Explanation:
Okay, so I need to find the remainder when the 100th term of the Fibonacci sequence is divided by 4. Hmm, let me think. The Fibonacci sequence starts with 1, 1, 2, 3, 5, and so on, right? Each term is the sum of the two previous ones. But calculating the 100th term directly sounds like a huge number. I remember something about modular arithmetic and maybe patterns repeating? Maybe there's a cycle when you take Fibonacci numbers modulo 4? Let me try to recall.

First, let me write down the Fibonacci sequence terms and see what they are modulo 4. Maybe there's a repeating pattern. Let me start listing them:

Term 1: 1
Term 2: 1
Term 3: 2
Term 4: 3
Term 5: 5
Term 6: 8
Term 7: 13
Term 8: 21
Term 9: 34
Term 10: 55
Term 11: 89
Term 12: 144
... and so on.

But instead of the actual numbers, let me compute each term modulo 4:

Term 1: 1 mod 4 = 1
Term 2: 1 mod 4 = 1
Term 3: 2 mod 4 = 2
Term 4: 3 mod 4 = 3
Term 5: 5 mod 4 = 1
Term 6: 8 mod 4 = 0
Term 7: 13 mod 4 = 1 (since 13 divided by 4 is 3 with remainder 1)
Term 8: 21 mod 4 = 1 (21 divided by 4 is 5*4=20, remainder 1)
Term 9: 34 mod 4 = 2 (34 divided by 4 is 8*4=32, remainder 2)
Term 10: 55 mod 4 = 3 (55 divided by 4 is 13*4=52, remainder 3)
Term 11: 89 mod 4 = 1 (89 divided by 4 is 22*4=88, remainder 1)
Term 12: 144 mod 4 = 0 (since 144 is 36*4, so remainder 0)
Term 13: 233 mod 4. Let me compute that. 233 divided by 4: 58*4=232, so remainder 1.
Term 14: 377 mod 4. 377 divided by 4: 94*4=376, remainder 1.
Term 15: 610 mod 4. 610 divided by 4 is 152*4=608, remainder 2.
Term 16: 987 mod 4. 987 divided by 4: 246*4=984, remainder 3.
Term 17: 1597 mod 4. 1597 divided by 4: 399*4=1596, remainder 1.
Term 18: 2584 mod 4. 2584 divided by 4 is 646*4, so remainder 0.
Term 19: 4181 mod 4. Let me see, 4180 is divisible by 4, so 4181 mod 4 is 1.
Term 20: 6765 mod 4. 6764 is divisible by 4, so remainder 1.

Wait, so let me list these mod 4 results:

Term 1: 1

Term 2: 1

Term 3: 2

Term 4: 3

Term 5: 1

Term 6: 0

Term 7: 1

Term 8: 1

Term 9: 2

Term 10: 3

Term 11: 1

Term 12: 0

Term 13: 1

Term 14: 1

Term 15: 2

Term 16: 3

Term 17: 1

Term 18: 0

Term 19: 1

Term 20: 1

Hmm, so looking at this sequence of remainders: 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1,...

It seems like there's a repeating pattern here. Let me check. From term 1 to term 6: 1, 1, 2, 3, 1, 0

Then term 7 to term 12: 1, 1, 2, 3, 1, 0

Then term 13 to term 18: 1, 1, 2, 3, 1, 0

Term 19 to term 24 would be 1, 1, 2, 3, 1, 0 again?

So the cycle seems to be of length 6? Let me check:

Starting from term 1: 1, 1, 2, 3, 1, 0, then repeats 1, 1, 2, 3, 1, 0,...

Yes, so the Pisano period for modulo 4 is 6? Wait, Pisano period is the period with which the Fibonacci sequence repeats modulo n. So if the cycle is 6, then every 6 terms, the remainders repeat.

But let me confirm. Let me check term 5: 1, term 11: 1, term 17: 1, term 23: 1... So every 6 terms, the remainder cycles. So the Pisano period for modulo 4 is 6?

Wait, but I thought the Pisano period for modulo m is usually denoted π(m). Let me check for m=4. According to some references I remember, the Pisano period for 2 is 3, for 3 is 8, for 4 is 6. So yes, seems like 6. So if that's the case, then the sequence of Fibonacci numbers modulo 4 repeats every 6 terms.

So if that's the case, then to find the 100th term modulo 4, we can find 100 modulo 6, and then find the corresponding term in the cycle.

But let me check again. Let me list the cycle:

Positions 1-6: [1, 1, 2, 3, 1, 0]

Positions 7-12: [1, 1, 2, 3, 1, 0]

So the cycle is indeed 6 terms. Therefore, the Pisano period π(4) = 6.

Therefore, the remainder of the nth Fibonacci number modulo 4 is equal to the remainder of the (n mod 6)th term in the cycle. However, we need to be careful with the offset. Let me check:

For example, term 1: n=1, 1 mod 6 = 1, which corresponds to the first term in the cycle, which is 1. Correct.

Term 6: n=6, 6 mod 6 = 0. But in our cycle, the 6th term is 0. However, if we take n mod 6, when n is a multiple of 6, we get 0, but in our cycle, the 6th term is 0. However, if we have n=6, which is 0 mod 6, but in the cycle, the 6th term is 0. So maybe we can adjust the formula. Let me think.

Alternatively, since the cycle is 6 terms, then for n, compute n mod 6. If the remainder is 0, then it corresponds to the 6th term. Otherwise, it corresponds to the remainder. Let me check:

For n=6: 6 mod 6 = 0, so we take the 6th term, which is 0. Correct.

For n=7: 7 mod 6 = 1, so take the 1st term, which is 1. Correct, term 7 is 1.

Similarly, n=12: 12 mod 6 = 0, so 6th term, which is 0. Correct.

Therefore, the formula is: if we let r = n mod 6, then if r = 0, the remainder is the 6th term; else, it's the rth term.

So, for n=100, compute 100 mod 6. Let me calculate that.

Divide 100 by 6: 6*16=96, so 100-96=4. Therefore, 100 mod 6 is 4. Therefore, the remainder is the 4th term in the cycle.

Looking back at the cycle: [1, 1, 2, 3, 1, 0]. The 4th term is 3. Therefore, the remainder when the 100th term is divided by 4 is 3?

Wait, but let me check with some known terms. For example, term 4 is 3, which mod 4 is 3. Term 10 is 55 mod 4 is 3. Term 10: 10 mod 6 is 4, so 4th term is 3. Correct. Term 16: 16 mod 6 is 4, so 4th term is 3. Term 16 mod 4 is 3. Correct. Term 22: 22 mod 6 is 4, so 3. Let me check term 22. Let me compute term 22.

But maybe I can trust the cycle. So if the cycle is 6, then term 100 corresponds to term 4 in the cycle, which is 3. Therefore, the answer is 3?

Wait, but let me check with another approach. Maybe I can compute Fibonacci numbers modulo 4 up to term 100 using the cycle. But since the cycle is 6, then term 100 is same as term 4, which is 3. So answer is 3. But let me check with another term. For example, term 5: 5 mod 4 is 1. Term 5: 5 mod 6 is 5, so 5th term in the cycle is 1. Correct. Term 6: 0. Term 6 mod 6 is 0, so 6th term is 0. Correct. Term 7: 1, which is 1 mod 6, so 1st term is 1. Correct. So seems consistent.

But let me check term 100 again. 100 divided by 6 is 16 with remainder 4. So term 100 is same as term 4, which is 3. Therefore, the remainder is 3. Therefore, the answer is 3?

Wait, but let me check with another method. Maybe using matrix exponentiation or some formula? But that might be complicated. Alternatively, maybe I can compute the Fibonacci numbers modulo 4 up to term 100 using the cycle. Since the cycle is 6, then term 100 is term 4 in the cycle. So 3. But let me check with term 4: 3, term 10: 3, term 16: 3, term 22: 3,... So term 6k + 4 would be 3 mod 4. Since 100 = 6*16 + 4, so yes, term 100 is 3 mod 4. Therefore, the answer is 3. So I think that's correct.

But just to be thorough, let me check with another approach. Let me try to compute Fibonacci numbers modulo 4 up to, say, term 12, and see if the cycle holds.

Term 1: 1

Term 2: 1

Term 3: 2

Term 4: 3

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

Term 6: (Term 4 + Term 5) mod 4 = (3 + 1) = 4 mod 4 = 0

Term 7: (Term 5 + Term 6) mod 4 = (1 + 0) = 1

Term 8: (Term 6 + Term 7) mod 4 = (0 + 1) = 1

Term 9: (Term 7 + Term 8) mod 4 = (1 + 1) = 2

Term 10: (Term 8 + Term 9) mod 4 = (1 + 2) = 3

Term 11: (Term 9 + Term 10) mod 4 = (2 + 3) = 5 mod 4 = 1

Term 12: (Term 10 + Term 11) mod 4 = (3 + 1) = 4 mod 4 = 0

So yes, the cycle is indeed [1, 1, 2, 3, 1, 0], repeating every 6 terms. Therefore, the cycle is confirmed. Therefore, term 100 is term 4 in the cycle, which is 3. Therefore, the remainder is 3.

But just to be 100% sure, let me check term 100 by another way. Let me compute the position in the cycle. Since the cycle is 6, then term number n corresponds to position (n-1) mod 6 + 1? Wait, maybe another way. Let me think.

Alternatively, since the cycle starts at term 1, so term 1 corresponds to position 1, term 2 position 2,... term 6 position 6, term 7 position 1 again. Therefore, the position in the cycle is (n-1) mod 6 + 1. Therefore, for n=100, position is (100-1) mod 6 + 1 = 99 mod 6 + 1. 99 divided by 6 is 16*6=96, remainder 3. Therefore, 99 mod 6 is 3, so position is 3 + 1 = 4. Therefore, position 4 in the cycle, which is 3. So same result. Therefore, the answer is 3.

Alternatively, if I think of the cycle as starting at term 1, then the cycle length is 6. Therefore, term n is equivalent to term (n mod 6). But if n mod 6 is 0, then it's term 6. So for n=6, 6 mod 6=0, term 6. For n=100, 100 mod 6 is 4, so term 4. Which is 3. So same answer.

Therefore, I think the answer is 3. But let me check with another example. Let me check term 4: 3, term 10: 3, term 16: 3, term 22: 3,... So term 6k + 4. For k=0, term 4: 3. For k=1, term 10: 3. For k=2, term 16: 3. For k=16, term 6*16 + 4 = term 100: 3. So yes, that seems consistent.

Alternatively, maybe I can compute the Fibonacci numbers modulo 4 using the formula for the nth term? But that would require knowing the formula, which is F(n) = (φ^n - ψ^n)/√5, where φ=(1+√5)/2 and ψ=(1-√5)/2. But since we are working modulo 4, maybe that's not helpful? Because dealing with irrational numbers modulo 4 is not straightforward. So probably not useful here.

Alternatively, maybe using recursion with modulo 4. Since we know the cycle is 6, we can just use that. So I think the answer is 3.

But just to be 100% sure, let me compute a few more terms manually. Let me compute term 13 to term 18:

Term 13: Term 11 + Term 12 = 1 + 0 = 1 mod 4.

Term 14: Term 12 + Term 13 = 0 + 1 = 1 mod 4.

Term 15: Term 13 + Term 14 = 1 + 1 = 2 mod 4.

Term 16: Term 14 + Term 15 = 1 + 2 = 3 mod 4.

Term 17: Term 15 + Term 16 = 2 + 3 = 5 mod 4 = 1.

Term 18: Term 16 + Term 17 = 3 + 1 = 4 mod 4 = 0.

So term 13 to 18: [1, 1, 2, 3, 1, 0], same as before. So the cycle continues. Therefore, the cycle is indeed 6. Therefore, term 100 is 3 mod 4.

Therefore, I think the answer is 3. Wait, but let me check with another approach. Let me try to compute the Fibonacci numbers modulo 4 up to term 100 using the cycle. Since the cycle is 6, then term 100 is term 4 in the cycle. Therefore, 3. So answer is 3. Therefore, I think that's correct.

But just to make sure, let me check term 4: 3, term 10: 3, term 16: 3, term 22: 3,... term 94: 3, term 100: 3. Since 100 - 4 = 96, which is 6*16, so yes, term 100 is same as term 4. Therefore, 3. So I think that's solid.

Therefore, the remainder when the 100th term is divided by 4 is 3. So the answer is 3. Therefore, I can write that as \boxed{3}.

**Final Answer**
\boxed{3}

To solve the problem of finding the remainder when the 100th term of the Fibonacci sequence is divided by 4, we can take advantage of a key property of the Fibonacci sequence: **modular periodicity**.

---

### Step 1: Understand the Fibonacci Sequence

The Fibonacci sequence is defined as:

$$
F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 3
$$

We are interested in the **100th term**, $ F_{100} $, and specifically its **remainder when divided by 4**, i.e., $ F_{100} \mod 4 $.

---

### Step 2: Use Modular Arithmetic to Find a Pattern

Rather than computing the full Fibonacci sequence up to the 100th term (which would be a very large number), we can compute the sequence **modulo 4** and look for a **repeating pattern**.

Let’s compute the first few Fibonacci numbers modulo 4:

$$
\begin{align*}
F_1 &\equiv 1 \mod 4 \\
F_2 &\equiv 1 \mod 4 \\
F_3 &\equiv 2 \mod 4 \\
F_4 &\equiv 3 \mod 4 \\
F_5 &\equiv 5 \equiv 1 \mod 4 \\
F_6 &\equiv 8 \equiv 0 \mod 4 \\
F_7 &\equiv 13 \equiv 1 \mod 4 \\
F_8 &\equiv 21 \equiv 1 \mod 4 \\
F_9 &\equiv 34 \equiv 2 \mod 4 \\
F_{10} &\equiv 55 \equiv 3 \mod 4 \\
F_{11} &\equiv 89 \equiv 1 \mod 4 \\
F_{12} &\equiv 144 \equiv 0 \mod 4 \\
\end{align*}
$$

We observe that the sequence of remainders modulo 4 is:

$$
[1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, \dots]
$$

This sequence **repeats every 6 terms**, forming a **cycle** of length 6:

$$
\text{Cycle: } [1, 1, 2, 3, 1, 0]
$$

This is known as the **Pisano period** for modulo 4, and it is denoted $ \pi(4) = 6 $.

---

### Step 3: Use the Cycle to Find $ F_{100} \mod 4 $

Since the sequence of remainders repeats every 6 terms, we can find the position of the 100th term within the cycle by computing:

$$
100 \mod 6 = 4
$$

This means that the 100th term corresponds to the **4th term** in the cycle.

From the cycle $[1, 1, 2, 3, 1, 0]$, the 4th term is:

$$
\boxed{3}
$$

---

### 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, let me think. The Fibonacci sequence starts with 1, 1, 2, 3, 5, and so on, right? Each term is the sum of the two previous ones. But calculating the 100th term directly sounds like a huge number. I remember something about modular arithmetic and maybe patterns repeating? Maybe there's a cycle when you take Fibonacci numbers modulo 4? Let me try to recall.

First, let me write down the Fibonacci sequence terms and see what they are modulo 4. Maybe there's a repeating pattern. Let me start listing them:

Term 1: 1
Term 2: 1
Term 3: 2
Term 4: 3
Term 5: 5
Term 6: 8
Term 7: 13
Term 8: 21
Term 9: 34
Term 10: 55
Term 11: 89
Term 12: 144
... and so on.

But instead of the actual numbers, let me compute each term modulo 4:

Term 1: 1 mod 4 = 1
Term 2: 1 mod 4 = 1
Term 3: 2 mod 4 = 2
Term 4: 3 mod 4 = 3
Term 5: 5 mod 4 = 1
Term 6: 8 mod 4 = 0
Term 7: 13 mod 4 = 1 (since 13 divided by 4 is 3 with remainder 1)
Term 8: 21 mod 4 = 1 (21 divided by 4 is 5*4=20, remainder 1)
Term 9: 34 mod 4 = 2 (34 divided by 4 is 8*4=32, remainder 2)
Term 10: 55 mod 4 = 3 (55 divided by 4 is 13*4=52, remainder 3)
Term 11: 89 mod 4 = 1 (89 divided by 4 is 22*4=88, remainder 1)
Term 12: 144 mod 4 = 0 (since 144 is 36*4, so remainder 0)
Term 13: 233 mod 4. Let me compute that. 233 divided by 4: 58*4=232, so remainder 1.
Term 14: 377 mod 4. 377 divided by 4: 94*4=376, remainder 1.
Term 15: 610 mod 4. 610 divided by 4 is 152*4=608, remainder 2.
Term 16: 987 mod 4. 987 divided by 4: 246*4=984, remainder 3.
Term 17: 1597 mod 4. 1597 divided by 4: 399*4=1596, remainder 1.
Term 18: 2584 mod 4. 2584 divided by 4 is 646*4, so remainder 0.
Term 19: 4181 mod 4. Let me see, 4180 is divisible by 4, so 4181 mod 4 is 1.
Term 20: 6765 mod 4. 6764 is divisible by 4, so remainder 1.

Wait, so let me list these mod 4 results:

Term 1: 1

Term 2: 1

Term 3: 2

Term 4: 3

Term 5: 1

Term 6: 0

Term 7: 1

Term 8: 1

Term 9: 2

Term 10: 3

Term 11: 1

Term 12: 0

Term 13: 1

Term 14: 1

Term 15: 2

Term 16: 3

Term 17: 1

Term 18: 0

Term 19: 1

Term 20: 1

Hmm, so looking at this sequence of remainders: 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1,...

It seems like there's a repeating pattern here. Let me check. From term 1 to term 6: 1, 1, 2, 3, 1, 0

Then term 7 to term 12: 1, 1, 2, 3, 1, 0

Then term 13 to term 18: 1, 1, 2, 3, 1, 0

Term 19 to term 24 would be 1, 1, 2, 3, 1, 0 again?

So the cycle seems to be of length 6? Let me check:

Starting from term 1: 1, 1, 2, 3, 1, 0, then repeats 1, 1, 2, 3, 1, 0,...

Yes, so the Pisano period for modulo 4 is 6? Wait, Pisano period is the period with which the Fibonacci sequence repeats modulo n. So if the cycle is 6, then every 6 terms, the remainders repeat.

But let me confirm. Let me check term 5: 1, term 11: 1, term 17: 1, term 23: 1... So every 6 terms, the remainder cycles. So the Pisano period for modulo 4 is 6?

Wait, but I thought the Pisano period for modulo m is usually denoted π(m). Let me check for m=4. According to some references I remember, the Pisano period for 2 is 3, for 3 is 8, for 4 is 6. So yes, seems like 6. So if that's the case, then the sequence of Fibonacci numbers modulo 4 repeats every 6 terms.

So if that's the case, then to find the 100th term modulo 4, we can find 100 modulo 6, and then find the corresponding term in the cycle.

But let me check again. Let me list the cycle:

Positions 1-6: [1, 1, 2, 3, 1, 0]

Positions 7-12: [1, 1, 2, 3, 1, 0]

So the cycle is indeed 6 terms. Therefore, the Pisano period π(4) = 6.

Therefore, the remainder of the nth Fibonacci number modulo 4 is equal to the remainder of the (n mod 6)th term in the cycle. However, we need to be careful with the offset. Let me check:

For example, term 1: n=1, 1 mod 6 = 1, which corresponds to the first term in the cycle, which is 1. Correct.

Term 6: n=6, 6 mod 6 = 0. But in our cycle, the 6th term is 0. However, if we take n mod 6, when n is a multiple of 6, we get 0, but in our cycle, the 6th term is 0. However, if we have n=6, which is 0 mod 6, but in the cycle, the 6th term is 0. So maybe we can adjust the formula. Let me think.

Alternatively, since the cycle is 6 terms, then for n, compute n mod 6. If the remainder is 0, then it corresponds to the 6th term. Otherwise, it corresponds to the remainder. Let me check:

For n=6: 6 mod 6 = 0, so we take the 6th term, which is 0. Correct.

For n=7: 7 mod 6 = 1, so take the 1st term, which is 1. Correct, term 7 is 1.

Similarly, n=12: 12 mod 6 = 0, so 6th term, which is 0. Correct.

Therefore, the formula is: if we let r = n mod 6, then if r = 0, the remainder is the 6th term; else, it's the rth term.

So, for n=100, compute 100 mod 6. Let me calculate that.

Divide 100 by 6: 6*16=96, so 100-96=4. Therefore, 100 mod 6 is 4. Therefore, the remainder is the 4th term in the cycle.

Looking back at the cycle: [1, 1, 2, 3, 1, 0]. The 4th term is 3. Therefore, the remainder when the 100th term is divided by 4 is 3?

Wait, but let me check with some known terms. For example, term 4 is 3, which mod 4 is 3. Term 10 is 55 mod 4 is 3. Term 10: 10 mod 6 is 4, so 4th term is 3. Correct. Term 16: 16 mod 6 is 4, so 4th term is 3. Term 16 mod 4 is 3. Correct. Term 22: 22 mod 6 is 4, so 3. Let me check term 22. Let me compute term 22.

But maybe I can trust the cycle. So if the cycle is 6, then term 100 corresponds to term 4 in the cycle, which is 3. Therefore, the answer is 3?

Wait, but let me check with another approach. Maybe I can compute Fibonacci numbers modulo 4 up to term 100 using the cycle. But since the cycle is 6, then term 100 is same as term 4, which is 3. So answer is 3. But let me check with another term. For example, term 5: 5 mod 4 is 1. Term 5: 5 mod 6 is 5, so 5th term in the cycle is 1. Correct. Term 6: 0. Term 6 mod 6 is 0, so 6th term is 0. Correct. Term 7: 1, which is 1 mod 6, so 1st term is 1. Correct. So seems consistent.

But let me check term 100 again. 100 divided by 6 is 16 with remainder 4. So term 100 is same as term 4, which is 3. Therefore, the remainder is 3. Therefore, the answer is 3?

Wait, but let me check with another method. Maybe using matrix exponentiation or some formula? But that might be complicated. Alternatively, maybe I can compute the Fibonacci numbers modulo 4 up to term 100 using the cycle. Since the cycle is 6, then term 100 is term 4 in the cycle. So 3. But let me check with term 4: 3, term 10: 3, term 16: 3, term 22: 3,... So term 6k + 4 would be 3 mod 4. Since 100 = 6*16 + 4, so yes, term 100 is 3 mod 4. Therefore, the answer is 3. So I think that's correct.

But just to be thorough, let me check with another approach. Let me try to compute Fibonacci numbers modulo 4 up to, say, term 12, and see if the cycle holds.

Term 1: 1

Term 2: 1

Term 3: 2

Term 4: 3

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

Term 6: (Term 4 + Term 5) mod 4 = (3 + 1) = 4 mod 4 = 0

Term 7: (Term 5 + Term 6) mod 4 = (1 + 0) = 1

Term 8: (Term 6 + Term 7) mod 4 = (0 + 1) = 1

Term 9: (Term 7 + Term 8) mod 4 = (1 + 1) = 2

Term 10: (Term 8 + Term 9) mod 4 = (1 + 2) = 3

Term 11: (Term 9 + Term 10) mod 4 = (2 + 3) = 5 mod 4 = 1

Term 12: (Term 10 + Term 11) mod 4 = (3 + 1) = 4 mod 4 = 0

So yes, the cycle is indeed [1, 1, 2, 3, 1, 0], repeating every 6 terms. Therefore, the cycle is confirmed. Therefore, term 100 is term 4 in the cycle, which is 3. Therefore, the remainder is 3.

But just to be 100% sure, let me check term 100 by another way. Let me compute the position in the cycle. Since the cycle is 6, then term number n corresponds to position (n-1) mod 6 + 1? Wait, maybe another way. Let me think.

Alternatively, since the cycle starts at term 1, so term 1 corresponds to position 1, term 2 position 2,... term 6 position 6, term 7 position 1 again. Therefore, the position in the cycle is (n-1) mod 6 + 1. Therefore, for n=100, position is (100-1) mod 6 + 1 = 99 mod 6 + 1. 99 divided by 6 is 16*6=96, remainder 3. Therefore, 99 mod 6 is 3, so position is 3 + 1 = 4. Therefore, position 4 in the cycle, which is 3. So same result. Therefore, the answer is 3.

Alternatively, if I think of the cycle as starting at term 1, then the cycle length is 6. Therefore, term n is equivalent to term (n mod 6). But if n mod 6 is 0, then it's term 6. So for n=6, 6 mod 6=0, term 6. For n=100, 100 mod 6 is 4, so term 4. Which is 3. So same answer.

Therefore, I think the answer is 3. But let me check with another example. Let me check term 4: 3, term 10: 3, term 16: 3, term 22: 3,... So term 6k + 4. For k=0, term 4: 3. For k=1, term 10: 3. For k=2, term 16: 3. For k=16, term 6*16 + 4 = term 100: 3. So yes, that seems consistent.

Alternatively, maybe I can compute the Fibonacci numbers modulo 4 using the formula for the nth term? But that would require knowing the formula, which is F(n) = (φ^n - ψ^n)/√5, where φ=(1+√5)/2 and ψ=(1-√5)/2. But since we are working modulo 4, maybe that's not helpful? Because dealing with irrational numbers modulo 4 is not straightforward. So probably not useful here.

Alternatively, maybe using recursion with modulo 4. Since we know the cycle is 6, we can just use that. So I think the answer is 3.

But just to be 100% sure, let me compute a few more terms manually. Let me compute term 13 to term 18:

Term 13: Term 11 + Term 12 = 1 + 0 = 1 mod 4.

Term 14: Term 12 + Term 13 = 0 + 1 = 1 mod 4.

Term 15: Term 13 + Term 14 = 1 + 1 = 2 mod 4.

Term 16: Term 14 + Term 15 = 1 + 2 = 3 mod 4.

Term 17: Term 15 + Term 16 = 2 + 3 = 5 mod 4 = 1.

Term 18: Term 16 + Term 17 = 3 + 1 = 4 mod 4 = 0.

So term 13 to 18: [1, 1, 2, 3, 1, 0], same as before. So the cycle continues. Therefore, the cycle is indeed 6. Therefore, term 100 is 3 mod 4.

Therefore, I think the answer is 3. Wait, but let me check with another approach. Let me try to compute the Fibonacci numbers modulo 4 up to term 100 using the cycle. Since the cycle is 6, then term 100 is term 4 in the cycle. Therefore, 3. So answer is 3. Therefore, I think that's correct.

But just to make sure, let me check term 4: 3, term 10: 3, term 16: 3, term 22: 3,... term 94: 3, term 100: 3. Since 100 - 4 = 96, which is 6*16, so yes, term 100 is same as term 4. Therefore, 3. So I think that's solid.

Therefore, the remainder when the 100th term is divided by 4 is 3. So the answer is 3. Therefore, I can write that as \boxed{3}.

**Final Answer**
\boxed{3}

To solve the problem of finding the remainder when the 100th term of the Fibonacci sequence is divided by 4, we can take advantage of a key property of the Fibonacci sequence: **modular periodicity**.

---

### Step 1: Understand the Fibonacci Sequence

The Fibonacci sequence is defined as:

$$
F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 3
$$

We are interested in the **100th term**, $ F_{100} $, and specifically its **remainder when divided by 4**, i.e., $ F_{100} \mod 4 $.

---

### Step 2: Use Modular Arithmetic to Find a Pattern

Rather than computing the full Fibonacci sequence up to the 100th term (which would be a very large number), we can compute the sequence **modulo 4** and look for a **repeating pattern**.

Let’s compute the first few Fibonacci numbers modulo 4:

$$
\begin{align*}
F_1 &\equiv 1 \mod 4 \\
F_2 &\equiv 1 \mod 4 \\
F_3 &\equiv 2 \mod 4 \\
F_4 &\equiv 3 \mod 4 \\
F_5 &\equiv 5 \equiv 1 \mod 4 \\
F_6 &\equiv 8 \equiv 0 \mod 4 \\
F_7 &\equiv 13 \equiv 1 \mod 4 \\
F_8 &\equiv 21 \equiv 1 \mod 4 \\
F_9 &\equiv 34 \equiv 2 \mod 4 \\
F_{10} &\equiv 55 \equiv 3 \mod 4 \\
F_{11} &\equiv 89 \equiv 1 \mod 4 \\
F_{12} &\equiv 144 \equiv 0 \mod 4 \\
\end{align*}
$$

We observe that the sequence of remainders modulo 4 is:

$$
[1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, \dots]
$$

This sequence **repeats every 6 terms**, forming a **cycle** of length 6:

$$
\text{Cycle: } [1, 1, 2, 3, 1, 0]
$$

This is known as the **Pisano period** for modulo 4, and it is denoted $ \pi(4) = 6 $.

---

### Step 3: Use the Cycle to Find $ F_{100} \mod 4 $

Since the sequence of remainders repeats every 6 terms, we can find the position of the 100th term within the cycle by computing:

$$
100 \mod 6 = 4
$$

This means that the 100th term corresponds to the **4th term** in the cycle.

From the cycle $[1, 1, 2, 3, 1, 0]$, the 4th term is:

$$
\boxed{3}
$$

---

### 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 which is one less than twice its reciprocal. Hmm, let me try to understand the problem first. Let me rephrase it in my own words. The number, let's call it x, is equal to one less than twice its reciprocal. The reciprocal of x is 1/x, right? So twice the reciprocal would be 2*(1/x), and then one less than that would be 2*(1/x) - 1. So according to the problem, x equals that expression. So, mathematically, that would be:

x = 2*(1/x) - 1

Wait, let me check that again. The problem says "one less than twice its reciprocal." So twice the reciprocal is 2*(1/x), and then subtract one. So yes, x = 2*(1/x) - 1. That seems right.

Alternatively, maybe I should write it as x = 2*(1/x) - 1. Let me confirm. If I have a number, say, 2, then its reciprocal is 1/2, twice that is 1, and one less than that is 0. So 2 is not equal to 0. So that example doesn't work, but that's okay. Let me just stick with the equation.

So, the equation is x = 2/x - 1. To solve this equation, I need to get rid of the fraction. Let me multiply both sides by x to eliminate the denominator. Let me do that step by step.

Multiplying both sides by x:

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

Which simplifies to:

x² = 2 - x

Wait, because (2/x)*x is 2, and (-1)*x is -x. So, x² = 2 - x.

Now, let me rearrange the equation to bring all terms to one side. Let me add x to both sides:

x² + x = 2

Then subtract 2 from both sides:

x² + x - 2 = 0

So, now I have a quadratic equation: x² + x - 2 = 0. Let me try to factor this quadratic. To factor x² + x - 2, I need two numbers that multiply to -2 and add up to 1. Let me think. The factors of -2 could be 2 and -1. Let me check: 2 * (-1) = -2, and 2 + (-1) = 1. Yes! Perfect. So the quadratic factors as:

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

Therefore, the solutions are x = -2 and x = 1. Now, the problem asks for the smallest number which satisfies this condition. So between -2 and 1, which is smaller? Well, -2 is smaller than 1. So is -2 the answer? Wait, but let me check if both solutions actually satisfy the original equation. Sometimes when you multiply both sides by x, you might introduce extraneous solutions, especially if x is zero, but in this case, x can't be zero because we have 1/x in the original equation. So x can't be zero, but our solutions are -2 and 1, neither of which is zero, so that's okay.

Let me check x = -2 first. Let's compute twice its reciprocal minus one. The reciprocal of -2 is -1/2. Twice that is -1. Then subtract one: -1 - 1 = -2. Which is equal to x. So that works. Now check x = 1. The reciprocal of 1 is 1. Twice that is 2. Subtract one: 2 - 1 = 1. Which is equal to x. So that also works. So both solutions are valid.

But the question asks for the smallest number. So between -2 and 1, -2 is smaller. Therefore, the answer is -2. Wait, but let me make sure there are no other solutions. The quadratic equation only has two solutions, so those are the only ones. Therefore, -2 is the smallest number. But let me think again. Is there a possibility that there's a smaller number? For example, if there are negative numbers smaller than -2? But according to the equation, the solutions are only -2 and 1. So unless there's a mistake in my solving, which I need to check again.

Original equation: x = 2*(1/x) - 1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Factored as (x + 2)(x - 1) = 0. So solutions x = -2 and x = 1. That seems correct. So unless I made a mistake in setting up the equation.

Wait, let me check the original problem again. "What is the smallest number which is one less than twice its reciprocal?" So, the number is one less than twice its reciprocal. So, if the number is x, then x = 2*(reciprocal of x) - 1. Which is x = 2*(1/x) - 1. So that's correct. So the equation is correct. Therefore, the solutions are indeed -2 and 1. Therefore, the smallest is -2.

But wait, maybe the problem is expecting a positive number? Sometimes, depending on the context, but the problem doesn't specify that the number has to be positive. So unless there's a restriction, -2 is valid. Let me check once more with x = -2. Twice its reciprocal is 2*(-1/2) = -1. Then one less than that is -1 - 1 = -2. Which is equal to x. So that works. So yes, -2 is a valid solution.

Alternatively, if someone interprets the problem as "one less than twice its reciprocal" as (twice the number) minus one, but that would be different. Wait, let me check that. If someone misreads the problem as "twice the number is one less than its reciprocal," that would be different. But the problem says "the number is one less than twice its reciprocal." So the number equals (twice the reciprocal) minus one. So the equation is correct.

Alternatively, maybe the problem is in the wording. Let me parse it again. "The smallest number which is one less than twice its reciprocal." So, "is one less than twice its reciprocal." So, number = (twice reciprocal) - 1. So yes, that's how I set it up. So I think my equation is correct.

Therefore, the answer is -2. But let me check if there's a different interpretation. For example, maybe "one less than twice its reciprocal" is interpreted as twice (the reciprocal of one less than the number). But that would be different. Let me check that. If that were the case, then the equation would be x = 2*(1/(x - 1)). But that's a different equation. However, the original problem says "one less than twice its reciprocal," which is more naturally read as (twice the reciprocal) minus one, not twice (reciprocal of (one less than the number)). So I think my original interpretation is correct.

But just to be thorough, let me check both interpretations. First, my original interpretation: x = 2*(1/x) - 1, leading to x = -2 or 1. Second, if it's twice the reciprocal of (x - 1), then x = 2*(1/(x - 1)). Let me solve that equation as well, just in case.

So, x = 2/(x - 1). Multiply both sides by (x - 1):

x(x - 1) = 2

Which is x² - x - 2 = 0. Then factoring: (x - 2)(x + 1) = 0. So solutions x = 2 and x = -1. Then check if they satisfy the original equation. For x = 2: reciprocal of (x - 1) is 1/(2 - 1) = 1. Twice that is 2. Which equals x. So that works. For x = -1: reciprocal of (-1 - 1) = reciprocal of -2 is -1/2. Twice that is -1. Which is equal to x = -1. So that works too. So in this case, the solutions would be -1 and 2. Then the smallest number would be -1. But this is under a different interpretation of the problem. However, since the original problem says "one less than twice its reciprocal," not "twice the reciprocal of one less than the number," I think the first interpretation is correct. But just to be safe, maybe the problem is ambiguous? However, in standard mathematical phrasing, "one less than twice its reciprocal" would mean twice the reciprocal minus one, not twice the reciprocal of one less. So I think my original answer of -2 is correct.

But let me check if there's another way. Maybe the problem is from a source where numbers are assumed to be positive? For example, if we consider only positive numbers, then the answer would be 1. But the problem says "smallest number," and if negative numbers are allowed, then -2 is smaller. However, maybe the problem is expecting a positive integer? But the problem doesn't specify. So unless there's a constraint, -2 is the answer.

Alternatively, maybe I made a mistake in the sign when solving. Let me check again. Original equation: x = 2*(1/x) - 1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Then factoring: looking for two numbers that multiply to -2 and add to 1. 2 and -1. So (x + 2)(x - 1) = 0. Correct. So solutions x = -2 and x = 1. So that's correct.

Alternatively, maybe the problem is in the translation? The original problem says "the smallest number which is one less than twice its reciprocal." If "smallest" refers to the least positive number, then 1 is the answer. But if "smallest" refers to the least real number, then -2 is the answer. Since the problem doesn't specify, but in mathematics, "smallest" usually refers to the least in value, so -2 is smaller than 1. However, sometimes in certain contexts, people might refer to the smallest positive number. But since the problem doesn't specify, I think -2 is the correct answer.

But let me check once more with the original problem. If I plug in x = -2 into the original condition: Is -2 one less than twice its reciprocal? Twice its reciprocal is 2*(-1/2) = -1. Then one less than that is -1 - 1 = -2. Which is equal to x. So yes, that works. So -2 satisfies the condition. Therefore, unless there's a restriction, -2 is the answer.

Alternatively, maybe the problem is from a math competition or a textbook chapter that deals with positive numbers only? But since the user hasn't specified, I can't assume that. Therefore, I think the answer is -2.

But let me check if there's another possible solution. For example, suppose someone thought that "one less than twice its reciprocal" is 2*(x - 1). Wait, that would be different. Let me see. If someone misreads it as "twice (one less than its reciprocal)", which would be 2*(1/x - 1). Then the equation would be x = 2*(1/x - 1). Let me solve that. x = 2/x - 2. Multiply both sides by x: x² = 2 - 2x. Then x² + 2x - 2 = 0. Using quadratic formula: x = [-2 ± sqrt(4 + 8)] / 2 = [-2 ± sqrt(12)] / 2 = [-2 ± 2*sqrt(3)] / 2 = -1 ± sqrt(3). So approximately -1 + 1.732 = 0.732 or -1 - 1.732 = -2.732. Then the smallest number would be -2.732. But this is under a different misinterpretation. However, this is not the original problem's wording. So again, unless there's a misinterpretation, this is not the case.

Therefore, going back, the original equation is x = 2*(1/x) - 1, leading to x = -2 and x = 1. Therefore, the smallest number is -2. Therefore, I think that is the answer.

But just to make sure, let me check if there's a different way to interpret "one less than twice its reciprocal." For example, maybe "twice (one less than its reciprocal)"? Which would be 2*(1/x - 1). But that would be different. Let me check that. If the problem was "twice (one less than its reciprocal)", then the equation would be x = 2*(1/x - 1). Which is different from what I had before. Let me solve that equation again. x = 2*(1/x - 1). Multiply both sides by x: x² = 2*(1 - x). So x² = 2 - 2x. Then x² + 2x - 2 = 0. Which is the same equation as before, leading to solutions -1 ± sqrt(3). But again, this is under a different interpretation. However, the original problem says "one less than twice its reciprocal," which is different from "twice (one less than its reciprocal)." So the correct interpretation is the first one.

Therefore, I think my original answer of -2 is correct. Therefore, the smallest number is -2.

But just to be absolutely thorough, let me check if there's a possibility that the problem is in another language, and "reciprocal" might have a different meaning? No, reciprocal is standard. The reciprocal of x is 1/x. So that's correct.

Alternatively, maybe the problem is referring to the multiplicative inverse, which is the same as reciprocal. So that's not an issue.

Therefore, after all these checks, I think the answer 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: Define the Equation

Let the number be $ x $. The **reciprocal** of $ x $ is $ \frac{1}{x} $. Then, **twice the reciprocal** is $ 2 \cdot \frac{1}{x} $, and **one less than that** is:

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

According to the problem, this expression is equal to the number $ x $. So we write the equation:

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

---

### Step 2: Eliminate the Fraction

To eliminate the denominator, multiply both sides of the equation by $ x $ (note: $ x \neq 0 $, since we are dealing with reciprocals):

$$
x \cdot x = x \cdot \left( \frac{2}{x} - 1 \right)
$$

$$
x^2 = 2 - x
$$

---

### Step 3: Rearrange into a Standard Quadratic Form

Bring all terms to one side:

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

---

### Step 4: Solve the Quadratic Equation

We factor the quadratic:

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

So the solutions are:

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

---

### Step 5: Determine the Smallest Number

We are asked for the **smallest number** that satisfies the condition. Comparing the two solutions:

- $ -2 < 1 $

So, the **smallest number** is:

$$
\boxed{-2}
$$

---

### Step 6: Verify the Solution

Let’s verify that $ x = -2 $ satisfies the original condition:

- Reciprocal of $ -2 $ is $ \frac{1}{-2} = -\frac{1}{2} $
- Twice that is $ 2 \cdot (-\frac{1}{2}) = -1 $
- One less than that is $ -1 - 1 = -2 $

So, $ x = -2 $ satisfies the condition.

Similarly, $ x = 1 $ also satisfies the condition, but it is **not** the smallest.

---

### Final Answer

$$
\boxed{-2}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest number which is one less than twice its reciprocal. Hmm, let me try to understand the problem first. Let me rephrase it in my own words. The number, let's call it x, is equal to one less than twice its reciprocal. The reciprocal of x is 1/x, right? So twice the reciprocal would be 2*(1/x), and then one less than that would be 2*(1/x) - 1. So according to the problem, x equals that expression. So, mathematically, that would be:

x = 2*(1/x) - 1

Wait, let me check that again. The problem says "one less than twice its reciprocal." So twice the reciprocal is 2*(1/x), and then subtract one. So yes, x = 2*(1/x) - 1. That seems right.

Alternatively, maybe I should write it as x = 2*(1/x) - 1. Let me confirm. If I have a number, say, 2, then its reciprocal is 1/2, twice that is 1, and one less than that is 0. So 2 is not equal to 0. So that example doesn't work, but that's okay. Let me just stick with the equation.

So, the equation is x = 2/x - 1. To solve this equation, I need to get rid of the fraction. Let me multiply both sides by x to eliminate the denominator. Let me do that step by step.

Multiplying both sides by x:

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

Which simplifies to:

x² = 2 - x

Wait, because (2/x)*x is 2, and (-1)*x is -x. So, x² = 2 - x.

Now, let me rearrange the equation to bring all terms to one side. Let me add x to both sides:

x² + x = 2

Then subtract 2 from both sides:

x² + x - 2 = 0

So, now I have a quadratic equation: x² + x - 2 = 0. Let me try to factor this quadratic. To factor x² + x - 2, I need two numbers that multiply to -2 and add up to 1. Let me think. The factors of -2 could be 2 and -1. Let me check: 2 * (-1) = -2, and 2 + (-1) = 1. Yes! Perfect. So the quadratic factors as:

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

Therefore, the solutions are x = -2 and x = 1. Now, the problem asks for the smallest number which satisfies this condition. So between -2 and 1, which is smaller? Well, -2 is smaller than 1. So is -2 the answer? Wait, but let me check if both solutions actually satisfy the original equation. Sometimes when you multiply both sides by x, you might introduce extraneous solutions, especially if x is zero, but in this case, x can't be zero because we have 1/x in the original equation. So x can't be zero, but our solutions are -2 and 1, neither of which is zero, so that's okay.

Let me check x = -2 first. Let's compute twice its reciprocal minus one. The reciprocal of -2 is -1/2. Twice that is -1. Then subtract one: -1 - 1 = -2. Which is equal to x. So that works. Now check x = 1. The reciprocal of 1 is 1. Twice that is 2. Subtract one: 2 - 1 = 1. Which is equal to x. So that also works. So both solutions are valid.

But the question asks for the smallest number. So between -2 and 1, -2 is smaller. Therefore, the answer is -2. Wait, but let me make sure there are no other solutions. The quadratic equation only has two solutions, so those are the only ones. Therefore, -2 is the smallest number. But let me think again. Is there a possibility that there's a smaller number? For example, if there are negative numbers smaller than -2? But according to the equation, the solutions are only -2 and 1. So unless there's a mistake in my solving, which I need to check again.

Original equation: x = 2*(1/x) - 1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Factored as (x + 2)(x - 1) = 0. So solutions x = -2 and x = 1. That seems correct. So unless I made a mistake in setting up the equation.

Wait, let me check the original problem again. "What is the smallest number which is one less than twice its reciprocal?" So, the number is one less than twice its reciprocal. So, if the number is x, then x = 2*(reciprocal of x) - 1. Which is x = 2*(1/x) - 1. So that's correct. So the equation is correct. Therefore, the solutions are indeed -2 and 1. Therefore, the smallest is -2.

But wait, maybe the problem is expecting a positive number? Sometimes, depending on the context, but the problem doesn't specify that the number has to be positive. So unless there's a restriction, -2 is valid. Let me check once more with x = -2. Twice its reciprocal is 2*(-1/2) = -1. Then one less than that is -1 - 1 = -2. Which is equal to x. So that works. So yes, -2 is a valid solution.

Alternatively, if someone interprets the problem as "one less than twice its reciprocal" as (twice the number) minus one, but that would be different. Wait, let me check that. If someone misreads the problem as "twice the number is one less than its reciprocal," that would be different. But the problem says "the number is one less than twice its reciprocal." So the number equals (twice the reciprocal) minus one. So the equation is correct.

Alternatively, maybe the problem is in the wording. Let me parse it again. "The smallest number which is one less than twice its reciprocal." So, "is one less than twice its reciprocal." So, number = (twice reciprocal) - 1. So yes, that's how I set it up. So I think my equation is correct.

Therefore, the answer is -2. But let me check if there's a different interpretation. For example, maybe "one less than twice its reciprocal" is interpreted as twice (the reciprocal of one less than the number). But that would be different. Let me check that. If that were the case, then the equation would be x = 2*(1/(x - 1)). But that's a different equation. However, the original problem says "one less than twice its reciprocal," which is more naturally read as (twice the reciprocal) minus one, not twice (reciprocal of (one less than the number)). So I think my original interpretation is correct.

But just to be thorough, let me check both interpretations. First, my original interpretation: x = 2*(1/x) - 1, leading to x = -2 or 1. Second, if it's twice the reciprocal of (x - 1), then x = 2*(1/(x - 1)). Let me solve that equation as well, just in case.

So, x = 2/(x - 1). Multiply both sides by (x - 1):

x(x - 1) = 2

Which is x² - x - 2 = 0. Then factoring: (x - 2)(x + 1) = 0. So solutions x = 2 and x = -1. Then check if they satisfy the original equation. For x = 2: reciprocal of (x - 1) is 1/(2 - 1) = 1. Twice that is 2. Which equals x. So that works. For x = -1: reciprocal of (-1 - 1) = reciprocal of -2 is -1/2. Twice that is -1. Which is equal to x = -1. So that works too. So in this case, the solutions would be -1 and 2. Then the smallest number would be -1. But this is under a different interpretation of the problem. However, since the original problem says "one less than twice its reciprocal," not "twice the reciprocal of one less than the number," I think the first interpretation is correct. But just to be safe, maybe the problem is ambiguous? However, in standard mathematical phrasing, "one less than twice its reciprocal" would mean twice the reciprocal minus one, not twice the reciprocal of one less. So I think my original answer of -2 is correct.

But let me check if there's another way. Maybe the problem is from a source where numbers are assumed to be positive? For example, if we consider only positive numbers, then the answer would be 1. But the problem says "smallest number," and if negative numbers are allowed, then -2 is smaller. However, maybe the problem is expecting a positive integer? But the problem doesn't specify. So unless there's a constraint, -2 is the answer.

Alternatively, maybe I made a mistake in the sign when solving. Let me check again. Original equation: x = 2*(1/x) - 1. Multiply both sides by x: x² = 2 - x. Then x² + x - 2 = 0. Then factoring: looking for two numbers that multiply to -2 and add to 1. 2 and -1. So (x + 2)(x - 1) = 0. Correct. So solutions x = -2 and x = 1. So that's correct.

Alternatively, maybe the problem is in the translation? The original problem says "the smallest number which is one less than twice its reciprocal." If "smallest" refers to the least positive number, then 1 is the answer. But if "smallest" refers to the least real number, then -2 is the answer. Since the problem doesn't specify, but in mathematics, "smallest" usually refers to the least in value, so -2 is smaller than 1. However, sometimes in certain contexts, people might refer to the smallest positive number. But since the problem doesn't specify, I think -2 is the correct answer.

But let me check once more with the original problem. If I plug in x = -2 into the original condition: Is -2 one less than twice its reciprocal? Twice its reciprocal is 2*(-1/2) = -1. Then one less than that is -1 - 1 = -2. Which is equal to x. So yes, that works. So -2 satisfies the condition. Therefore, unless there's a restriction, -2 is the answer.

Alternatively, maybe the problem is from a math competition or a textbook chapter that deals with positive numbers only? But since the user hasn't specified, I can't assume that. Therefore, I think the answer is -2.

But let me check if there's another possible solution. For example, suppose someone thought that "one less than twice its reciprocal" is 2*(x - 1). Wait, that would be different. Let me see. If someone misreads it as "twice (one less than its reciprocal)", which would be 2*(1/x - 1). Then the equation would be x = 2*(1/x - 1). Let me solve that. x = 2/x - 2. Multiply both sides by x: x² = 2 - 2x. Then x² + 2x - 2 = 0. Using quadratic formula: x = [-2 ± sqrt(4 + 8)] / 2 = [-2 ± sqrt(12)] / 2 = [-2 ± 2*sqrt(3)] / 2 = -1 ± sqrt(3). So approximately -1 + 1.732 = 0.732 or -1 - 1.732 = -2.732. Then the smallest number would be -2.732. But this is under a different misinterpretation. However, this is not the original problem's wording. So again, unless there's a misinterpretation, this is not the case.

Therefore, going back, the original equation is x = 2*(1/x) - 1, leading to x = -2 and x = 1. Therefore, the smallest number is -2. Therefore, I think that is the answer.

But just to make sure, let me check if there's a different way to interpret "one less than twice its reciprocal." For example, maybe "twice (one less than its reciprocal)"? Which would be 2*(1/x - 1). But that would be different. Let me check that. If the problem was "twice (one less than its reciprocal)", then the equation would be x = 2*(1/x - 1). Which is different from what I had before. Let me solve that equation again. x = 2*(1/x - 1). Multiply both sides by x: x² = 2*(1 - x). So x² = 2 - 2x. Then x² + 2x - 2 = 0. Which is the same equation as before, leading to solutions -1 ± sqrt(3). But again, this is under a different interpretation. However, the original problem says "one less than twice its reciprocal," which is different from "twice (one less than its reciprocal)." So the correct interpretation is the first one.

Therefore, I think my original answer of -2 is correct. Therefore, the smallest number is -2.

But just to be absolutely thorough, let me check if there's a possibility that the problem is in another language, and "reciprocal" might have a different meaning? No, reciprocal is standard. The reciprocal of x is 1/x. So that's correct.

Alternatively, maybe the problem is referring to the multiplicative inverse, which is the same as reciprocal. So that's not an issue.

Therefore, after all these checks, I think the answer 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: Define the Equation

Let the number be $ x $. The **reciprocal** of $ x $ is $ \frac{1}{x} $. Then, **twice the reciprocal** is $ 2 \cdot \frac{1}{x} $, and **one less than that** is:

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

According to the problem, this expression is equal to the number $ x $. So we write the equation:

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

---

### Step 2: Eliminate the Fraction

To eliminate the denominator, multiply both sides of the equation by $ x $ (note: $ x \neq 0 $, since we are dealing with reciprocals):

$$
x \cdot x = x \cdot \left( \frac{2}{x} - 1 \right)
$$

$$
x^2 = 2 - x
$$

---

### Step 3: Rearrange into a Standard Quadratic Form

Bring all terms to one side:

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

---

### Step 4: Solve the Quadratic Equation

We factor the quadratic:

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

So the solutions are:

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

---

### Step 5: Determine the Smallest Number

We are asked for the **smallest number** that satisfies the condition. Comparing the two solutions:

- $ -2 < 1 $

So, the **smallest number** is:

$$
\boxed{-2}
$$

---

### Step 6: Verify the Solution

Let’s verify that $ x = -2 $ satisfies the original condition:

- Reciprocal of $ -2 $ is $ \frac{1}{-2} = -\frac{1}{2} $
- Twice that is $ 2 \cdot (-\frac{1}{2}) = -1 $
- One less than that is $ -1 - 1 = -2 $

So, $ x = -2 $ satisfies the condition.

Similarly, $ x = 1 $ also satisfies the condition, but it is **not** the smallest.

---

### 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 $\boxed{-2}$.


Explanation:
Okay, so I need to solve this equation: the cube root of (x squared minus 4x plus 4) equals 16. Let me write that down first to make sure I have it right.

The equation is ∛(x² - 4x + 4) = 16. Alright, cube root of something equals 16. To get rid of the cube root, I should cube both sides of the equation. That way, the cube root and the cube will cancel each other out. Let me try that.

So, if I cube both sides, I get:

(∛(x² - 4x + 4))³ = 16³

Which simplifies to:

x² - 4x + 4 = 16³

Now, I need to calculate 16 cubed. Let me do that. 16 times 16 is 256, and then 256 times 16. Hmm, let me compute that step by step. 256 times 10 is 2560, 256 times 6 is 1536. Adding those together: 2560 + 1536. Let me add them. 2560 + 1500 is 4060, and then +36 is 4096. So 16³ is 4096.

So now the equation is:

x² - 4x + 4 = 4096

Okay, so I need to solve this quadratic equation. Let me rearrange it to standard form. Subtract 4096 from both sides:

x² - 4x + 4 - 4096 = 0

Simplify the constants:

4 - 4096 is... let me calculate that. 4096 - 4 is 4092, so with the negative sign, it's -4092. So the equation becomes:

x² - 4x - 4092 = 0

Now, I need to solve this quadratic equation. Let me check if I can factor it, but given the numbers, it might not be straightforward. Let me check the discriminant first to see if it's a perfect square or not. The discriminant D is b² - 4ac. Here, a = 1, b = -4, c = -4092.

So D = (-4)² - 4(1)(-4092) = 16 + 4*4092. Let me compute that. 4*4092. Let me do 4*4000 = 16000, and 4*92 = 368. So total is 16000 + 368 = 16368. Then add 16: 16368 + 16 = 16384.

Hmm, 16384. Wait, that number seems familiar. Isn't that 128 squared? Let me check. 128 squared is (120 + 8)^2 = 120² + 2*120*8 + 8² = 14400 + 1920 + 64 = 14400 + 1920 is 16320, plus 64 is 16384. Yes! So the discriminant is 128². That's a perfect square, so the equation can be solved with integer solutions.

So, using the quadratic formula: x = [-b ± √D]/(2a). Here, b is -4, so -b is 4. So:

x = [4 ± √16384]/2

Since √16384 is 128, as we saw earlier:

x = [4 ± 128]/2

So that gives two solutions:

First solution: (4 + 128)/2 = 132/2 = 66

Second solution: (4 - 128)/2 = (-124)/2 = -62

But the problem asks for the positive value of x. So between 66 and -62, the positive one is 66.

Wait, but let me check if both solutions are valid in the original equation. Sometimes when you cube both sides or do other operations, you might introduce extraneous solutions. Let me verify both solutions in the original equation.

First, let's check x = 66.

Original equation: ∛(x² - 4x + 4) = 16

Compute x² - 4x + 4 when x = 66:

First, x² is 66². Let me calculate that. 60² is 3600, 6² is 36, and then cross term 2*60*6 = 720. So (60 + 6)^2 = 3600 + 720 + 36 = 4356.

Then subtract 4x: 4*66 = 264. So 4356 - 264 = 4092. Then add 4: 4092 + 4 = 4096.

So inside the cube root, we have 4096. Then ∛4096. Let me check what is 16³. Earlier, we had 16³ = 4096. So ∛4096 is 16. Which matches the right-hand side. So x = 66 is a valid solution.

Now check x = -62.

Compute x² - 4x + 4 when x = -62.

First, x² is (-62)^2 = 62². Let me compute that. 60² is 3600, 2² is 4, and cross term 2*60*2 = 240. So (60 + 2)^2 = 3600 + 240 + 4 = 3844.

Then subtract 4x: 4*(-62) = -248. So subtracting 4x is the same as adding 248. So 3844 + 248 = 4092. Then add 4: 4092 + 4 = 4096.

So again, inside the cube root is 4096, so ∛4096 = 16. Which is equal to the right-hand side. So x = -62 also satisfies the original equation.

But the problem asks for the positive value of x, so the answer is 66.

Wait, but let me check if there's any mistake in my calculation. Let me verify the quadratic equation again.

Original equation after cubing both sides: x² - 4x + 4 = 4096. Then moving 4096 to the left: x² - 4x + 4 - 4096 = x² - 4x - 4092 = 0. Correct. Then discriminant D = (-4)^2 - 4*1*(-4092) = 16 + 16368 = 16384. Correct. Square root of 16384 is 128. Then solutions (4 ± 128)/2. Correct. So 66 and -62. Both check out.

But maybe there's another way to approach the original equation? Let me think. The expression inside the cube root is x² - 4x + 4. Wait, that looks like a perfect square. Let me check.

x² - 4x + 4. Let me factor it. It's x² - 4x + 4. Which is (x - 2)^2. Because (x - 2)^2 = x² - 4x + 4. Yes! So the original equation can be rewritten as:

∛[(x - 2)^2] = 16

So that's another way to write it. Maybe this helps? Let me see.

So, if I have the cube root of (x - 2)^2 equals 16. Then, cubing both sides gives (x - 2)^2 = 16³ = 4096. Then, taking square roots of both sides, we get |x - 2| = √4096.

Wait, but √4096 is 64, since 64² is 4096. So |x - 2| = 64. Therefore, x - 2 = 64 or x - 2 = -64. Therefore, x = 64 + 2 = 66 or x = -64 + 2 = -62.

So same solutions as before. So that's another way to approach it, recognizing the perfect square. So that's a good check.

Therefore, the positive solution is 66.

So, I think that's solid. Both methods lead to the same answer, and both solutions check out in the original equation. Therefore, the positive value of x is 66.

**Final Answer**
The positive value of $ x $ is \boxed{66}.

To solve the equation:

$$
\sqrt[3]{x^2 - 4x + 4} = 16
$$

we begin by eliminating the cube root by cubing both sides of the equation:

$$
\left(\sqrt[3]{x^2 - 4x + 4}\right)^3 = 16^3
$$

This simplifies to:

$$
x^2 - 4x + 4 = 4096
$$

---

### Step 1: Recognize the Perfect Square

Observe that the expression $ x^2 - 4x + 4 $ is a perfect square:

$$
x^2 - 4x + 4 = (x - 2)^2
$$

So the equation becomes:

$$
\sqrt[3]{(x - 2)^2} = 16
$$

Cubing both sides again:

$$
(x - 2)^2 = 16^3 = 4096
$$

---

### Step 2: Solve the Equation

Take the square root of both sides:

$$
|x - 2| = \sqrt{4096} = 64
$$

This gives two possible solutions:

$$
x - 2 = 64 \quad \text{or} \quad x - 2 = -64
$$

Solving for $ x $:

$$
x = 66 \quad \text{or} \quad x = -62
$$

---

### Step 3: Identify the Positive Solution

The problem asks for the **positive** value of $ x $, so the answer is:

$$
\boxed{66}
$$

---

### Final Verification

To ensure correctness, substitute $ x = 66 $ into the original equation:

$$
\sqrt[3]{(66 - 2)^2} = \sqrt[3]{64^2} = \sqrt[3]{4096} = 16
$$

This confirms that $ x = 66 $ is a valid solution.

---

### Final Answer

$$
\boxed{66}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: the cube root of (x squared minus 4x plus 4) equals 16. Let me write that down first to make sure I have it right.

The equation is ∛(x² - 4x + 4) = 16. Alright, cube root of something equals 16. To get rid of the cube root, I should cube both sides of the equation. That way, the cube root and the cube will cancel each other out. Let me try that.

So, if I cube both sides, I get:

(∛(x² - 4x + 4))³ = 16³

Which simplifies to:

x² - 4x + 4 = 16³

Now, I need to calculate 16 cubed. Let me do that. 16 times 16 is 256, and then 256 times 16. Hmm, let me compute that step by step. 256 times 10 is 2560, 256 times 6 is 1536. Adding those together: 2560 + 1536. Let me add them. 2560 + 1500 is 4060, and then +36 is 4096. So 16³ is 4096.

So now the equation is:

x² - 4x + 4 = 4096

Okay, so I need to solve this quadratic equation. Let me rearrange it to standard form. Subtract 4096 from both sides:

x² - 4x + 4 - 4096 = 0

Simplify the constants:

4 - 4096 is... let me calculate that. 4096 - 4 is 4092, so with the negative sign, it's -4092. So the equation becomes:

x² - 4x - 4092 = 0

Now, I need to solve this quadratic equation. Let me check if I can factor it, but given the numbers, it might not be straightforward. Let me check the discriminant first to see if it's a perfect square or not. The discriminant D is b² - 4ac. Here, a = 1, b = -4, c = -4092.

So D = (-4)² - 4(1)(-4092) = 16 + 4*4092. Let me compute that. 4*4092. Let me do 4*4000 = 16000, and 4*92 = 368. So total is 16000 + 368 = 16368. Then add 16: 16368 + 16 = 16384.

Hmm, 16384. Wait, that number seems familiar. Isn't that 128 squared? Let me check. 128 squared is (120 + 8)^2 = 120² + 2*120*8 + 8² = 14400 + 1920 + 64 = 14400 + 1920 is 16320, plus 64 is 16384. Yes! So the discriminant is 128². That's a perfect square, so the equation can be solved with integer solutions.

So, using the quadratic formula: x = [-b ± √D]/(2a). Here, b is -4, so -b is 4. So:

x = [4 ± √16384]/2

Since √16384 is 128, as we saw earlier:

x = [4 ± 128]/2

So that gives two solutions:

First solution: (4 + 128)/2 = 132/2 = 66

Second solution: (4 - 128)/2 = (-124)/2 = -62

But the problem asks for the positive value of x. So between 66 and -62, the positive one is 66.

Wait, but let me check if both solutions are valid in the original equation. Sometimes when you cube both sides or do other operations, you might introduce extraneous solutions. Let me verify both solutions in the original equation.

First, let's check x = 66.

Original equation: ∛(x² - 4x + 4) = 16

Compute x² - 4x + 4 when x = 66:

First, x² is 66². Let me calculate that. 60² is 3600, 6² is 36, and then cross term 2*60*6 = 720. So (60 + 6)^2 = 3600 + 720 + 36 = 4356.

Then subtract 4x: 4*66 = 264. So 4356 - 264 = 4092. Then add 4: 4092 + 4 = 4096.

So inside the cube root, we have 4096. Then ∛4096. Let me check what is 16³. Earlier, we had 16³ = 4096. So ∛4096 is 16. Which matches the right-hand side. So x = 66 is a valid solution.

Now check x = -62.

Compute x² - 4x + 4 when x = -62.

First, x² is (-62)^2 = 62². Let me compute that. 60² is 3600, 2² is 4, and cross term 2*60*2 = 240. So (60 + 2)^2 = 3600 + 240 + 4 = 3844.

Then subtract 4x: 4*(-62) = -248. So subtracting 4x is the same as adding 248. So 3844 + 248 = 4092. Then add 4: 4092 + 4 = 4096.

So again, inside the cube root is 4096, so ∛4096 = 16. Which is equal to the right-hand side. So x = -62 also satisfies the original equation.

But the problem asks for the positive value of x, so the answer is 66.

Wait, but let me check if there's any mistake in my calculation. Let me verify the quadratic equation again.

Original equation after cubing both sides: x² - 4x + 4 = 4096. Then moving 4096 to the left: x² - 4x + 4 - 4096 = x² - 4x - 4092 = 0. Correct. Then discriminant D = (-4)^2 - 4*1*(-4092) = 16 + 16368 = 16384. Correct. Square root of 16384 is 128. Then solutions (4 ± 128)/2. Correct. So 66 and -62. Both check out.

But maybe there's another way to approach the original equation? Let me think. The expression inside the cube root is x² - 4x + 4. Wait, that looks like a perfect square. Let me check.

x² - 4x + 4. Let me factor it. It's x² - 4x + 4. Which is (x - 2)^2. Because (x - 2)^2 = x² - 4x + 4. Yes! So the original equation can be rewritten as:

∛[(x - 2)^2] = 16

So that's another way to write it. Maybe this helps? Let me see.

So, if I have the cube root of (x - 2)^2 equals 16. Then, cubing both sides gives (x - 2)^2 = 16³ = 4096. Then, taking square roots of both sides, we get |x - 2| = √4096.

Wait, but √4096 is 64, since 64² is 4096. So |x - 2| = 64. Therefore, x - 2 = 64 or x - 2 = -64. Therefore, x = 64 + 2 = 66 or x = -64 + 2 = -62.

So same solutions as before. So that's another way to approach it, recognizing the perfect square. So that's a good check.

Therefore, the positive solution is 66.

So, I think that's solid. Both methods lead to the same answer, and both solutions check out in the original equation. Therefore, the positive value of x is 66.

**Final Answer**
The positive value of $ x $ is \boxed{66}.

To solve the equation:

$$
\sqrt[3]{x^2 - 4x + 4} = 16
$$

we begin by eliminating the cube root by cubing both sides of the equation:

$$
\left(\sqrt[3]{x^2 - 4x + 4}\right)^3 = 16^3
$$

This simplifies to:

$$
x^2 - 4x + 4 = 4096
$$

---

### Step 1: Recognize the Perfect Square

Observe that the expression $ x^2 - 4x + 4 $ is a perfect square:

$$
x^2 - 4x + 4 = (x - 2)^2
$$

So the equation becomes:

$$
\sqrt[3]{(x - 2)^2} = 16
$$

Cubing both sides again:

$$
(x - 2)^2 = 16^3 = 4096
$$

---

### Step 2: Solve the Equation

Take the square root of both sides:

$$
|x - 2| = \sqrt{4096} = 64
$$

This gives two possible solutions:

$$
x - 2 = 64 \quad \text{or} \quad x - 2 = -64
$$

Solving for $ x $:

$$
x = 66 \quad \text{or} \quad x = -62
$$

---

### Step 3: Identify the Positive Solution

The problem asks for the **positive** value of $ x $, so the answer is:

$$
\boxed{66}
$$

---

### Final Verification

To ensure correctness, substitute $ x = 66 $ into the original equation:

$$
\sqrt[3]{(66 - 2)^2} = \sqrt[3]{64^2} = \sqrt[3]{4096} = 16
$$

This confirms that $ x = 66 $ is a valid solution.

---

### Final Answer

$$
\boxed{66}
$$

Solution:
We first cube each side of the equation to get $x^2 - 4x + 4 = 16^3$.  Notice that $x^2 - 4x + 4 = (x-2)^2.$

Therefore, we have that $x-2 = \pm 16^{3/2} = \pm 64$.  Therefore, the possible values of $x$ are $-62$ and $66,$ and the only positive value is therefore $\boxed{66}$.