by the conjugate can be a bit complex. The denominator is $ \sqrt[3]{2} - 1 $. Let me denote $ x = \sqrt[3]{2} $. Then, the denominator is $ x - 1 $, and the expression is $ \frac{1}{x - 1} $.

Since $ x^3 = 2 $, maybe I can express $ \frac{1}{x - 1} $ in terms of $ x $. Let me try to rationalize $ \frac{1}{x - 1} $.

Multiply numerator and denominator by $ x^2 + x + 1 $, because $ (x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1 $.

Oh! That's brilliant! Because $ x^3 - 1 = (x - 1)(x^2 + x + 1) = 1 $, so indeed, multiplying numerator and denominator by $ x^2 + x + 1 $ gives:

$$
\frac{1}{x - 1} \times \frac{x^2 + x + 1}{x^2 + x + 1} = \frac{x^2 + x + 1}{1} = x^2 + x + 1
$$

Therefore, $ \frac{1}{x - 1} = x^2 + x + 1 $. Since $ x = \sqrt[3]{2} $, substituting back:

$$
\frac{1}{\sqrt[3]{2} - 1} = (\sqrt[3]{2})^2 + \sqrt[3]{2} + 1
$$

So, this expression simplifies to $ (\sqrt[3]{2})^2 + \sqrt[3]{2} + 1 $, which is approximately $ 1.5874 + 1.2599 + 1 \approx 3.8473 $, which matches the decimal value I obtained earlier.

But now, how does this help me? The expression $ \frac{1}{\sqrt[3]{2} - 1} $ is equal to $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $. But how does this relate to the continued fraction?

Well, from earlier steps, we had:

$$
\frac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

Therefore, we can write:

$$
\sqrt[3]{4} + \sqrt[3]{2} + 1 = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

Since $ b $ is the integer part of $ \frac{1}{\sqrt[3]{2} - 1} $, which is approximately 3.8473, so $ b = 3 $. But maybe I can confirm this algebraically as well.

But let me check if $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $ is indeed equal to approximately 3.8473. Let me compute each term:

- $ \sqrt[3]{4} \approx 1.5874 $
- $ \sqrt[3]{2} \approx 1.2599 $
- So adding them up: 1.5874 + 1.2599 + 1 ≈ 1.5874 + 1.2599 = 2.8473 + 1 = 3.8473. Yep, that's correct.

Therefore, the value of $ b $ is 3? But wait, is there a way to confirm that the continued fraction expansion after 3 is indeed... Let me think.

Wait, so if $ \frac{1}{\sqrt[3]{2} - 1} = 3 + \frac{1}{c + \dotsb} $, then perhaps we can write:

Let me denote $ y = \frac{1}{\sqrt[3]{2} - 1} \approx 3.8473 $, so $ y = 3 + \frac{1}{c + \dotsb} $. Therefore, the fractional part is $ y - 3 \approx 0.8473 $, so $ \frac{1}{c + \dotsb} \approx 0.8473 $, so $ c + \dotsb \approx \frac{1}{0.8473} \approx 1.18 \ldots $. Therefore, $ c $ must be 1, because the integer part is 1. But I need to check this more accurately.

But maybe there's a better way to compute $ c $ by repeating the process. Let me try to formalize this.

Let me denote $ y_0 = \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 3.8473 $.

So, $ b = \lfloor y_0 \rfloor = 3 $, so then the next term is $ \frac{1}{y_0 - b} = \frac{1}{y_0 - 3} $.

So let me compute $ y_1 = \frac{1}{y_0 - 3} $. Since $ y_0 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, then $ y_0 - 3 = \sqrt[3]{4} + \sqrt[3]{2} + 1 - 3 = \sqrt[3]{4} + \sqrt[3]{2} - 2 $.

So $ y_1 = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} - 2} $. Let me compute this expression.

Again, perhaps rationalizing or manipulating algebraically. Let me denote $ x = \sqrt[3]{2} $. Then $ \sqrt[3]{4} = x^2 $, and $ y_0 - 3 = x^2 + x - 2 $. So $ y_1 = \frac{1}{x^2 + x - 2} $.

Let me factor the denominator. Let me see: $ x^2 + x - 2 $. If we treat this as a quadratic in x, maybe we can factor it? Let me check.

Let me try to factor $ x^2 + x - 2 $. The discriminant is $ 1^2 - 4(1)(-2) = 1 + 8 = 9 $. So the roots are $ \frac{-1 \pm 3}{2} $. So, $ \frac{-1 + 3}{2} = 1 $, and $ \frac{-1 - 3}{2} = -2 $. Therefore, the quadratic factors as $ (x - 1)(x + 2) $. Wait:

$ (x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2 $. Yes, correct.

Therefore, $ y_1 = \frac{1}{(x - 1)(x + 2)} $, where $ x = \sqrt[3]{2} $. Interesting.

But how does this help me? Maybe I can perform partial fractions or find another way to simplify this. Alternatively, maybe express this in terms of cube roots. Let me see.

Alternatively, perhaps I can express $ \frac{1}{(x - 1)(x + 2)} $ as $ \frac{A}{x - 1} + \frac{B}{x + 2} $. Let me try partial fractions.

Let me set:

$$
\frac{1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}
$$

Multiply both sides by $ (x - 1)(x + 2) $:

$$
1 = A(x + 2) + B(x - 1)
$$

Expanding:

$$
1 = A x + 2A + B x - B = (A + B)x + (2A - B)
$$

Therefore, equating coefficients:

- Coefficient of x: $ A + B = 0 $
- Constant term: $ 2A - B = 1 $

From the first equation: $ B = -A $. Substitute into the second equation:

$ 2A - (-A) = 3A = 1 \implies A = 1/3 $, so $ B = -1/3 $.

Therefore,

$$
\frac{1}{(x - 1)(x + 2)} = \frac{1}{3(x - 1)} - \frac{1}{3(x + 2)}
$$

Therefore, $ y_1 = \frac{1}{3(x - 1)} - \frac{1}{3(x + 2)} $.

But we already know that $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore:

$$
\frac{1}{3(x - 1)} = \frac{x^2 + x + 1}{3}
$$

So substituting back into $ y_1 $:

$$
y_1 = \frac{x^2 + x + 1}{3} - \frac{1}{3(x + 2)}
$$

Hmm. Hmm. Maybe this isn't helpful. Alternatively, let me compute $ y_1 = \frac{1}{(x - 1)(x + 2)} $. Let me plug in the value of $ x \approx 1.2599 $:

So, $ x + 2 \approx 3.2599 $, and $ x - 1 \approx 0.2599 $. So, denominator is approximately 0.2599 * 3.2599 ≈ 0.2599 * 3.2599 ≈ let me compute:

First, approximate 0.2599 * 3.2599:

Well, 0.25 * 3.25 = 0.8125, but since it's 0.2599 * 3.2599, which is a bit more. Let me compute:

Let me approximate 3.2599 ≈ 3.26, 0.2599 ≈ 0.26. So 0.26 * 3.26 ≈ 0.26 * 3 + 0.26 * 0.26 ≈ 0.78 + 0.0676 ≈ 0.8476. So the denominator is approximately 0.8476, so $ y_1 = 1 / 0.8476 ≈ 1.18 $.

Therefore, $ y_1 ≈ 1.18 $. Therefore, the next term $ c $ is the integer part of $ y_1 $, which is 1. So $ c = 1 $, and then the next term is $ \frac{1}{y_1 - 1} \approx \frac{1}{0.18} ≈ 5.555... $. Therefore, the next term $ d = 5 $. But perhaps this is getting too deep into the continued fraction. However, since the problem only asks for $ b $, maybe I can stop here? But wait, the problem says that $ a, b, c, d $ are positive integers. So I need to confirm that $ b = 3 $, but maybe there's something wrong here? Let me think again.

Wait, but I had earlier that $ \frac{1}{\sqrt[3]{2} - 1} = 3 + \frac{1}{c + \dotsb} $. So if I have that $ \frac{1}{\sqrt[3]{2} - 1} $ equals $ y_0 = \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is exactly equal to 3.847... So the integer part is indeed 3, so $ b = 3 $. Therefore, perhaps the answer is 3? But maybe I need to check if there's a mistake here?

Alternatively, maybe there's a mistake in my assumption that $ \frac{1}{\sqrt[3]{2} - 1} = x^2 + x + 1 $. Let me check that again.

Let me compute $ (x - 1)(x^2 + x + 1) $. Expanding:

$ x*(x^2 + x + 1) - 1*(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1 $. Since $ x^3 = 2 $, so this equals $ 2 - 1 = 1 $. Therefore, yes, $ (x - 1)(x^2 + x + 1) = 1 $, so $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore, that's correct. Therefore, the previous calculation is correct.

Therefore, $ y_0 = x^2 + x + 1 = \sqrt[3]{4} + \sqrt[3]{2} + 1 \approx 1.5874 + 1.2599 + 1 = approx 3.8473. So floor of that is 3, so $ b = 3 $.

But maybe I should check some references or known continued fractions for cube roots? Wait, but since I don't have references here, I need to verify this another way.

Alternatively, perhaps I can use the definition of continued fractions. The continued fraction expansion of a number is found by taking the integer part, then taking reciprocal of the fractional part, and repeating.

So starting with $ \sqrt[3]{2} \approx 1.2599 $, so $ a = 1 $. Then, the fractional part is $ \sqrt[3]{2} - 1 \approx 0.2599 $. Taking reciprocal: $ 1 / 0.2599 ≈ 3.847 $, so the next term $ b = 3 $. Then, take the fractional part of that, which is $ 0.847 $, reciprocal is approx 1.18, so next term is 1. Then fractional part is 0.18, reciprocal approx 5.55, next term 5, etc.

But this seems to confirm that $ b = 3 $. So perhaps the answer is 3?

But wait, let me check with another approach. Maybe using the continued fraction expansion algorithm.

Let me recall that the continued fraction algorithm for a real number $ \alpha $ is:

- Let $ a_0 = \lfloor \alpha \rfloor $
- Let $ \alpha_1 = 1 / (\alpha - a_0) $
- Then $ a_1 = \lfloor \alpha_1 \rfloor $, $ \alpha_2 = 1 / (\alpha_1 - a_1) $, etc.

Applying this to $ \alpha = \sqrt[3]{2} \approx 1.2599 $:

- $ a_0 = 1 $
- $ \alpha_1 = 1 / (\sqrt[3]{2} - 1) \approx 1 / 0.2599 ≈ 3.847 $
- $ a_1 = \lfloor 3.847 \rfloor = 3 $
- $ \alpha_2 = 1 / (3.847 - 3) ≈ 1 / 0.847 ≈ 1.18 $
- $ a_2 = \lfloor 1.18 \rfloor = 1 $
- $ \alpha_3 = 1 / (1.18 - 1) = 1 / 0.18 ≈ 5.555 $
- $ a_3 = 5 $
- And so on.

So, according to this, the continued fraction expansion of $ \sqrt[3]{2} $ starts as [1; 3, 1, 5, ...], so $ a = 1 $, $ b = 3 $, $ c = 1 $, $ d = 5 $, etc. Therefore, the value of $ b $ is 3.

Therefore, I think the answer is 3. However, just to be thorough, let me check if there's a possibility that my calculation of $ \alpha_1 $ is incorrect. Let me recast the problem using algebra.

Given that $ \sqrt[3]{2} = a + \frac{1}{b + \frac{1}{c + \dots}} $. We found $ a = 1 $, and then $ \frac{1}{\sqrt[3]{2} - 1} = b + \frac{1}{c + \dots} $.

Earlier, we found that $ \frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 $. Let me compute $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $:

Since $ \sqrt[3]{4} = (\sqrt[3]{2})^2 $. Let me denote $ x = \sqrt[3]{2} $, so expression becomes $ x^2 + x + 1 $. Let me check if this is equal to $ \frac{1}{x - 1} $:

Earlier, we saw that $ (x - 1)(x^2 + x + 1) = 1 $, so yes, $ \frac{1}{x - 1} = x^2 + x + 1 $. Therefore, that's correct. Therefore, the value of $ \frac{1}{x - 1} $ is indeed $ x^2 + x + 1 $, which is approximately 3.847 as we saw. Therefore, the continued fraction expansion gives us $ b = 3 $, since the integer part is 3.

Therefore, unless there's a mistake in this entire reasoning, $ b = 3 $.

But to be 100% sure, let me check if someone else has computed the continued fraction of cube root of 2. Wait, since I can't access external resources, I need to rely on my own calculations. Let me try another approach.

Suppose we assume that the continued fraction for $ \sqrt[3]{2} $ is [1; 3, 1, 5, ...], as per previous steps, then let me see if the convergents approach the value.

First convergent is 1.

Second convergent is 1 + 1/3 = 4/3 ≈ 1.333...

Third convergent is 1 + 1/(3 + 1/1) = 1 + 1/(4) = 5/4 = 1.25

Fourth convergent is 1 + 1/(3 + 1/(1 + 1/5)) = 1 + 1/(3 + 1/(6/5)) = 1 + 1/(3 + 5/6) = 1 + 1/(23/6) = 1 + 6/23 ≈ 1 + 0.2609 ≈ 1.2609

Compare with $ \sqrt[3]{2} \approx 1.2599 $. So, the fourth convergent is approx 1.2609, which is slightly larger than the actual value. Then the next convergent would involve more terms. So, as we can see, the convergents approach the actual value. Therefore, the continued fraction expansion seems to be accurate.

Alternatively, maybe I can compute more terms. Let me try.

We had up to $ \alpha_3 ≈ 5.555 $, so $ a_3 = 5 $, then $ \alpha_4 = 1 / (5.555 - 5) = 1 / 0.555 ≈ 1.8 $, so $ a_4 = 1 $, and so on.

Therefore, the continued fraction is [1; 3, 1, 5, 1, ...]. So the sequence is 1; 3, 1, 5, 1, ...

Therefore, in this case, $ b = 3 $, so I think that the answer is 3.

Alternatively, perhaps there is a mistake in assuming that all steps are correct. Let me check once more the first step.

Original value: $ \sqrt[3]{2} \approx 1.2599 $. Then, subtract integer part 1: approx 0.2599. Take reciprocal: approx 3.847. So, the continued fraction is 1 + 1/(3 + ...). Therefore, yes, b = 3.

Therefore, unless there is some hidden trick or something I'm missing, I think the answer is 3.

But just to be thorough, let me try to see if there's an alternative expression or identity that might suggest otherwise. For example, could there be a continued fraction representation where b is different?

Alternatively, perhaps I made a mistake in the algebra earlier when rationalizing. Let me check that again.

We had $ \frac{1}{\sqrt[3]{2} - 1} $. Let me rationalize it again. Let me denote $ x = \sqrt[3]{2} $, then $ x^3 = 2 $. Multiply numerator and denominator by $ x^2 + x + 1 $:

Numerator becomes $ x^2 + x + 1 $. Denominator becomes $ (x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1 $. Therefore, yes, the expression is indeed $ x^2 + x + 1 $. Therefore, correct. Therefore, the value is $ \sqrt[3]{4} + \sqrt[3]{2} + 1 $, which is approximately 3.847.

Hence, the continued fraction term b is 3. Therefore, I think this is the correct answer.

Wait, but maybe I should check with another method. For example, suppose I assume that the continued fraction is [1; 3, 1, 5, ...], then compute the value of the continued fraction up to certain terms and check if it approximates $ \sqrt[3]{2} $.

So, first convergent: 1.

Second convergent: 1 + 1/3 = 4/3 ≈ 1.333...

Third convergent: 1 + 1/(3 + 1/1) = 1 + 1/4 = 5/4 = 1.25

Fourth convergent: 1 + 1/(3 + 1/(1 + 1/5)) = 1 + 1/(3 + 1/(6/5)) = 1 + 1/(3 + 5/6) = 1 + 1/(23/6) = 1 + 6/23 ≈ 1 + 0.26087 ≈ 1.26087

Fifth convergent: Let me compute it. The fifth term is 1, so we have:

Continued fraction up to [1; 3, 1, 5, 1]. So, the convergent is:

Starting from the end:

Let me denote the continued fraction as:

1 + 1/(3 + 1/(1 + 1/(5 + 1/1)))

Compute innermost first:

5 + 1/1 = 6

Then, 1 + 1/6 = 7/6

Then, 3 + 1/(7/6) = 3 + 6/7 = 27/7

Then, 1 + 1/(27/7) = 1 + 7/27 = 34/27 ≈ 1.259259...

Compare with $ \sqrt[3]{2} ≈ 1.259921 $. So, the fifth convergent is 34/27 ≈ 1.259259, which is slightly less than the actual value.

Then, the next term would be... Let me compute the next term. Let me go further.

Previously, after the fifth term (which was 1), the next term $ \alpha_4 $ was approx 1.8, so $ a_4 = 1 $, then $ \alpha_5 = 1 / (1.8 - 1) = 1 / 0.8 = 1.25 $. Therefore, the sixth term is 1, and so on.

So, sixth convergent would be:

Continued fraction [1; 3, 1, 5, 1, 1]. So, compute:

Starting from the end:

1 + 1/1 = 2

Then, 5 + 1/2 = 11/2

Then, 1 + 1/(11/2) = 1 + 2/11 = 13/11

Then, 3 + 1/(13/11) = 3 + 11/13 = 50/13 ≈ 3.846...

Then, 1 + 1/(50/13) = 1 + 13/50 = 63/50 = 1.26

Wait, that doesn't seem to align. Maybe I need to compute it step by step.

Alternatively, perhaps it's getting too complex, but the point is that with each convergent, we get closer to the actual value. The fifth convergent is approx 1.259259, and the actual value is approx 1.259921. The difference is about 0.00066, which is small.

If we take more terms, we would approach closer. Therefore, this further supports that the continued fraction is [1; 3, 1, 5, 1, ...], so $ b = 3 $.

Therefore, after all these verifications, I think that the value of $ b $ is indeed 3.

But just to ensure, let me think if there is a mathematical reason why $ b $ cannot be something else. For example, suppose I thought maybe $ b = 2 $, would that be possible?

If $ b = 2 $, then the continued fraction would be $ 1 + 1/(2 + ...) $, so the value would be $ 1 + 1/(something) $. Let me compute $ 1 + 1/(2 + ...) $. If the rest of the continued fraction is such that $ 2 + ... $ is about 3.847 / 1? Wait, no. Wait, the value after subtracting 1 is approx 0.2599, which is 1 / 3.847. If $ b = 2 $, then the next term would be 2 + ..., which would make the reciprocal $ 1/(2 + ...) = 0.2599 $, so $ 2 + ... = 1 / 0.2599 ≈ 3.847 $. Therefore, the continued fraction after $ b = 2 $ would be $ 2 + ... = 3.847 $, which would mean that the ... part is $ 3.847 - 2 = 1.847 $, and then taking reciprocal would be approx 0.541... So, then $ c $ would be 0? But $ c $ must be a positive integer. Therefore, this is impossible. Therefore, if we had $ b = 2 $, then the next term would require a fractional part that leads to $ c $ being zero or negative, which is not allowed. Therefore, $ b $ cannot be 2. Similarly, if $ b = 4 $, then the reciprocal would be 1 / (something). Let me check:

If $ b = 4 $, then the value after subtracting 1 would be approx 0.2599 = 1/(4 + ...). Therefore, 4 + ... = 1 / 0.2599 ≈ 3.847, so ... = 3.847 - 4 = -0.153, which is negative. But since continued fractions have positive integers, this is impossible. Therefore, $ b = 4 $ is invalid. Hence, the only possible value is $ b = 3 $.

Therefore, this further confirms that $ b = 3 $ is the correct answer.

Thus, after all these steps and checks, I can be confident that the value of $ b $ is 3.

**Final Answer**
\boxed{3}

To determine the value of $ b $ in the continued fraction expansion of $ \sqrt[3]{2} $, we begin by examining the structure of the provided identity:

$$
\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

where $ a, b, c, d, \ldots $ are positive integers.

---

### Step 1: Determine the Integer Part

We start by approximating the cube root:

$$
\sqrt[3]{2} \approx 1.25992105
$$

Thus, the integer part $ a = \lfloor \sqrt[3]{2} \rfloor = 1 $, and the fractional part is:

$$
\sqrt[3]{2} - 1 \approx 0.25992105
$$

Rewriting the identity:

$$
\sqrt[3]{2} = 1 + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Subtracting 1 from both sides:

$$
\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}}
$$

Taking the reciprocal:

$$
\cfrac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}
$$

So, the task reduces to evaluating this reciprocal expression.

---

### Step 2: Algebraic Rationalization

Let $ x = \sqrt[3]{2} $. Then:

$$
\frac{1}{\sqrt[3]{2} - 1} = \frac{1}{x - 1}
$$

To rationalize, we multiply the numerator and denominator by the factor $ x^2 + x + 1 $, which is the conjugate expression for $ x^3 - 1 $:

$$
(x - 1)(x^2 + x + 1) = x^3 - 1 = 2 - 1 = 1
$$

Thus:

$$
\frac{1}{x - 1} = x^2 + x + 1
$$

Now substitute $ x = \sqrt[3]{2} $:

$$
\frac{1}{\sqrt[3]{2} - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1
$$

Numerically:

- $ \sqrt[3]{4} = (\sqrt[3]{2})^2 \approx 1.5874 $
- $ \sqrt[3]{2} \approx 1.259921 $
- Sum: $ 1.5874 + 1.259921 + 1 \approx 3.8473 $

Thus:

$$
\frac{1}{\sqrt[3]{2} - 1} \approx 3.8473
$$

---

### Step 3: Extract the First Continued Fraction Term

From:

$$
\frac{1}{\sqrt[3]{2} - 1} = b + \text{(continued fraction)}
$$

we find the integer part $ b = \lfloor 3.8473 \rfloor = 3 $

Hence, the value of $ b $ is:

$$
\boxed{3}
$$

Solution:
We know that
\[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} > a,\]and
\[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} < a + 1.\]The integer $a$ that satisfies $a < \sqrt[3]{2} < a + 1$ is $a = 1.$

Then
\[\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}},\]so
\[\frac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}.\]As before, $b$ must satisfy
\[b < \frac{1}{\sqrt[3]{2} - 1} < b + 1.\]Rationalizing the denominator, we get
\[\frac{1}{\sqrt[3]{2} - 1} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{(\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1)} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{2 - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1.\]We have that
\[\sqrt[3]{4} + \sqrt[3]{2} + 1 > 1 + 1 + 1 = 3.\]Also, $1.3^3 = 2.197 > 2$ and $1.6^3 = 4.096 > 4,$ so
\[\sqrt[3]{4} + \sqrt[3]{2} + 1 < 1.3 + 1.6 + 1 = 3.9 < 4,\]so $b = \boxed{3}.$


Explanation:
Okay, so I need to find the remainder when the 100th term of the Fibonacci sequence is divided by 4. Hmm, Fibonacci sequence... right, each term is the sum of the two previous ones. Let me recall: the first two terms are 1 and 1, then 2, 3, 5, 8, etc. So, the problem is asking for Fib(100) mod 4.

First thought: calculating the 100th term directly would be a huge number, and that seems impractical. There must be a pattern or a cycle when Fibonacci numbers are divided by 4. I remember something about Pisano periods, which are the periods with which the Fibonacci sequence repeats modulo some number. Maybe I can find the Pisano period for modulo 4 and then use that to find Fib(100) mod 4?

Let me check. The Pisano period for modulo m is the period after which the Fibonacci sequence repeats when taken modulo m. So, if I can find the Pisano period for 4, then I can compute 100 modulo that period to find the equivalent smaller term, which would be easier to compute.

Alright, let's try to find the Pisano period for modulo 4. Let me list out the Fibonacci numbers modulo 4 and see when the sequence starts repeating.

Starting with Fib(1) = 1, Fib(2) = 1:

Term 1: 1 mod 4 = 1

Term 2: 1 mod 4 = 1

Term 3: (1 + 1) = 2 mod 4 = 2

Term 4: (1 + 2) = 3 mod 4 = 3

Term 5: (2 + 3) = 5 mod 4 = 1

Term 6: (3 + 1) = 4 mod 4 = 0

Term 7: (1 + 0) = 1 mod 4 = 1

Term 8: (0 + 1) = 1 mod 4 = 1

Wait a second, term 7 and term 8 are 1 and 1, which are the same as term 1 and term 2. So does that mean the Pisano period is 6? Because after term 6, the sequence starts repeating?

Let me check the next terms to confirm:

Term 9: (1 + 1) = 2 mod 4 = 2 (same as term 3)

Term 10: (1 + 2) = 3 mod 4 = 3 (same as term 4)

Term 11: (2 + 3) = 5 mod 4 = 1 (same as term 5)

Term 12: (3 + 1) = 4 mod 4 = 0 (same as term 6)

Term 13: (1 + 0) = 1 (same as term 7)

So yes, the sequence repeats every 6 terms. So the Pisano period for modulo 4 is 6. Therefore, Fib(n) mod 4 repeats every 6 terms. Therefore, to find Fib(100) mod 4, I can compute 100 mod 6, find the equivalent term in the first period, and then take that term's value mod 4.

Let me compute 100 divided by 6. 6*16 = 96, so 100 - 96 = 4. Therefore, 100 mod 6 is 4. Therefore, Fib(100) mod 4 is equal to Fib(4) mod 4. Looking back at the list:

Term 4: 3 mod 4 = 3. Therefore, the remainder is 3?

Wait, but let me confirm with another approach. Maybe I made a mistake in the Pisano period?

Alternatively, perhaps I should check the Pisano period again. Let me list the terms again:

n: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16...

Fib(n): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987...

Now mod 4:

1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3...

Yes, so starting from term 1, the sequence mod 4 is: 1, 1, 2, 3, 1, 0, then repeats 1, 1, 2, 3, 1, 0... So the Pisano period here is indeed 6. So the cycle is 6 terms long. Therefore, the remainder when dividing n by 6 gives the position in the cycle.

So, n=100. Let me compute 100 divided by 6. 6*16=96, remainder 4. So it's equivalent to term 4 in the cycle. Term 4 mod 4 is 3. Therefore, Fib(100) mod 4 is 3.

But wait, let me check with another method to be sure. Maybe compute Fib(100) mod 4 using matrix exponentiation or recursion with mod?

Alternatively, perhaps compute some terms manually and see if the pattern holds.

Let me compute Fib(n) mod 4 for n up to, say, 12 as before:

n : Fib(n) : Fib(n) mod 4

1 : 1 : 1

2 : 1 : 1

3 : 2 : 2

4 : 3 : 3

5 : 5 : 1

6 : 8 : 0

7 : 13 : 1

8 : 21 : 1

9 : 34 : 2

10: 55 : 3

11: 89 : 1

12: 144 : 0

So, the cycle is indeed 6: 1, 1, 2, 3, 1, 0 repeating. Therefore, for n=100, since 100 mod 6 = 4, Fib(100) mod 4 = Fib(4) mod 4 = 3.

But let me check with n=6, which is 0, and n=12 is also 0, which is 6*2. So for n=6k, Fib(6k) mod 4 is 0. Similarly, n=6k+1: Fib(6k+1) mod 4 = 1. n=6k+2: 1, n=6k+3: 2, n=6k+4: 3, n=6k+5: 1.

Therefore, 100 divided by 6: 6*16=96, remainder 4, so 100=6*16+4, so Fib(100) mod 4 is same as Fib(4) mod 4 = 3. Therefore, answer is 3.

But just to be thorough, let me check another term. For example, n=7: 6*1 +1=7, Fib(7)=13 mod4=1, which matches. n=8=6*1+2: 21 mod4=1, which matches. n=9=6*1+3: 34 mod4=2, yes. n=10=6*1+4: 55 mod4=3, yes. n=11=6*1+5: 89 mod4=1, yes. n=12=6*2+0: 144 mod4=0. So the pattern holds. Therefore, confident that the pattern is correct.

Alternatively, maybe I can compute Fib(100) mod 4 using the property of Pisano period. Since Pisano period for 4 is 6, then Fib(n) mod4 = Fib(n mod 6) mod4? Wait, not exactly. Because if the period is 6, then Fib(n) mod4 = Fib(n mod 6) mod4? Wait, no. Wait, for example, Fib(7) mod4 = Fib(1) mod4 = 1, since 7 mod6=1. Similarly Fib(8)=Fib(2)=1. Fib(9)=Fib(3)=2. Fib(10)=Fib(4)=3. Fib(11)=Fib(5)=1. Fib(12)=Fib(6)=0. Fib(13)=Fib(7)=Fib(1)=1. So yes, it seems that Fib(n) mod4 = Fib(n mod 6) mod4, except when n mod6 is 0, in which case it's Fib(6) mod4=0. So if n mod6=0, then it's Fib(6)=0. So for n=6, 12, etc.

Therefore, in general, for n, compute n mod6. If the remainder is 0, then Fib(n) mod4=0. Else, Fib(n) mod4=Fib(remainder) mod4.

So, in our case, n=100. 100 divided by 6 is 16 with remainder 4. Therefore, Fib(100) mod4=Fib(4) mod4=3. Therefore, the remainder is 3.

Alternatively, perhaps I can compute Fib(100) mod4 by using matrix exponentiation or some recursive formula with modulus?

But given that the Pisano period is 6, it's much easier to use that. But just to check, let me try to compute Fib(100) mod4 using another approach.

Alternatively, maybe use the formula for Fibonacci numbers? But that involves irrational numbers and might not be helpful here. Alternatively, perhaps use the fact that the Fibonacci sequence modulo m is periodic, so we can compute Fib(100) mod4 by finding the cycle.

But since we already know the Pisano period is 6, we can just use that. However, just to make sure that my calculation of the Pisano period is correct, let me check again. The Pisano period for 4 is known?

Wait, actually, I think the Pisano period for modulo m can be calculated, but I need to make sure that for modulo 4, the Pisano period is indeed 6. Let me check a reference in my mind... The Pisano period for m=2 is 3, for m=3 it's 8, for m=4 it's 6. Yes, I think that's right. For example, Pisano periods for small m:

m | Pisano period
1 | 1
2 | 3
3 | 8
4 | 6
5 | 20
...

So for 4, 6 is correct. So that's good. So the Pisano period is 6, so the cycle is 6 terms. Therefore, Fib(n) mod4 repeats every 6 terms. Therefore, Fib(100) mod4 is Fib(100 mod6) mod4. Since 100 mod6 is 4, so Fib(4)=3 mod4=3. So the answer is 3.

But let me check with another example. For instance, Fib(6)=8, 8 mod4=0. Let me check Fib(6+6)=Fib(12)=144. 144 mod4=0. Same as 8 mod4. Similarly Fib(12+6)=Fib(18)=? Let me compute Fib(13)=233, Fib(14)=377, Fib(15)=610, Fib(16)=987, Fib(17)=1597, Fib(18)=2584. 2584 divided by 4: 2584/4=646, so remainder 0. So yes, it's 0 again. So the pattern holds. So that seems valid.

Alternatively, maybe I can use the concept of modular Fibonacci numbers. Let me try to compute Fib(n) mod4 up to n=100, but that would take too long. However, since the period is 6, and 100 is a multiple of 6 plus 4, we can be confident that it's 3.

Alternatively, maybe compute Fib(100) mod4 by using the recursive formula with mod4 at each step. But since we know the Pisano period is 6, we can just figure out which term in the cycle it is. However, if I wanted to do it manually, perhaps I can compute the terms up to the 100th term mod4, but that would take time.

Alternatively, since the period is 6, we can think of the sequence as repeating every 6 terms:

Positions:

1: 1

2: 1

3: 2

4: 3

5: 1

6: 0

Then repeats:

7: 1 (same as 1)

8: 1 (same as 2)

9: 2 (same as 3)

10: 3 (same as 4)

11: 1 (same as 5)

12: 0 (same as 6)

and so on.

Therefore, to find Fib(100) mod4, since 100 divided by 6 is 16 with remainder 4, so it's the same as the 4th term in the cycle, which is 3. So answer is 3.

But let me think if there is any possibility that the Pisano period is longer. For example, sometimes the Pisano period for m is 2*something? For example, for m=2, Pisano period is 3, which is not 2. For m=4, Pisano period is 6, which is 2*3. So maybe it's different. However, in our case, we have confirmed that the cycle is 6. So unless there is an error in my calculation, the answer should be 3.

Wait, let me check Fib(4). Fib(4) is 3, 3 mod4=3. Correct. So Fib(100) mod4=3. Therefore, the remainder is 3.

But let me test with another value. Let me compute Fib(10) mod4. According to our previous list, Fib(10)=55, 55 mod4 is 55 - 13*4=55-52=3. Which matches Fib(4)=3. Which is consistent with the 10 mod6=4. So Fib(10) is same as Fib(4). So same result.

Similarly, Fib(16). 16 mod6=4. Fib(16)=987. 987 divided by 4: 4*246=984, remainder 3. Which is also 3. So same as Fib(4). Therefore, seems consistent.

Therefore, confident that answer is 3.

Wait, but hold on. Let me check Fib(7). 7 mod6=1. Fib(7)=13, 13 mod4=1. Which matches Fib(1)=1. Yes. So that works.

Alternatively, if I wanted to do this without knowing the Pisano period, maybe I can find the cycle by computing the terms mod4 until I see the repetition of 1,1 at the start. Let me try that approach again:

Start with Fib(1) mod4=1, Fib(2)=1.

Then:

Fib(3)=(1+1)=2 mod4=2

Fib(4)=(1+2)=3 mod4=3

Fib(5)=(2+3)=5 mod4=1

Fib(6)=(3+1)=4 mod4=0

Fib(7)=(1+0)=1 mod4=1

Fib(8)=(0+1)=1 mod4=1

So here, Fib(7)=1 and Fib(8)=1, which are the same as Fib(1) and Fib(2). Therefore, the cycle length is 6 because it took 6 steps to get back to 1,1. So the Pisano period is 6. Therefore, same conclusion.

So, there's no mistake here. Therefore, answer is 3. Therefore, when the 100th term is divided by 4, the remainder is 3.

But just to make sure that there isn't an off-by-one error. Let me check the indexing. The problem says the Fibonacci sequence is 1, 1, 2, 3, 5,... so term 1 is 1, term 2 is 1, term 3 is 2, etc. So the first term is Fib(1)=1. So when I calculated the Pisano period, I considered n starting at 1, so when n=1, it's 1, n=2, it's 1, etc. Therefore, the calculation is correct.

Alternatively, sometimes people index Fibonacci numbers starting at 0, but in this problem, they start at 1. For example, the problem says "the Fibonacci sequence is the sequence 1, 1, 2, 3, 5,..." so term 1 is 1, term 2 is 1. So the indexing here is 1-based. Therefore, the calculation above is correct.

Therefore, the remainder when the 100th term is divided by 4 is 3.

But just to make sure, let me try to compute Fib(100) mod4 using another method, like matrix exponentiation with modulus. Since that's a method to compute Fibonacci numbers efficiently.

The Fibonacci numbers can be computed using the matrix exponentiation method:

[ F(n+1)  F(n) ] = [ [1, 1], [1, 0] ]^n multiplied by [F(1), F(0)]^T

But since we are working mod4, we can perform the exponentiation with modulus.

But computing [ [1, 1], [1, 0] ]^99 (since n=100, need to raise to the 99th power?) Wait, maybe I need to check the exact formula.

Alternatively, using the formula that Fib(n) = (phi^n - psi^n)/sqrt(5), where phi=(1+sqrt(5))/2 and psi=(1-sqrt(5))/2. But since we are working mod4, and sqrt(5) is not an integer, this might not be helpful here.

Alternatively, since we have already established the Pisano period is 6, and 100 mod6=4, so Fib(100) mod4=Fib(4) mod4=3. So maybe it's safe.

Alternatively, let me compute Fib(n) mod4 up to n=10, as before:

n : Fib(n) : Fib(n) mod4

1 : 1 : 1

2 : 1 : 1

3 : 2 : 2

4 : 3 : 3

5 : 5 : 1

6 : 8 : 0

7 : 13 : 1

8 : 21 : 1

9 : 34 : 2

10: 55 : 3

So n=10, mod4=3 which is Fib(4)=3. Therefore, same result. If I compute Fib(10) mod4=3, which is same as Fib(4) mod4=3. So if I compute Fib(100) mod4, since 100 mod6=4, same as Fib(4)=3.

Therefore, answer is 3.

Alternatively, let me think if there is any chance that the Pisano period for 4 is different. For example, maybe 6 is not the correct Pisano period?

Wait, according to the OEIS or known results, the Pisano period for 4 is indeed 6. Let me check with another source in my mind. For example, Pisano periods for m=4:

The Pisano period for m=4 is 6. For example, the sequence of Fibonacci numbers modulo 4 is:

1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3,...

which repeats every 6 terms. Hence, Pisano period is 6. Therefore, correct.

Therefore, conclusion is 3. Therefore, answer is 3.

But just to be thorough, let me try to compute Fib(100) mod4 by using the fact that Fib(n) mod4 is periodic with period 6, and that 100 mod6=4, so Fib(100) mod4=Fib(4) mod4=3. Therefore, the remainder is 3.

Therefore, I think the answer is 3.

**Final Answer**
The remainder when the $100^{\text{th}}$ term of the Fibonacci sequence is divided by 4 is \boxed{3}.

To solve the problem, we are tasked with finding the **remainder when the 100th term of the Fibonacci sequence is divided by 4**, i.e., compute $ F_{100} \mod 4 $, where $ F_n $ denotes the $ n $-th term of the Fibonacci sequence.

---

### Step 1: Understanding the Fibonacci Sequence Modulo 4

The Fibonacci sequence is defined recursively:

$$
F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n > 2
$$

However, computing the 100th term directly is impractical due to its size. A more effective method involves **modular computation** and **Pisano periods**.

---

### Step 2: Pisano Period for Modulo 4

The **Pisano period** $ \pi(m) $ is the period after which the Fibonacci sequence **mod $ m $** repeats. That is, the cycle of remainders when the Fibonacci numbers are divided by $ m $ repeats every $ \pi(m) $ terms.

For $ m = 4 $, the Pisano period is known to be:

$$
\pi(4) = 6
$$

We can verify this by computing the Fibonacci sequence modulo 4:

$$
\begin{align*}
F_1 &\equiv 1 \mod 4 \\
F_2 &\equiv 1 \mod 4 \\
F_3 &\equiv 2 \mod 4 \\
F_4 &\equiv 3 \mod 4 \\
F_5 &\equiv 1 \mod 4 \\
F_6 &\equiv 0 \mod 4 \\
F_7 &\equiv 1 \mod 4 \\
F_8 &\equiv 1 \mod 4 \\
\end{align*}
$$

From this, the cycle $ [1, 1, 2, 3, 1, 0] $ repeats every 6 terms. So:

$$
F_n \mod 4 = F_{n \mod 6} \mod 4
$$

---

### Step 3: Finding the Equivalent Term

We are interested in $ F_{100} \mod 4 $. Using the Pisano period:

$$
100 \mod 6 = 4
$$

Thus:

$$
F_{100} \mod 4 = F_4 \mod 4 = 3
$$

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when the 100th term of the Fibonacci sequence is divided by 4. Hmm, Fibonacci sequence... right, each term is the sum of the two previous ones. Let me recall: the first two terms are 1 and 1, then 2, 3, 5, 8, etc. So, the problem is asking for Fib(100) mod 4.

First thought: calculating the 100th term directly would be a huge number, and that seems impractical. There must be a pattern or a cycle when Fibonacci numbers are divided by 4. I remember something about Pisano periods, which are the periods with which the Fibonacci sequence repeats modulo some number. Maybe I can find the Pisano period for modulo 4 and then use that to find Fib(100) mod 4?

Let me check. The Pisano period for modulo m is the period after which the Fibonacci sequence repeats when taken modulo m. So, if I can find the Pisano period for 4, then I can compute 100 modulo that period to find the equivalent smaller term, which would be easier to compute.

Alright, let's try to find the Pisano period for modulo 4. Let me list out the Fibonacci numbers modulo 4 and see when the sequence starts repeating.

Starting with Fib(1) = 1, Fib(2) = 1:

Term 1: 1 mod 4 = 1

Term 2: 1 mod 4 = 1

Term 3: (1 + 1) = 2 mod 4 = 2

Term 4: (1 + 2) = 3 mod 4 = 3

Term 5: (2 + 3) = 5 mod 4 = 1

Term 6: (3 + 1) = 4 mod 4 = 0

Term 7: (1 + 0) = 1 mod 4 = 1

Term 8: (0 + 1) = 1 mod 4 = 1

Wait a second, term 7 and term 8 are 1 and 1, which are the same as term 1 and term 2. So does that mean the Pisano period is 6? Because after term 6, the sequence starts repeating?

Let me check the next terms to confirm:

Term 9: (1 + 1) = 2 mod 4 = 2 (same as term 3)

Term 10: (1 + 2) = 3 mod 4 = 3 (same as term 4)

Term 11: (2 + 3) = 5 mod 4 = 1 (same as term 5)

Term 12: (3 + 1) = 4 mod 4 = 0 (same as term 6)

Term 13: (1 + 0) = 1 (same as term 7)

So yes, the sequence repeats every 6 terms. So the Pisano period for modulo 4 is 6. Therefore, Fib(n) mod 4 repeats every 6 terms. Therefore, to find Fib(100) mod 4, I can compute 100 mod 6, find the equivalent term in the first period, and then take that term's value mod 4.

Let me compute 100 divided by 6. 6*16 = 96, so 100 - 96 = 4. Therefore, 100 mod 6 is 4. Therefore, Fib(100) mod 4 is equal to Fib(4) mod 4. Looking back at the list:

Term 4: 3 mod 4 = 3. Therefore, the remainder is 3?

Wait, but let me confirm with another approach. Maybe I made a mistake in the Pisano period?

Alternatively, perhaps I should check the Pisano period again. Let me list the terms again:

n: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16...

Fib(n): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987...

Now mod 4:

1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3...

Yes, so starting from term 1, the sequence mod 4 is: 1, 1, 2, 3, 1, 0, then repeats 1, 1, 2, 3, 1, 0... So the Pisano period here is indeed 6. So the cycle is 6 terms long. Therefore, the remainder when dividing n by 6 gives the position in the cycle.

So, n=100. Let me compute 100 divided by 6. 6*16=96, remainder 4. So it's equivalent to term 4 in the cycle. Term 4 mod 4 is 3. Therefore, Fib(100) mod 4 is 3.

But wait, let me check with another method to be sure. Maybe compute Fib(100) mod 4 using matrix exponentiation or recursion with mod?

Alternatively, perhaps compute some terms manually and see if the pattern holds.

Let me compute Fib(n) mod 4 for n up to, say, 12 as before:

n : Fib(n) : Fib(n) mod 4

1 : 1 : 1

2 : 1 : 1

3 : 2 : 2

4 : 3 : 3

5 : 5 : 1

6 : 8 : 0

7 : 13 : 1

8 : 21 : 1

9 : 34 : 2

10: 55 : 3

11: 89 : 1

12: 144 : 0

So, the cycle is indeed 6: 1, 1, 2, 3, 1, 0 repeating. Therefore, for n=100, since 100 mod 6 = 4, Fib(100) mod 4 = Fib(4) mod 4 = 3.

But let me check with n=6, which is 0, and n=12 is also 0, which is 6*2. So for n=6k, Fib(6k) mod 4 is 0. Similarly, n=6k+1: Fib(6k+1) mod 4 = 1. n=6k+2: 1, n=6k+3: 2, n=6k+4: 3, n=6k+5: 1.

Therefore, 100 divided by 6: 6*16=96, remainder 4, so 100=6*16+4, so Fib(100) mod 4 is same as Fib(4) mod 4 = 3. Therefore, answer is 3.

But just to be thorough, let me check another term. For example, n=7: 6*1 +1=7, Fib(7)=13 mod4=1, which matches. n=8=6*1+2: 21 mod4=1, which matches. n=9=6*1+3: 34 mod4=2, yes. n=10=6*1+4: 55 mod4=3, yes. n=11=6*1+5: 89 mod4=1, yes. n=12=6*2+0: 144 mod4=0. So the pattern holds. Therefore, confident that the pattern is correct.

Alternatively, maybe I can compute Fib(100) mod 4 using the property of Pisano period. Since Pisano period for 4 is 6, then Fib(n) mod4 = Fib(n mod 6) mod4? Wait, not exactly. Because if the period is 6, then Fib(n) mod4 = Fib(n mod 6) mod4? Wait, no. Wait, for example, Fib(7) mod4 = Fib(1) mod4 = 1, since 7 mod6=1. Similarly Fib(8)=Fib(2)=1. Fib(9)=Fib(3)=2. Fib(10)=Fib(4)=3. Fib(11)=Fib(5)=1. Fib(12)=Fib(6)=0. Fib(13)=Fib(7)=Fib(1)=1. So yes, it seems that Fib(n) mod4 = Fib(n mod 6) mod4, except when n mod6 is 0, in which case it's Fib(6) mod4=0. So if n mod6=0, then it's Fib(6)=0. So for n=6, 12, etc.

Therefore, in general, for n, compute n mod6. If the remainder is 0, then Fib(n) mod4=0. Else, Fib(n) mod4=Fib(remainder) mod4.

So, in our case, n=100. 100 divided by 6 is 16 with remainder 4. Therefore, Fib(100) mod4=Fib(4) mod4=3. Therefore, the remainder is 3.

Alternatively, perhaps I can compute Fib(100) mod4 by using matrix exponentiation or some recursive formula with modulus?

But given that the Pisano period is 6, it's much easier to use that. But just to check, let me try to compute Fib(100) mod4 using another approach.

Alternatively, maybe use the formula for Fibonacci numbers? But that involves irrational numbers and might not be helpful here. Alternatively, perhaps use the fact that the Fibonacci sequence modulo m is periodic, so we can compute Fib(100) mod4 by finding the cycle.

But since we already know the Pisano period is 6, we can just use that. However, just to make sure that my calculation of the Pisano period is correct, let me check again. The Pisano period for 4 is known?

Wait, actually, I think the Pisano period for modulo m can be calculated, but I need to make sure that for modulo 4, the Pisano period is indeed 6. Let me check a reference in my mind... The Pisano period for m=2 is 3, for m=3 it's 8, for m=4 it's 6. Yes, I think that's right. For example, Pisano periods for small m:

m | Pisano period
1 | 1
2 | 3
3 | 8
4 | 6
5 | 20
...

So for 4, 6 is correct. So that's good. So the Pisano period is 6, so the cycle is 6 terms. Therefore, Fib(n) mod4 repeats every 6 terms. Therefore, Fib(100) mod4 is Fib(100 mod6) mod4. Since 100 mod6 is 4, so Fib(4)=3 mod4=3. So the answer is 3.

But let me check with another example. For instance, Fib(6)=8, 8 mod4=0. Let me check Fib(6+6)=Fib(12)=144. 144 mod4=0. Same as 8 mod4. Similarly Fib(12+6)=Fib(18)=? Let me compute Fib(13)=233, Fib(14)=377, Fib(15)=610, Fib(16)=987, Fib(17)=1597, Fib(18)=2584. 2584 divided by 4: 2584/4=646, so remainder 0. So yes, it's 0 again. So the pattern holds. So that seems valid.

Alternatively, maybe I can use the concept of modular Fibonacci numbers. Let me try to compute Fib(n) mod4 up to n=100, but that would take too long. However, since the period is 6, and 100 is a multiple of 6 plus 4, we can be confident that it's 3.

Alternatively, maybe compute Fib(100) mod4 by using the recursive formula with mod4 at each step. But since we know the Pisano period is 6, we can just figure out which term in the cycle it is. However, if I wanted to do it manually, perhaps I can compute the terms up to the 100th term mod4, but that would take time.

Alternatively, since the period is 6, we can think of the sequence as repeating every 6 terms:

Positions:

1: 1

2: 1

3: 2

4: 3

5: 1

6: 0

Then repeats:

7: 1 (same as 1)

8: 1 (same as 2)

9: 2 (same as 3)

10: 3 (same as 4)

11: 1 (same as 5)

12: 0 (same as 6)

and so on.

Therefore, to find Fib(100) mod4, since 100 divided by 6 is 16 with remainder 4, so it's the same as the 4th term in the cycle, which is 3. So answer is 3.

But let me think if there is any possibility that the Pisano period is longer. For example, sometimes the Pisano period for m is 2*something? For example, for m=2, Pisano period is 3, which is not 2. For m=4, Pisano period is 6, which is 2*3. So maybe it's different. However, in our case, we have confirmed that the cycle is 6. So unless there is an error in my calculation, the answer should be 3.

Wait, let me check Fib(4). Fib(4) is 3, 3 mod4=3. Correct. So Fib(100) mod4=3. Therefore, the remainder is 3.

But let me test with another value. Let me compute Fib(10) mod4. According to our previous list, Fib(10)=55, 55 mod4 is 55 - 13*4=55-52=3. Which matches Fib(4)=3. Which is consistent with the 10 mod6=4. So Fib(10) is same as Fib(4). So same result.

Similarly, Fib(16). 16 mod6=4. Fib(16)=987. 987 divided by 4: 4*246=984, remainder 3. Which is also 3. So same as Fib(4). Therefore, seems consistent.

Therefore, confident that answer is 3.

Wait, but hold on. Let me check Fib(7). 7 mod6=1. Fib(7)=13, 13 mod4=1. Which matches Fib(1)=1. Yes. So that works.

Alternatively, if I wanted to do this without knowing the Pisano period, maybe I can find the cycle by computing the terms mod4 until I see the repetition of 1,1 at the start. Let me try that approach again:

Start with Fib(1) mod4=1, Fib(2)=1.

Then:

Fib(3)=(1+1)=2 mod4=2

Fib(4)=(1+2)=3 mod4=3

Fib(5)=(2+3)=5 mod4=1

Fib(6)=(3+1)=4 mod4=0

Fib(7)=(1+0)=1 mod4=1

Fib(8)=(0+1)=1 mod4=1

So here, Fib(7)=1 and Fib(8)=1, which are the same as Fib(1) and Fib(2). Therefore, the cycle length is 6 because it took 6 steps to get back to 1,1. So the Pisano period is 6. Therefore, same conclusion.

So, there's no mistake here. Therefore, answer is 3. Therefore, when the 100th term is divided by 4, the remainder is 3.

But just to make sure that there isn't an off-by-one error. Let me check the indexing. The problem says the Fibonacci sequence is 1, 1, 2, 3, 5,... so term 1 is 1, term 2 is 1, term 3 is 2, etc. So the first term is Fib(1)=1. So when I calculated the Pisano period, I considered n starting at 1, so when n=1, it's 1, n=2, it's 1, etc. Therefore, the calculation is correct.

Alternatively, sometimes people index Fibonacci numbers starting at 0, but in this problem, they start at 1. For example, the problem says "the Fibonacci sequence is the sequence 1, 1, 2, 3, 5,..." so term 1 is 1, term 2 is 1. So the indexing here is 1-based. Therefore, the calculation above is correct.

Therefore, the remainder when the 100th term is divided by 4 is 3.

But just to make sure, let me try to compute Fib(100) mod4 using another method, like matrix exponentiation with modulus. Since that's a method to compute Fibonacci numbers efficiently.

The Fibonacci numbers can be computed using the matrix exponentiation method:

[ F(n+1)  F(n) ] = [ [1, 1], [1, 0] ]^n multiplied by [F(1), F(0)]^T

But since we are working mod4, we can perform the exponentiation with modulus.

But computing [ [1, 1], [1, 0] ]^99 (since n=100, need to raise to the 99th power?) Wait, maybe I need to check the exact formula.

Alternatively, using the formula that Fib(n) = (phi^n - psi^n)/sqrt(5), where phi=(1+sqrt(5))/2 and psi=(1-sqrt(5))/2. But since we are working mod4, and sqrt(5) is not an integer, this might not be helpful here.

Alternatively, since we have already established the Pisano period is 6, and 100 mod6=4, so Fib(100) mod4=Fib(4) mod4=3. So maybe it's safe.

Alternatively, let me compute Fib(n) mod4 up to n=10, as before:

n : Fib(n) : Fib(n) mod4

1 : 1 : 1

2 : 1 : 1

3 : 2 : 2

4 : 3 : 3

5 : 5 : 1

6 : 8 : 0

7 : 13 : 1

8 : 21 : 1

9 : 34 : 2

10: 55 : 3

So n=10, mod4=3 which is Fib(4)=3. Therefore, same result. If I compute Fib(10) mod4=3, which is same as Fib(4) mod4=3. So if I compute Fib(100) mod4, since 100 mod6=4, same as Fib(4)=3.

Therefore, answer is 3.

Alternatively, let me think if there is any chance that the Pisano period for 4 is different. For example, maybe 6 is not the correct Pisano period?

Wait, according to the OEIS or known results, the Pisano period for 4 is indeed 6. Let me check with another source in my mind. For example, Pisano periods for m=4:

The Pisano period for m=4 is 6. For example, the sequence of Fibonacci numbers modulo 4 is:

1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3,...

which repeats every 6 terms. Hence, Pisano period is 6. Therefore, correct.

Therefore, conclusion is 3. Therefore, answer is 3.

But just to be thorough, let me try to compute Fib(100) mod4 by using the fact that Fib(n) mod4 is periodic with period 6, and that 100 mod6=4, so Fib(100) mod4=Fib(4) mod4=3. Therefore, the remainder is 3.

Therefore, I think the answer is 3.

**Final Answer**
The remainder when the $100^{\text{th}}$ term of the Fibonacci sequence is divided by 4 is \boxed{3}.

To solve the problem, we are tasked with finding the **remainder when the 100th term of the Fibonacci sequence is divided by 4**, i.e., compute $ F_{100} \mod 4 $, where $ F_n $ denotes the $ n $-th term of the Fibonacci sequence.

---

### Step 1: Understanding the Fibonacci Sequence Modulo 4

The Fibonacci sequence is defined recursively:

$$
F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n > 2
$$

However, computing the 100th term directly is impractical due to its size. A more effective method involves **modular computation** and **Pisano periods**.

---

### Step 2: Pisano Period for Modulo 4

The **Pisano period** $ \pi(m) $ is the period after which the Fibonacci sequence **mod $ m $** repeats. That is, the cycle of remainders when the Fibonacci numbers are divided by $ m $ repeats every $ \pi(m) $ terms.

For $ m = 4 $, the Pisano period is known to be:

$$
\pi(4) = 6
$$

We can verify this by computing the Fibonacci sequence modulo 4:

$$
\begin{align*}
F_1 &\equiv 1 \mod 4 \\
F_2 &\equiv 1 \mod 4 \\
F_3 &\equiv 2 \mod 4 \\
F_4 &\equiv 3 \mod 4 \\
F_5 &\equiv 1 \mod 4 \\
F_6 &\equiv 0 \mod 4 \\
F_7 &\equiv 1 \mod 4 \\
F_8 &\equiv 1 \mod 4 \\
\end{align*}
$$

From this, the cycle $ [1, 1, 2, 3, 1, 0] $ repeats every 6 terms. So:

$$
F_n \mod 4 = F_{n \mod 6} \mod 4
$$

---

### Step 3: Finding the Equivalent Term

We are interested in $ F_{100} \mod 4 $. Using the Pisano period:

$$
100 \mod 6 = 4
$$

Thus:

$$
F_{100} \mod 4 = F_4 \mod 4 = 3
$$

---

### Final Answer

$$
\boxed{3}
$$

Solution:
If we look at the terms of the sequence mod 4, we see that they follow a pattern of period 6: \begin{align*}
F_1 &\equiv 1\pmod{4}, \\
F_2 &\equiv 1\pmod{4}, \\
F_3 &\equiv 2\pmod{4}, \\
F_4 &\equiv 3\pmod{4}, \\
F_5 &\equiv 1\pmod{4}, \\
F_6 &\equiv 0\pmod{4}, \\
F_7 &\equiv 1\pmod{4}, \\
F_8 &\equiv 1\pmod{4},~\ldots
\end{align*} Then we see that the terms repeat.  Therefore, the $100^{\text{th}}$ term is the same as the $4^{\text{th}}$ term, and thus has a remainder of $\boxed{3}$ when divided by 4.


Explanation:
Okay, so I need to find the smallest number that is one less than twice its reciprocal. Hmm, let me try to understand the problem first. The number we're looking for, let's call it x, has this property: x is equal to one less than twice its reciprocal. Wait, reciprocal of x is 1/x, right? So twice the reciprocal would be 2*(1/x). Then one less than that would be 2*(1/x) - 1. So the equation should be x = 2*(1/x) - 1? Let me check that again.

The problem says "one less than twice its reciprocal." So twice its reciprocal is 2*(1/x), and then subtracting one gives 2/x - 1. And this is equal to the original number x. So the equation is x = 2/x - 1. Is that right? Let me make sure I didn't mix up the order. "One less than twice its reciprocal" – that would mean take twice the reciprocal and subtract one, so yes, x = (2*(1/x)) - 1. Okay, that seems correct.

Now, I need to solve this equation for x. Let me write it down:

x = (2/x) - 1.

To solve this equation, I can multiply both sides by x to eliminate the denominator. Let's do that. Multiply every term by x:

x * x = (2/x) * x - 1 * x

Simplifying each term:

x² = 2 - x

Now, let's bring all terms to one side to form a quadratic equation:

x² + x - 2 = 0

Wait, let me check that. If I have x² = 2 - x, then subtract 2 and add x to both sides:

x² + x - 2 = 0. Yes, that seems right.

Now, we need to solve the quadratic equation x² + x - 2 = 0. Let me try factoring first. Looking for two numbers that multiply to -2 and add to 1. Hmm. Let's see... 2 and -1? Because 2 * (-1) = -2 and 2 + (-1) = 1. Yes! So the equation factors as:

(x + 2)(x - 1) = 0

Therefore, the solutions are x = -2 and x = 1.

But the question asks for the smallest number. So between -2 and 1, the smaller one is -2. But wait, let me check if both solutions are valid in the original equation because sometimes when you multiply both sides by x, you might introduce extraneous solutions. Let me verify both.

First, test x = 1. The original equation was x = 2/x - 1. Plug in x=1: left side is 1. Right side is 2/1 -1 = 2 -1 = 1. Okay, so 1 = 1, which works.

Now test x = -2. Left side is -2. Right side is 2/(-2) -1 = -1 -1 = -2. So -2 = -2, which also works. So both solutions are valid.

But the question says "the smallest number". Now, between -2 and 1, the smaller number is -2. So is -2 the answer? Wait, but maybe there's a trick here? Let me make sure I didn't miss anything.

Wait, the problem says "the smallest number which is one less than twice its reciprocal". So, if we consider real numbers, then -2 is smaller than 1, so -2 is the answer. However, sometimes people might think about positive numbers only, but the problem doesn't specify that. So unless there's a restriction to positive numbers, -2 is valid.

But let me check again the original problem statement. It just says "the smallest number", so unless there's context implying positive numbers, I think both solutions are valid. Therefore, -2 is smaller, so that's the answer.

Wait, but maybe there's a different interpretation of the problem? Let me read again: "What is the smallest number which is one less than twice its reciprocal?"

So, "one less than twice its reciprocal" – another way to interpret this could be that the number is equal to twice its reciprocal minus one, which is how I initially set it up. So, x = 2*(1/x) - 1. Which leads to the quadratic equation, and solutions -2 and 1. So, yeah, I think that's correct.

But maybe the problem is expecting a positive number? Let me think. If the problem is from a context where numbers are assumed to be positive (like in some math problems), maybe they want 1? But the question says "smallest number", not "smallest positive number". So unless there's a restriction, -2 is smaller. However, maybe there's an error in my calculation?

Let me check my quadratic equation again. Starting from x = 2/x -1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Factoring: (x + 2)(x - 1) = 0. Yep, that's correct. So solutions are -2 and 1. Therefore, the smallest is -2.

Alternatively, maybe the problem is phrased differently? Let me think. If someone says "one less than twice its reciprocal", could this be interpreted as twice (one less than its reciprocal)? But that would be different. Let me check.

If it were twice (one less than its reciprocal), that would be 2*(1/x - 1). But the original problem says "one less than twice its reciprocal", which is different. The wording is important. "One less than twice its reciprocal" is (twice its reciprocal) minus one, which is 2*(1/x) - 1. Whereas "twice (one less than its reciprocal)" would be 2*(1/x -1). So the original equation is correct as x = 2/x -1.

Therefore, my answer should be -2. But to make sure, let me think again. Suppose the number is -2. Let's compute twice its reciprocal. Reciprocal of -2 is -1/2. Twice that is -1. Then one less than that would be -1 -1 = -2. Which is equal to the original number. So that works.

Similarly, for 1: reciprocal is 1, twice that is 2, one less is 1. Which matches. So both are valid. Hence, -2 is indeed smaller. Therefore, the answer is -2.

But maybe the problem is expecting a positive number? Maybe there's a mistake in the problem's phrasing? Or maybe I made a wrong equation?

Wait, let me try another approach. Suppose the number is x. Then, according to the problem, x = 2*(1/x) - 1. So that equation. Let me rearrange:

x + 1 = 2*(1/x)

Multiply both sides by x:

x(x + 1) = 2

Which is x² + x - 2 = 0, same equation. So same solutions. So no mistake here.

Alternatively, if the problem had said "one less than twice its reciprocal is the number", same thing.

Alternatively, if someone misreads it as "the number is one less than twice its reciprocal", which is same as above.

So, yeah, the answer is -2. Therefore, the smallest number is -2.

But let me check if there are any other solutions or if I missed something. For example, maybe complex numbers? But the problem probably refers to real numbers.

Therefore, my conclusion is that the smallest number is -2.

**Final Answer**
The smallest number is \boxed{-2}.

To solve the problem, we are asked to find the **smallest number** that satisfies the condition:

> "It is one less than twice its reciprocal."

---

### Step 1: Translate the words into an algebraic equation

Let the number be $ x $.
The **reciprocal** of $ x $ is $ \frac{1}{x} $.
"Twice its reciprocal" is $ 2 \cdot \frac{1}{x} = \frac{2}{x} $.
"One less than that" is then $ \frac{2}{x} - 1 $.

So, the number $ x $ is equal to $ \frac{2}{x} - 1 $, giving the equation:

$$
x = \frac{2}{x} - 1
$$

---

### Step 2: Eliminate the denominator

To simplify, we multiply both sides of the equation by $ x $ (assuming $ x \neq 0 $):

$$
x^2 = 2 - x
$$

Rewriting in standard quadratic form:

$$
x^2 + x - 2 = 0
$$

---

### Step 3: Solve the quadratic equation

We solve $ x^2 + x - 2 = 0 $ by factoring.
We look for two numbers that multiply to $ -2 $ and add to $ 1 $. These are $ 2 $ and $ -1 $.

$$
(x + 2)(x - 1) = 0
$$

So the solutions are:

$$
x = -2 \quad \text{or} \quad x = 1
$$

---

### Step 4: Verify both solutions

We verify whether both solutions satisfy the original condition.

- **For $ x = 1 $:**

  Reciprocal is $ \frac{1}{1} = 1 $
  Twice that is $ 2 \cdot 1 = 2 $
  One less is $ 2 - 1 = 1 $
  So $ x = 1 $ satisfies the condition.

- **For $ x = -2 $:**

  Reciprocal is $ \frac{1}{-2} = -\frac{1}{2} $
  Twice that is $ 2 \cdot (-\frac{1}{2}) = -1 $
  One less is $ -1 - 1 = -2 $
  So $ x = -2 $ also satisfies the condition.

---

### Step 5: Determine the smallest number

From the solutions $ x = 1 $ and $ x = -2 $, the **smallest number** is clearly:

$$
\boxed{-2}
$$

---

### Final Answer

$$
\boxed{-2}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest number that is one less than twice its reciprocal. Hmm, let me try to understand the problem first. The number we're looking for, let's call it x, has this property: x is equal to one less than twice its reciprocal. Wait, reciprocal of x is 1/x, right? So twice the reciprocal would be 2*(1/x). Then one less than that would be 2*(1/x) - 1. So the equation should be x = 2*(1/x) - 1? Let me check that again.

The problem says "one less than twice its reciprocal." So twice its reciprocal is 2*(1/x), and then subtracting one gives 2/x - 1. And this is equal to the original number x. So the equation is x = 2/x - 1. Is that right? Let me make sure I didn't mix up the order. "One less than twice its reciprocal" – that would mean take twice the reciprocal and subtract one, so yes, x = (2*(1/x)) - 1. Okay, that seems correct.

Now, I need to solve this equation for x. Let me write it down:

x = (2/x) - 1.

To solve this equation, I can multiply both sides by x to eliminate the denominator. Let's do that. Multiply every term by x:

x * x = (2/x) * x - 1 * x

Simplifying each term:

x² = 2 - x

Now, let's bring all terms to one side to form a quadratic equation:

x² + x - 2 = 0

Wait, let me check that. If I have x² = 2 - x, then subtract 2 and add x to both sides:

x² + x - 2 = 0. Yes, that seems right.

Now, we need to solve the quadratic equation x² + x - 2 = 0. Let me try factoring first. Looking for two numbers that multiply to -2 and add to 1. Hmm. Let's see... 2 and -1? Because 2 * (-1) = -2 and 2 + (-1) = 1. Yes! So the equation factors as:

(x + 2)(x - 1) = 0

Therefore, the solutions are x = -2 and x = 1.

But the question asks for the smallest number. So between -2 and 1, the smaller one is -2. But wait, let me check if both solutions are valid in the original equation because sometimes when you multiply both sides by x, you might introduce extraneous solutions. Let me verify both.

First, test x = 1. The original equation was x = 2/x - 1. Plug in x=1: left side is 1. Right side is 2/1 -1 = 2 -1 = 1. Okay, so 1 = 1, which works.

Now test x = -2. Left side is -2. Right side is 2/(-2) -1 = -1 -1 = -2. So -2 = -2, which also works. So both solutions are valid.

But the question says "the smallest number". Now, between -2 and 1, the smaller number is -2. So is -2 the answer? Wait, but maybe there's a trick here? Let me make sure I didn't miss anything.

Wait, the problem says "the smallest number which is one less than twice its reciprocal". So, if we consider real numbers, then -2 is smaller than 1, so -2 is the answer. However, sometimes people might think about positive numbers only, but the problem doesn't specify that. So unless there's a restriction to positive numbers, -2 is valid.

But let me check again the original problem statement. It just says "the smallest number", so unless there's context implying positive numbers, I think both solutions are valid. Therefore, -2 is smaller, so that's the answer.

Wait, but maybe there's a different interpretation of the problem? Let me read again: "What is the smallest number which is one less than twice its reciprocal?"

So, "one less than twice its reciprocal" – another way to interpret this could be that the number is equal to twice its reciprocal minus one, which is how I initially set it up. So, x = 2*(1/x) - 1. Which leads to the quadratic equation, and solutions -2 and 1. So, yeah, I think that's correct.

But maybe the problem is expecting a positive number? Let me think. If the problem is from a context where numbers are assumed to be positive (like in some math problems), maybe they want 1? But the question says "smallest number", not "smallest positive number". So unless there's a restriction, -2 is smaller. However, maybe there's an error in my calculation?

Let me check my quadratic equation again. Starting from x = 2/x -1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Factoring: (x + 2)(x - 1) = 0. Yep, that's correct. So solutions are -2 and 1. Therefore, the smallest is -2.

Alternatively, maybe the problem is phrased differently? Let me think. If someone says "one less than twice its reciprocal", could this be interpreted as twice (one less than its reciprocal)? But that would be different. Let me check.

If it were twice (one less than its reciprocal), that would be 2*(1/x - 1). But the original problem says "one less than twice its reciprocal", which is different. The wording is important. "One less than twice its reciprocal" is (twice its reciprocal) minus one, which is 2*(1/x) - 1. Whereas "twice (one less than its reciprocal)" would be 2*(1/x -1). So the original equation is correct as x = 2/x -1.

Therefore, my answer should be -2. But to make sure, let me think again. Suppose the number is -2. Let's compute twice its reciprocal. Reciprocal of -2 is -1/2. Twice that is -1. Then one less than that would be -1 -1 = -2. Which is equal to the original number. So that works.

Similarly, for 1: reciprocal is 1, twice that is 2, one less is 1. Which matches. So both are valid. Hence, -2 is indeed smaller. Therefore, the answer is -2.

But maybe the problem is expecting a positive number? Maybe there's a mistake in the problem's phrasing? Or maybe I made a wrong equation?

Wait, let me try another approach. Suppose the number is x. Then, according to the problem, x = 2*(1/x) - 1. So that equation. Let me rearrange:

x + 1 = 2*(1/x)

Multiply both sides by x:

x(x + 1) = 2

Which is x² + x - 2 = 0, same equation. So same solutions. So no mistake here.

Alternatively, if the problem had said "one less than twice its reciprocal is the number", same thing.

Alternatively, if someone misreads it as "the number is one less than twice its reciprocal", which is same as above.

So, yeah, the answer is -2. Therefore, the smallest number is -2.

But let me check if there are any other solutions or if I missed something. For example, maybe complex numbers? But the problem probably refers to real numbers.

Therefore, my conclusion is that the smallest number is -2.

**Final Answer**
The smallest number is \boxed{-2}.

To solve the problem, we are asked to find the **smallest number** that satisfies the condition:

> "It is one less than twice its reciprocal."

---

### Step 1: Translate the words into an algebraic equation

Let the number be $ x $.
The **reciprocal** of $ x $ is $ \frac{1}{x} $.
"Twice its reciprocal" is $ 2 \cdot \frac{1}{x} = \frac{2}{x} $.
"One less than that" is then $ \frac{2}{x} - 1 $.

So, the number $ x $ is equal to $ \frac{2}{x} - 1 $, giving the equation:

$$
x = \frac{2}{x} - 1
$$

---

### Step 2: Eliminate the denominator

To simplify, we multiply both sides of the equation by $ x $ (assuming $ x \neq 0 $):

$$
x^2 = 2 - x
$$

Rewriting in standard quadratic form:

$$
x^2 + x - 2 = 0
$$

---

### Step 3: Solve the quadratic equation

We solve $ x^2 + x - 2 = 0 $ by factoring.
We look for two numbers that multiply to $ -2 $ and add to $ 1 $. These are $ 2 $ and $ -1 $.

$$
(x + 2)(x - 1) = 0
$$

So the solutions are:

$$
x = -2 \quad \text{or} \quad x = 1
$$

---

### Step 4: Verify both solutions

We verify whether both solutions satisfy the original condition.

- **For $ x = 1 $:**

  Reciprocal is $ \frac{1}{1} = 1 $
  Twice that is $ 2 \cdot 1 = 2 $
  One less is $ 2 - 1 = 1 $
  So $ x = 1 $ satisfies the condition.

- **For $ x = -2 $:**

  Reciprocal is $ \frac{1}{-2} = -\frac{1}{2} $
  Twice that is $ 2 \cdot (-\frac{1}{2}) = -1 $
  One less is $ -1 - 1 = -2 $
  So $ x = -2 $ also satisfies the condition.

---

### Step 5: Determine the smallest number

From the solutions $ x = 1 $ and $ x = -2 $, the **smallest number** is clearly:

$$
\boxed{-2}
$$

---

### Final Answer

$$
\boxed{-2}
$$

Solution: The question is asking us to find the smallest value of $x$ such that $x = 2\cdot\frac{1}{x} - 1$. We multiply through by $x$ to clear the fraction, then rearrange terms: $x^2 + x - 2 = 0$. This can be factored as $(x + 2)(x - 1) = 0$. We could also use the quadratic formula to find $x$: $$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2}.$$ Either way, we find that $x = 1$ or $x = -2$. Since we want the smallest value of $x$, our answer is $\boxed{-2}$.