Explanation:
Okay, so I need to find the units digit of 4$, which is defined as a superfactorial. The problem says that n$ is equal to a tower of n! exponents, with n! levels. So, for n=4, that would be 4! raised to the power of 4! raised to the power of ... and so on, with a total of 4! levels. Let me write that down to make sure I understand.

First, let me compute 4! because that's going to be the number of exponents in the tower. 4! is 4×3×2×1 = 24. So, 4$ is a tower of 24 copies of 4! (which is 24). So, the expression is 24^24^24^...^24 with 24 levels. That's a massive number! Obviously, I can't compute it directly, so I need to find a way to determine just the units digit.

The units digit of a number is the same as the number modulo 10. So, I need to compute this tower modulo 10. However, since the tower is so tall, I need to find some patterns or properties that can help me simplify this.

First, let me recall that for exponents, Euler's theorem might be useful. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since we're working modulo 10, φ(10) = φ(2×5) = φ(2)×φ(5) = 1×4 = 4. So, φ(10) = 4. However, Euler's theorem requires that a and n are coprime. So, if the base of the exponent is coprime with 10, then we can use this theorem. But if not, we need to handle it differently.

But first, let me note that the tower is 24^24^24^...^24 with 24 levels. Let me denote this tower as T. So, T = 24^T', where T' is the tower of 24s with 23 levels. But since we need to compute T mod 10, maybe we can find a pattern or use properties of exponents modulo 10.

But before that, let me check if 24 and 10 are coprime. 24 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem doesn't apply directly here. However, maybe I can break down the modulus into prime power components using the Chinese Remainder Theorem. Since 10 = 2 × 5, and 2 and 5 are coprime, if I can find T mod 2 and T mod 5, then I can combine them to get T mod 10.

Let me try that approach. Let's compute T mod 2 and T mod 5 separately.

First, T mod 2:

Since 24 is even, any power of 24 will also be even. Therefore, 24^anything is congruent to 0 mod 2. Therefore, T mod 2 = 0.

Now, T mod 5:

This is more complicated. Let me think. Since 24 mod 5 is 4, because 24 divided by 5 is 4 with a remainder of 4. So, T mod 5 is equivalent to 4^T' mod 5, where T' is the tower of 24s with 23 levels. So, we need to compute 4^T' mod 5.

But again, 4 and 5 are coprime, so Euler's theorem applies here. φ(5) = 4, so 4^4 ≡ 1 mod 5. Therefore, 4^k mod 5 cycles every 4. Therefore, if we can find T' mod φ(5) = 4, then we can compute 4^(T' mod 4) mod 5. However, if T' mod 4 is 0, then it would be 4^4 ≡ 1 mod 5, so we need to adjust accordingly.

But T' is the tower of 24s with 23 levels. Let me denote T' as 24^T'' where T'' is the tower of 24s with 22 levels. So, T' = 24^T''. Therefore, to compute T' mod 4, we need to compute 24^T'' mod 4.

But 24 mod 4 is 0, since 24 is divisible by 4. Therefore, 24^anything is 0 mod 4, as long as the exponent is at least 1. Since T'' is a tower of 24s with 22 levels, which is certainly at least 1, so T' mod 4 = 0.

Therefore, going back, T' mod 4 = 0, so 4^T' mod 5 = 4^(0) mod 5? Wait, no. Wait, Euler's theorem says that 4^φ(5) = 4^4 ≡ 1 mod 5. Therefore, 4^k ≡ 4^(k mod 4) mod 5, but only if we adjust for the exponent. Wait, actually, more precisely, if we have 4^k mod 5, since φ(5)=4, then 4^k ≡ 4^(k mod 4) * 4^(4 * m) ≡ 4^(k mod 4) * (1)^m ≡ 4^(k mod 4) mod 5. However, if k mod 4 is 0, then it's 4^0 = 1 mod 5. Wait, but 4^4 ≡ 1 mod 5, so 4^(4m) ≡ 1^m = 1 mod 5. Therefore, if k is a multiple of 4, then 4^k ≡ 1 mod 5. Therefore, since T' mod 4 = 0, then 4^T' ≡ 1 mod 5. Therefore, T mod 5 = 1.

Therefore, combining the two results:

T ≡ 0 mod 2

T ≡ 1 mod 5

We need to find a number x such that x ≡ 0 mod 2 and x ≡ 1 mod 5. Let's solve this system.

Numbers congruent to 1 mod 5 are 1, 6, 11, 16, 21, 26,... Now, which of these are even? 6, 16, 26,... So, the solutions are numbers congruent to 6 mod 10. Therefore, x ≡ 6 mod 10. Therefore, the units digit is 6.

Wait, but hold on. Let me check my steps again because this seems straightforward, but maybe I made a mistake in the exponents.

First, let me recap:

We have T = 24^T' where T' is a tower of 24s with 23 levels.

We need T mod 10. We split into mod 2 and mod 5.

For mod 2: 24 is 0 mod 2, so any power is 0 mod 2. So T ≡ 0 mod 2.

For mod 5: T ≡ 4^T' mod 5. Then, since φ(5)=4, we need T' mod 4. T' is 24^T'' mod 4. Since 24 ≡ 0 mod 4, so 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' ≡ 4^0 ≡ 1 mod 5? Wait, but 4^T' mod 5. If T' ≡ 0 mod 4, then 4^T' ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. So yes, that's correct. Therefore, T ≡ 1 mod 5.

Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, the units digit is 6. So, the answer is 6?

Wait, but let me check with smaller numbers to see if my approach is correct. Let me take a smaller case where I can compute manually.

For example, let's take n=1. Then 1$ is a tower of 1! = 1 level of 1! = 1. So, 1$ = 1. Units digit is 1. Let's check with our method.

Compute 1$ mod 10. 1! = 1. So, tower of 1 level: 1. So, 1 mod 10 is 1. Correct.

Another example: n=2. 2$ is a tower of 2! = 2 levels of 2! = 2. So, 2^2 = 4. Units digit 4. Let's check with our method.

Compute T mod 2 and mod 5.

T = 2^2 = 4. T mod 2 = 0, T mod 5 = 4. Then solving x ≡ 0 mod 2 and x ≡ 4 mod 5. The solution is 4 mod 10. Correct.

Another example: n=3. 3$ is a tower of 3! = 6 levels of 3! = 6. So, 6^6^6^6^6^6. Let's compute its units digit.

But maybe I can compute it step by step. However, since it's a tower of 6s, let's try to compute mod 10.

But let's try our method. First, compute T mod 2 and mod 5.

First, T mod 2: 6 is even, so any power is 0 mod 2. So T ≡ 0 mod 2.

T mod 5: 6 mod 5 = 1. So T = 1^T' mod 5. Since 1^anything is 1 mod 5. Therefore, T ≡ 1 mod 5. Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, units digit 6. Let me check with actual computation.

Compute the tower 6^6^6^6^6^6. Let's compute the units digit. Since the units digit of 6^anything is always 6. Because 6^1=6, 6^2=36, 6^3=216, etc. So regardless of the exponent, the units digit is 6. Therefore, the units digit is 6. Which matches our result. So our method works here.

Another example: n=4. Wait, that's the original problem. But let me check with n=4. Wait, but according to our previous calculation, the units digit is 6. But let me see if that's correct.

But let me think again. Wait, for n=4, the tower is 24^24^...^24 with 24 levels. But when we calculated mod 5, we found that T ≡ 1 mod 5 and T ≡ 0 mod 2, leading to 6 mod 10. However, let me check if there's a mistake in the step where we calculated T' mod 4.

Wait, T' is the tower of 24s with 23 levels. So, T' = 24^T''. T'' is the tower with 22 levels. So, to compute T' mod 4, since 24 ≡ 0 mod 4, then 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' mod 5 is 4^0 mod 5? Wait, no. Wait, 4^T' mod 5. Since φ(5)=4, and T' ≡ 0 mod 4, then 4^T' ≡ 4^(4k) ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. Therefore, yes, that's correct. Therefore, T ≡ 1 mod 5. Therefore, combining with T ≡ 0 mod 2 gives 6 mod 10. So, the units digit is 6. But let me check with another approach.

Alternatively, maybe I can compute the tower modulo 10 directly, but considering that exponents can be simplified.

But given that the tower is 24^24^...^24 with 24 levels. Let me think about the exponents. Since we need to compute this modulo 10, and since 10 = 2 * 5, and we already did the Chinese Remainder Theorem approach, which gave 6. But let me check if there's another way.

Alternatively, since 24 is congruent to 4 mod 5 and 0 mod 2. But maybe I can think about the tower modulo 10. However, since the tower is so tall, maybe we can find a pattern in the exponents.

But let me think again. Let me try to compute the tower modulo 10. However, since the tower is 24^T', and T' is a tower of 24s with 23 levels. Let me note that 24 mod 10 is 4, so the entire tower is equivalent to 4^T' mod 10. Wait, but earlier I split into mod 2 and mod 5. However, if I try to compute 4^T' mod 10 directly, maybe I can find a pattern.

But 4^k mod 10 cycles every 2: 4^1=4, 4^2=16→6, 4^3=64→4, 4^4=256→6, etc. So, the cycle is 4,6,4,6,... So, if the exponent is odd, it's 4 mod 10; if even, 6 mod 10. Therefore, if I can determine whether T' is odd or even, then I can find 4^T' mod 10.

But T' is the tower of 24s with 23 levels. Let me check if T' is even or odd. Since 24 is even, and any power of an even number is even. Therefore, T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Which matches our previous result. So, this seems to confirm that the answer is 6.

Wait, so this approach is simpler. Let me explain again:

Since the entire tower T is 24^T', and 24 mod 10 is 4, so T ≡ 4^T' mod 10. Then, since 4^k mod 10 cycles every 2, with period 2: 4,6,4,6,... So, if T' is even, then 4^T' ≡ 6 mod 10; if T' is odd, then 4^T' ≡ 4 mod 10. Therefore, we need to check if T' is even or odd.

But T' is a tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even. Therefore, regardless of the number of levels (as long as it's at least 1), the tower T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Therefore, the answer is 6. So, this seems to confirm the previous result.

But why did we go through the Chinese Remainder Theorem earlier? Because when I first thought about mod 5, I considered that 4^T' mod 5 is 1, and mod 2 is 0, leading to 6 mod 10. But here, by directly looking at mod 10, we can see that since T' is even, the result is 6. So, both methods agree.

Therefore, the units digit is 6. However, let me check once more with another example. Let's take n=2 again. The tower is 2^2 = 4. The units digit is 4. According to the second method: T = 2^T', where T' is the tower with 1 level (since n! = 2, so 2 levels? Wait, n=2, n! = 2, so the tower has 2 levels. So, T = 2^2 = 4. Then, T' is the tower with 1 level, which is 2. Therefore, T' = 2. Then, 4^T' mod 10 is 4^2 = 16 mod 10 = 6? Wait, but the actual value is 4. Wait, this contradicts. Wait, so there's a problem here.

Wait, hold on. Wait, for n=2, the tower is 2!^2! = 2^2 = 4. So, T is 4. But according to the second method, I said that T = 4^T', where T' is the tower with n! -1 levels. Wait, but in this case, n=2, n! = 2, so T is 2^T', where T' is the tower with 1 level, which is 2. Therefore, T = 2^2 = 4. However, when I tried to compute 4^T' mod 10, I was wrong. Wait, why?

Because in the case of n=4, the tower is 24^T', where T' is the tower with 23 levels. However, in the case of n=2, the tower is 2^T', where T' is the tower with 1 level (which is 2). Therefore, T = 2^T', not 4^T'. Wait, so my mistake was in the second approach when I said that T ≡ 4^T' mod 10. That was incorrect. Because in reality, T is 24^T', but 24 mod 10 is 4, so T ≡ 4^T' mod 10. However, in the case of n=2, the tower is 2^T', where T' is 2, so T = 2^2 = 4. However, 2 mod 10 is 2, not 4. Therefore, my mistake was in the second approach when I thought that T ≡ 4^T' mod 10. That was only true for n=4, where the base is 24, which is 4 mod 10. For n=2, the base is 2, which is 2 mod 10. Therefore, my second approach was specific to n=4. Therefore, in the case of n=4, since the base is 24 ≡ 4 mod 10, then T ≡ 4^T' mod 10. However, in the case of n=2, the base is 2, so T ≡ 2^T' mod 10. Therefore, my second approach was correct for n=4, but I confused it with n=2. Therefore, in the case of n=4, since the base is 4 mod 10, and T' is even, then 4^T' mod 10 is 6. Therefore, the answer is 6. However, in the case of n=2, since the base is 2 mod 10, and T' is 2, which is even, then 2^T' mod 10 is 2^2 = 4 mod 10, which is correct. Therefore, my second approach was correct for n=4. Therefore, the answer is 6.

But let me check another example where the base is 4 mod 10. Let's take n=3. Wait, n=3, n! = 6. So, the tower is 6^6^...^6 with 6 levels. Let me compute its units digit. As before, since 6^anything has units digit 6. Therefore, the units digit is 6. Let me check with the second approach. The tower is 6^T', where T' is the tower with 5 levels. Since 6 mod 10 is 6, and 6^k mod 10 is always 6. Therefore, regardless of T', the units digit is 6. Therefore, that works. So, if the base is 6, the units digit is always 6. Similarly, if the base is 4, then depending on the exponent's parity, it's 4 or 6. Therefore, in our case, for n=4, since the base is 4 mod 10, and the exponent T' is even, then the units digit is 6. Therefore, the answer is 6.

Therefore, after checking with multiple examples and approaches, I think the answer is 6. However, let me just make sure that T' is indeed even. T' is the tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even, so regardless of the number of levels (as long as it's at least 1), the tower is even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

But let me think again about the tower. For example, if we have a tower of 24s with 23 levels, which is 24^24^...^24. Since the first exponent is 24, which is even, so the entire tower is 24 raised to an even number. Therefore, 24^even. But 24 is even, so 24^even is divisible by 2^even, which is certainly even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

Therefore, I think the answer is 6. However, let me check if there's a possibility that T' is 0 mod φ(10) or something else. Wait, but in the first approach, we used Chinese Remainder Theorem and found that T ≡ 6 mod 10. In the second approach, we found the same. Therefore, both methods confirm that the units digit is 6.

But just to be thorough, let me check with another angle. Let me try to compute the tower modulo 10, considering that exponents can be simplified.

We have T = 24^T', where T' is the tower with 23 levels. Since 24 ≡ 4 mod 10, T ≡ 4^T' mod 10. As mentioned, 4^k mod 10 cycles every 2: 4, 6, 4, 6,... So, if T' is even, it's 6; if odd, 4. Since T' is even, as established, then T ≡ 6 mod 10. Therefore, units digit is 6.

Alternatively, if I wanted to compute T' mod 2, since the exponent in 4^T' mod 10 depends on T' mod 2. Wait, but T' is even, so T' mod 2 = 0. Therefore, 4^T' mod 10 = 4^0 mod 10? No, that's not correct. Wait, no. The cycle is every 2. So, if T' is even, then T' = 2k, so 4^{2k} = (4^2)^k = 16^k. 16 mod 10 is 6, so 16^k mod 10 is 6^k mod 10. But 6^k mod 10 is always 6. Therefore, 4^{even} mod 10 is 6. Therefore, regardless of the exponent, as long as it's even, it's 6. Therefore, that's another way to see it. Therefore, since T' is even, the result is 6. Therefore, the units digit is 6.

Therefore, after multiple approaches and checks, I feel confident that the units digit of 4$ is 6.

**Final Answer**
\boxed{6}

To determine the **units digit** of the superfactorial $ 4\$ $, we begin by understanding the definition:

$$
n\$ = \underbrace{n!^{{n!}^{{\cdot}^{{\cdot}^{{\cdot}^{n!}}}}}}_{n!}
$$

This means that $ 4\$ $ is a **tower of exponents** with $ 4! = 24 $ levels, each level being $ 24 $. So:

$$
4\$ = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(with 24 levels of 24)}
$$

We are asked to find the **units digit** of this number, which is equivalent to computing:

$$
4\$ \mod 10
$$

---

### Step 1: Reduce the base modulo 10

Since the base of the tower is 24, and we are interested in the **units digit**, we reduce 24 modulo 10:

$$
24 \equiv 4 \mod 10
$$

So, the entire tower becomes:

$$
4^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \mod 10
$$

Let’s denote the tower of 24s with 23 levels as $ T' $. Then:

$$
4\$ = 4^{T'} \mod 10
$$

---

### Step 2: Analyze the exponent $ T' \mod 2 $

We now need to determine the **parity** of the exponent $ T' $, because the powers of 4 modulo 10 cycle every 2:

- $ 4^1 = 4 $
- $ 4^2 = 16 \equiv 6 \mod 10 $
- $ 4^3 = 64 \equiv 4 \mod 10 $
- $ 4^4 = 256 \equiv 6 \mod 10 $

So, the pattern is: **4, 6, 4, 6, ...**

Thus, if the exponent is **even**, the result is **6**; if **odd**, it is **4**.

So, we need to determine whether $ T' $ is even or odd.

---

### Step 3: Determine the parity of $ T' $

Recall that $ T' $ is a tower of 24s with 23 levels:

$$
T' = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(23 levels)}
$$

Since 24 is even, and any power of an even number is even, the entire tower $ T' $ is **even**. Therefore:

$$
T' \equiv 0 \mod 2
$$

---

### Step 4: Compute $ 4^{T'} \mod 10 $

Since $ T' $ is even, we use the pattern from earlier:

$$
4^{\text{even}} \equiv 6 \mod 10
$$

Thus:

$$
4\$ \equiv 6 \mod 10
$$

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the units digit of 4$, which is defined as a superfactorial. The problem says that n$ is equal to a tower of n! exponents, with n! levels. So, for n=4, that would be 4! raised to the power of 4! raised to the power of ... and so on, with a total of 4! levels. Let me write that down to make sure I understand.

First, let me compute 4! because that's going to be the number of exponents in the tower. 4! is 4×3×2×1 = 24. So, 4$ is a tower of 24 copies of 4! (which is 24). So, the expression is 24^24^24^...^24 with 24 levels. That's a massive number! Obviously, I can't compute it directly, so I need to find a way to determine just the units digit.

The units digit of a number is the same as the number modulo 10. So, I need to compute this tower modulo 10. However, since the tower is so tall, I need to find some patterns or properties that can help me simplify this.

First, let me recall that for exponents, Euler's theorem might be useful. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since we're working modulo 10, φ(10) = φ(2×5) = φ(2)×φ(5) = 1×4 = 4. So, φ(10) = 4. However, Euler's theorem requires that a and n are coprime. So, if the base of the exponent is coprime with 10, then we can use this theorem. But if not, we need to handle it differently.

But first, let me note that the tower is 24^24^24^...^24 with 24 levels. Let me denote this tower as T. So, T = 24^T', where T' is the tower of 24s with 23 levels. But since we need to compute T mod 10, maybe we can find a pattern or use properties of exponents modulo 10.

But before that, let me check if 24 and 10 are coprime. 24 and 10 share a common factor of 2, so they are not coprime. Therefore, Euler's theorem doesn't apply directly here. However, maybe I can break down the modulus into prime power components using the Chinese Remainder Theorem. Since 10 = 2 × 5, and 2 and 5 are coprime, if I can find T mod 2 and T mod 5, then I can combine them to get T mod 10.

Let me try that approach. Let's compute T mod 2 and T mod 5 separately.

First, T mod 2:

Since 24 is even, any power of 24 will also be even. Therefore, 24^anything is congruent to 0 mod 2. Therefore, T mod 2 = 0.

Now, T mod 5:

This is more complicated. Let me think. Since 24 mod 5 is 4, because 24 divided by 5 is 4 with a remainder of 4. So, T mod 5 is equivalent to 4^T' mod 5, where T' is the tower of 24s with 23 levels. So, we need to compute 4^T' mod 5.

But again, 4 and 5 are coprime, so Euler's theorem applies here. φ(5) = 4, so 4^4 ≡ 1 mod 5. Therefore, 4^k mod 5 cycles every 4. Therefore, if we can find T' mod φ(5) = 4, then we can compute 4^(T' mod 4) mod 5. However, if T' mod 4 is 0, then it would be 4^4 ≡ 1 mod 5, so we need to adjust accordingly.

But T' is the tower of 24s with 23 levels. Let me denote T' as 24^T'' where T'' is the tower of 24s with 22 levels. So, T' = 24^T''. Therefore, to compute T' mod 4, we need to compute 24^T'' mod 4.

But 24 mod 4 is 0, since 24 is divisible by 4. Therefore, 24^anything is 0 mod 4, as long as the exponent is at least 1. Since T'' is a tower of 24s with 22 levels, which is certainly at least 1, so T' mod 4 = 0.

Therefore, going back, T' mod 4 = 0, so 4^T' mod 5 = 4^(0) mod 5? Wait, no. Wait, Euler's theorem says that 4^φ(5) = 4^4 ≡ 1 mod 5. Therefore, 4^k ≡ 4^(k mod 4) mod 5, but only if we adjust for the exponent. Wait, actually, more precisely, if we have 4^k mod 5, since φ(5)=4, then 4^k ≡ 4^(k mod 4) * 4^(4 * m) ≡ 4^(k mod 4) * (1)^m ≡ 4^(k mod 4) mod 5. However, if k mod 4 is 0, then it's 4^0 = 1 mod 5. Wait, but 4^4 ≡ 1 mod 5, so 4^(4m) ≡ 1^m = 1 mod 5. Therefore, if k is a multiple of 4, then 4^k ≡ 1 mod 5. Therefore, since T' mod 4 = 0, then 4^T' ≡ 1 mod 5. Therefore, T mod 5 = 1.

Therefore, combining the two results:

T ≡ 0 mod 2

T ≡ 1 mod 5

We need to find a number x such that x ≡ 0 mod 2 and x ≡ 1 mod 5. Let's solve this system.

Numbers congruent to 1 mod 5 are 1, 6, 11, 16, 21, 26,... Now, which of these are even? 6, 16, 26,... So, the solutions are numbers congruent to 6 mod 10. Therefore, x ≡ 6 mod 10. Therefore, the units digit is 6.

Wait, but hold on. Let me check my steps again because this seems straightforward, but maybe I made a mistake in the exponents.

First, let me recap:

We have T = 24^T' where T' is a tower of 24s with 23 levels.

We need T mod 10. We split into mod 2 and mod 5.

For mod 2: 24 is 0 mod 2, so any power is 0 mod 2. So T ≡ 0 mod 2.

For mod 5: T ≡ 4^T' mod 5. Then, since φ(5)=4, we need T' mod 4. T' is 24^T'' mod 4. Since 24 ≡ 0 mod 4, so 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' ≡ 4^0 ≡ 1 mod 5? Wait, but 4^T' mod 5. If T' ≡ 0 mod 4, then 4^T' ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. So yes, that's correct. Therefore, T ≡ 1 mod 5.

Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, the units digit is 6. So, the answer is 6?

Wait, but let me check with smaller numbers to see if my approach is correct. Let me take a smaller case where I can compute manually.

For example, let's take n=1. Then 1$ is a tower of 1! = 1 level of 1! = 1. So, 1$ = 1. Units digit is 1. Let's check with our method.

Compute 1$ mod 10. 1! = 1. So, tower of 1 level: 1. So, 1 mod 10 is 1. Correct.

Another example: n=2. 2$ is a tower of 2! = 2 levels of 2! = 2. So, 2^2 = 4. Units digit 4. Let's check with our method.

Compute T mod 2 and mod 5.

T = 2^2 = 4. T mod 2 = 0, T mod 5 = 4. Then solving x ≡ 0 mod 2 and x ≡ 4 mod 5. The solution is 4 mod 10. Correct.

Another example: n=3. 3$ is a tower of 3! = 6 levels of 3! = 6. So, 6^6^6^6^6^6. Let's compute its units digit.

But maybe I can compute it step by step. However, since it's a tower of 6s, let's try to compute mod 10.

But let's try our method. First, compute T mod 2 and mod 5.

First, T mod 2: 6 is even, so any power is 0 mod 2. So T ≡ 0 mod 2.

T mod 5: 6 mod 5 = 1. So T = 1^T' mod 5. Since 1^anything is 1 mod 5. Therefore, T ≡ 1 mod 5. Therefore, solving x ≡ 0 mod 2 and x ≡ 1 mod 5 gives x ≡ 6 mod 10. Therefore, units digit 6. Let me check with actual computation.

Compute the tower 6^6^6^6^6^6. Let's compute the units digit. Since the units digit of 6^anything is always 6. Because 6^1=6, 6^2=36, 6^3=216, etc. So regardless of the exponent, the units digit is 6. Therefore, the units digit is 6. Which matches our result. So our method works here.

Another example: n=4. Wait, that's the original problem. But let me check with n=4. Wait, but according to our previous calculation, the units digit is 6. But let me see if that's correct.

But let me think again. Wait, for n=4, the tower is 24^24^...^24 with 24 levels. But when we calculated mod 5, we found that T ≡ 1 mod 5 and T ≡ 0 mod 2, leading to 6 mod 10. However, let me check if there's a mistake in the step where we calculated T' mod 4.

Wait, T' is the tower of 24s with 23 levels. So, T' = 24^T''. T'' is the tower with 22 levels. So, to compute T' mod 4, since 24 ≡ 0 mod 4, then 24^anything ≥1 is 0 mod 4. Therefore, T' ≡ 0 mod 4. Therefore, 4^T' mod 5 is 4^0 mod 5? Wait, no. Wait, 4^T' mod 5. Since φ(5)=4, and T' ≡ 0 mod 4, then 4^T' ≡ 4^(4k) ≡ (4^4)^k ≡ 1^k ≡ 1 mod 5. Therefore, yes, that's correct. Therefore, T ≡ 1 mod 5. Therefore, combining with T ≡ 0 mod 2 gives 6 mod 10. So, the units digit is 6. But let me check with another approach.

Alternatively, maybe I can compute the tower modulo 10 directly, but considering that exponents can be simplified.

But given that the tower is 24^24^...^24 with 24 levels. Let me think about the exponents. Since we need to compute this modulo 10, and since 10 = 2 * 5, and we already did the Chinese Remainder Theorem approach, which gave 6. But let me check if there's another way.

Alternatively, since 24 is congruent to 4 mod 5 and 0 mod 2. But maybe I can think about the tower modulo 10. However, since the tower is so tall, maybe we can find a pattern in the exponents.

But let me think again. Let me try to compute the tower modulo 10. However, since the tower is 24^T', and T' is a tower of 24s with 23 levels. Let me note that 24 mod 10 is 4, so the entire tower is equivalent to 4^T' mod 10. Wait, but earlier I split into mod 2 and mod 5. However, if I try to compute 4^T' mod 10 directly, maybe I can find a pattern.

But 4^k mod 10 cycles every 2: 4^1=4, 4^2=16→6, 4^3=64→4, 4^4=256→6, etc. So, the cycle is 4,6,4,6,... So, if the exponent is odd, it's 4 mod 10; if even, 6 mod 10. Therefore, if I can determine whether T' is odd or even, then I can find 4^T' mod 10.

But T' is the tower of 24s with 23 levels. Let me check if T' is even or odd. Since 24 is even, and any power of an even number is even. Therefore, T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Which matches our previous result. So, this seems to confirm that the answer is 6.

Wait, so this approach is simpler. Let me explain again:

Since the entire tower T is 24^T', and 24 mod 10 is 4, so T ≡ 4^T' mod 10. Then, since 4^k mod 10 cycles every 2, with period 2: 4,6,4,6,... So, if T' is even, then 4^T' ≡ 6 mod 10; if T' is odd, then 4^T' ≡ 4 mod 10. Therefore, we need to check if T' is even or odd.

But T' is a tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even. Therefore, regardless of the number of levels (as long as it's at least 1), the tower T' is even. Therefore, T' is even, so 4^T' mod 10 is 6. Therefore, the units digit is 6. Therefore, the answer is 6. So, this seems to confirm the previous result.

But why did we go through the Chinese Remainder Theorem earlier? Because when I first thought about mod 5, I considered that 4^T' mod 5 is 1, and mod 2 is 0, leading to 6 mod 10. But here, by directly looking at mod 10, we can see that since T' is even, the result is 6. So, both methods agree.

Therefore, the units digit is 6. However, let me check once more with another example. Let's take n=2 again. The tower is 2^2 = 4. The units digit is 4. According to the second method: T = 2^T', where T' is the tower with 1 level (since n! = 2, so 2 levels? Wait, n=2, n! = 2, so the tower has 2 levels. So, T = 2^2 = 4. Then, T' is the tower with 1 level, which is 2. Therefore, T' = 2. Then, 4^T' mod 10 is 4^2 = 16 mod 10 = 6? Wait, but the actual value is 4. Wait, this contradicts. Wait, so there's a problem here.

Wait, hold on. Wait, for n=2, the tower is 2!^2! = 2^2 = 4. So, T is 4. But according to the second method, I said that T = 4^T', where T' is the tower with n! -1 levels. Wait, but in this case, n=2, n! = 2, so T is 2^T', where T' is the tower with 1 level, which is 2. Therefore, T = 2^2 = 4. However, when I tried to compute 4^T' mod 10, I was wrong. Wait, why?

Because in the case of n=4, the tower is 24^T', where T' is the tower with 23 levels. However, in the case of n=2, the tower is 2^T', where T' is the tower with 1 level (which is 2). Therefore, T = 2^T', not 4^T'. Wait, so my mistake was in the second approach when I said that T ≡ 4^T' mod 10. That was incorrect. Because in reality, T is 24^T', but 24 mod 10 is 4, so T ≡ 4^T' mod 10. However, in the case of n=2, the tower is 2^T', where T' is 2, so T = 2^2 = 4. However, 2 mod 10 is 2, not 4. Therefore, my mistake was in the second approach when I thought that T ≡ 4^T' mod 10. That was only true for n=4, where the base is 24, which is 4 mod 10. For n=2, the base is 2, which is 2 mod 10. Therefore, my second approach was specific to n=4. Therefore, in the case of n=4, since the base is 24 ≡ 4 mod 10, then T ≡ 4^T' mod 10. However, in the case of n=2, the base is 2, so T ≡ 2^T' mod 10. Therefore, my second approach was correct for n=4, but I confused it with n=2. Therefore, in the case of n=4, since the base is 4 mod 10, and T' is even, then 4^T' mod 10 is 6. Therefore, the answer is 6. However, in the case of n=2, since the base is 2 mod 10, and T' is 2, which is even, then 2^T' mod 10 is 2^2 = 4 mod 10, which is correct. Therefore, my second approach was correct for n=4. Therefore, the answer is 6.

But let me check another example where the base is 4 mod 10. Let's take n=3. Wait, n=3, n! = 6. So, the tower is 6^6^...^6 with 6 levels. Let me compute its units digit. As before, since 6^anything has units digit 6. Therefore, the units digit is 6. Let me check with the second approach. The tower is 6^T', where T' is the tower with 5 levels. Since 6 mod 10 is 6, and 6^k mod 10 is always 6. Therefore, regardless of T', the units digit is 6. Therefore, that works. So, if the base is 6, the units digit is always 6. Similarly, if the base is 4, then depending on the exponent's parity, it's 4 or 6. Therefore, in our case, for n=4, since the base is 4 mod 10, and the exponent T' is even, then the units digit is 6. Therefore, the answer is 6.

Therefore, after checking with multiple examples and approaches, I think the answer is 6. However, let me just make sure that T' is indeed even. T' is the tower of 24s with 23 levels. Since 24 is even, and any power of an even number is even, so regardless of the number of levels (as long as it's at least 1), the tower is even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

But let me think again about the tower. For example, if we have a tower of 24s with 23 levels, which is 24^24^...^24. Since the first exponent is 24, which is even, so the entire tower is 24 raised to an even number. Therefore, 24^even. But 24 is even, so 24^even is divisible by 2^even, which is certainly even. Therefore, T' is even. Therefore, 4^T' mod 10 is 6. Therefore, the units digit is 6.

Therefore, I think the answer is 6. However, let me check if there's a possibility that T' is 0 mod φ(10) or something else. Wait, but in the first approach, we used Chinese Remainder Theorem and found that T ≡ 6 mod 10. In the second approach, we found the same. Therefore, both methods confirm that the units digit is 6.

But just to be thorough, let me check with another angle. Let me try to compute the tower modulo 10, considering that exponents can be simplified.

We have T = 24^T', where T' is the tower with 23 levels. Since 24 ≡ 4 mod 10, T ≡ 4^T' mod 10. As mentioned, 4^k mod 10 cycles every 2: 4, 6, 4, 6,... So, if T' is even, it's 6; if odd, 4. Since T' is even, as established, then T ≡ 6 mod 10. Therefore, units digit is 6.

Alternatively, if I wanted to compute T' mod 2, since the exponent in 4^T' mod 10 depends on T' mod 2. Wait, but T' is even, so T' mod 2 = 0. Therefore, 4^T' mod 10 = 4^0 mod 10? No, that's not correct. Wait, no. The cycle is every 2. So, if T' is even, then T' = 2k, so 4^{2k} = (4^2)^k = 16^k. 16 mod 10 is 6, so 16^k mod 10 is 6^k mod 10. But 6^k mod 10 is always 6. Therefore, 4^{even} mod 10 is 6. Therefore, regardless of the exponent, as long as it's even, it's 6. Therefore, that's another way to see it. Therefore, since T' is even, the result is 6. Therefore, the units digit is 6.

Therefore, after multiple approaches and checks, I feel confident that the units digit of 4$ is 6.

**Final Answer**
\boxed{6}

To determine the **units digit** of the superfactorial $ 4\$ $, we begin by understanding the definition:

$$
n\$ = \underbrace{n!^{{n!}^{{\cdot}^{{\cdot}^{{\cdot}^{n!}}}}}}_{n!}
$$

This means that $ 4\$ $ is a **tower of exponents** with $ 4! = 24 $ levels, each level being $ 24 $. So:

$$
4\$ = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(with 24 levels of 24)}
$$

We are asked to find the **units digit** of this number, which is equivalent to computing:

$$
4\$ \mod 10
$$

---

### Step 1: Reduce the base modulo 10

Since the base of the tower is 24, and we are interested in the **units digit**, we reduce 24 modulo 10:

$$
24 \equiv 4 \mod 10
$$

So, the entire tower becomes:

$$
4^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \mod 10
$$

Let’s denote the tower of 24s with 23 levels as $ T' $. Then:

$$
4\$ = 4^{T'} \mod 10
$$

---

### Step 2: Analyze the exponent $ T' \mod 2 $

We now need to determine the **parity** of the exponent $ T' $, because the powers of 4 modulo 10 cycle every 2:

- $ 4^1 = 4 $
- $ 4^2 = 16 \equiv 6 \mod 10 $
- $ 4^3 = 64 \equiv 4 \mod 10 $
- $ 4^4 = 256 \equiv 6 \mod 10 $

So, the pattern is: **4, 6, 4, 6, ...**

Thus, if the exponent is **even**, the result is **6**; if **odd**, it is **4**.

So, we need to determine whether $ T' $ is even or odd.

---

### Step 3: Determine the parity of $ T' $

Recall that $ T' $ is a tower of 24s with 23 levels:

$$
T' = 24^{24^{24^{\cdot^{\cdot^{\cdot^{24}}}}}} \quad \text{(23 levels)}
$$

Since 24 is even, and any power of an even number is even, the entire tower $ T' $ is **even**. Therefore:

$$
T' \equiv 0 \mod 2
$$

---

### Step 4: Compute $ 4^{T'} \mod 10 $

Since $ T' $ is even, we use the pattern from earlier:

$$
4^{\text{even}} \equiv 6 \mod 10
$$

Thus:

$$
4\$ \equiv 6 \mod 10
$$

---

### Final Answer

$$
\boxed{6}
$$

Solution: Since $4! = 4\times 3 \times 2 \times 1 = 24$, we need to evaluate the units digit of $4\$ = \underbrace{24^{24^{\cdot^{\cdot^\cdot}}}}_{24}$. The exponent of the base $24$ is an even number, say $2n$. Then, $4\$ = 24^{2n} = 576^n$. The units digit of the product of two numbers with units digit $6$ is equal to $6$. Thus, the desired answer is $\boxed{6}$.


Explanation:
Okay, so I need to figure out how many two-digit prime numbers have digits that add up to 8. Let me start by recalling what two-digit primes are. Two-digit numbers range from 10 to 99, right? And primes are numbers greater than 1 that have no divisors other than 1 and themselves.

First, maybe I should list all two-digit primes and then check which ones have digits summing to 8. But that might take a while. Maybe there's a smarter way. Let me think.

Alternatively, I can think about all two-digit numbers where the sum of the digits is 8 and then check which of those are prime. That might be more efficient. Let me try that approach.

So, a two-digit number can be represented as 10*a + b, where a is the tens digit and b is the units digit. Both a and b are digits, so a is from 1 to 9 (since it's a two-digit number) and b is from 0 to 9. The sum of the digits is a + b = 8. So, I need to find all pairs (a, b) where a is between 1 and 9, b is between 0 and 9, and a + b = 8. Then, for each such number, check if it's prime.

Let me list all possible pairs (a, b):

Starting with a = 1: then b = 8 - 1 = 7. So the number is 17.

a = 2: b = 6. Number is 26.

a = 3: b = 5. Number is 35.

a = 4: b = 4. Number is 44.

a = 5: b = 3. Number is 53.

a = 6: b = 2. Number is 62.

a = 7: b = 1. Number is 71.

a = 8: b = 0. Number is 80.

a = 9: b = -1. Wait, that's not possible because b can't be negative. So a can't be 9 here. So the possible numbers are from a=1 to a=8.

So the numbers are: 17, 26, 35, 44, 53, 62, 71, 80.

Now, I need to check which of these numbers are prime.

Let me recall that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So let's check each number:

1. 17: Let me check if it's prime. The square root of 17 is about 4.12, so I need to check divisibility by primes up to 4. Divided by 2? No, it's odd. Divided by 3? 1+7=8, which isn't divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*2=14, 17-14=3, not divisible by 7. So 17 is prime.

2. 26: Even number, ends with 6, so divisible by 2. Not prime.

3. 35: Ends with 5, so divisible by 5. Not prime.

4. 44: Even number, divisible by 2. Not prime.

5. 53: Let's check. Square root of 53 is about 7.28. Check divisibility by 2, 3, 5, 7. 53 is odd, not divisible by 2. 5+3=8, not divisible by 3. Doesn't end with 5, so not divisible by 5. Divided by 7? 7*7=49, 53-49=4, not divisible by 7. So 53 is prime.

6. 62: Even number, ends with 2. Divisible by 2. Not prime.

7. 71: Check if prime. Square root is about 8.43. Check divisibility by 2, 3, 5, 7. 71 is odd, not divisible by 2. 7+1=8, not divisible by 3. Doesn't end with 5. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Next prime is 11, but 11*6=66, 11*7=77, which is higher. So 71 is prime.

8. 80: Ends with 0, divisible by 10. Not prime.

So from the list, the primes are 17, 53, 71. That's three numbers. Wait, let me check again to make sure I didn't miss any.

Wait, let me check 17, 53, 71. Are there any others?

Wait, let me check 26: 26 is 2*13, not prime. 35 is 5*7. 44 is 4*11. 62 is 2*31. 80 is 8*10. So only 17, 53, 71. So three primes. Therefore, the answer is 3?

Wait, but let me check again. Maybe I missed some numbers? Let me check the list again.

Original numbers generated: 17, 26, 35, 44, 53, 62, 71, 80. That's 8 numbers. Checked all of them. Only 17, 53, 71 are primes. So answer is 3?

Wait, but let me think again. Maybe I made a mistake in generating the numbers? Let me check the pairs again.

a + b = 8, with a from 1-9, b from 0-9.

So:

a=1, b=7: 17

a=2, b=6: 26

a=3, b=5: 35

a=4, b=4: 44

a=5, b=3: 53

a=6, b=2: 62

a=7, b=1: 71

a=8, b=0: 80

a=9, b=-1: invalid. So that's correct. So 8 numbers. So only three primes. So answer is 3?

But wait, let me check if there are other two-digit primes with digits summing to 8. Maybe I missed some numbers?

Wait, for example, is there a two-digit prime where digits add to 8 that I didn't list? Let me think. For example, 17, 53, 71. Are there others?

Wait, let me think of primes with digits adding to 8. Let me think of primes between 10 and 99. Let me list all two-digit primes and check their digit sums. Maybe that's a better approach?

List of two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37, 39 (wait, 39 is not prime),

Wait, no, 39 is divisible by 3. Let me list them properly.

Two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37,

41, 43, 47,

53, 59,

61, 67,

71, 73, 79,

83, 89,

97.

Wait, let me check if I missed any. Let me go step by step:

From 10-19: 11, 13, 17, 19.

From 20-29: 23, 29.

From 30-39: 31, 37.

From 40-49: 41, 43, 47.

From 50-59: 53, 59.

From 60-69: 61, 67.

From 70-79: 71, 73, 79.

From 80-89: 83, 89.

From 90-99: 97.

So that's all two-digit primes. Now, let me check which of these have digits summing to 8.

Let me go through each:

11: 1+1=2. Not 8.

13: 1+3=4. No.

17: 1+7=8. Yes! That's one.

19: 1+9=10. No.

23: 2+3=5. No.

29: 2+9=11. No.

31: 3+1=4. No.

37: 3+7=10. No.

41: 4+1=5. No.

43: 4+3=7. No.

47: 4+7=11. No.

53: 5+3=8. Yes! Second one.

59: 5+9=14. No.

61: 6+1=7. No.

67: 6+7=13. No.

71: 7+1=8. Yes! Third one.

73: 7+3=10. No.

79: 7+9=16. No.

83: 8+3=11. No.

89: 8+9=17. No.

97: 9+7=16. No.

So only 17, 53, 71. So three primes. Therefore, the answer is 3. So that matches my previous result. Therefore, the answer is 3.

But let me check once more to make sure I didn't miss any primes. For example, is there a prime like 89? Wait, 8+9=17. Not 8. How about 73? 7+3=10. No. 19? 1+9=10. No. 13? 4. 17 is 8. 53 is 8. 71 is 8. So yeah, only three.

Alternatively, maybe I can think of another way. For example, the numbers we generated earlier were 17, 26, 35, 44, 53, 62, 71, 80. Then checking which are prime. So 17, 53, 71. So three numbers. So answer is 3.

Therefore, I think the answer is 3. But let me check once again if there's a possibility that I missed a prime. For example, is 89 a prime? Yes, but digits sum to 17. 73 is prime, digits sum to 10. 19 is prime, digits sum to 10. 13 is prime, digits sum to 4. 11 is prime, digits sum to 2. 23 is prime, digits sum to 5. 29 is prime, digits sum to 11. 31 is prime, digits sum to 4. 37 is prime, digits sum to 10. 41 is prime, digits sum to 5. 43 is prime, digits sum to 7. 47 is prime, digits sum to 11. 53 is prime, digits sum to 8. 59 is prime, digits sum to 14. 61 is prime, digits sum to 7. 67 is prime, digits sum to 13. 71 is prime, digits sum to 8. 73 is prime, digits sum to 10. 79 is prime, digits sum to 16. 83 is prime, digits sum to 11. 89 is prime, digits sum to 17. 97 is prime, digits sum to 16.

So yeah, only 17, 53, 71. So three primes. Therefore, the answer is 3.

But just to be thorough, let me check if there are any other numbers with digits summing to 8 that I might have missed. For example, numbers like 80, which is not prime. 80 is 8+0=8, but it's not prime. 71 is prime. 62 is 6+2=8, but 62 is not prime. 53 is prime. 44 is 4+4=8, not prime. 35 is 3+5=8, not prime. 26 is 2+6=8, not prime. 17 is prime. So yeah, only those three.

Therefore, I think the answer is 3.

**Final Answer**
The number of two-digit primes with digits summing to 8 is \boxed{3}.

To determine how many two-digit prime numbers have digits that sum to 8, we can approach the problem in a structured way.

---

### Step 1: Understand the Structure of Two-Digit Numbers

A two-digit number can be written as $10a + b$, where:
- $a$ is the tens digit (from 1 to 9),
- $b$ is the units digit (from 0 to 9).

We are looking for numbers where the **sum of the digits** is 8, i.e., $a + b = 8$.

---

### Step 2: Generate All Valid Two-Digit Numbers with Digit Sum 8

We iterate over all possible values of $a$ from 1 to 8 (since $a = 9$ would require $b = -1$, which is invalid), and compute $b = 8 - a$:

| $a$ | $b$ | Number |
|------|------|--------|
| 1    | 7    | 17     |
| 2    | 6    | 26     |
| 3    | 5    | 35     |
| 4    | 4    | 44     |
| 5    | 3    | 53     |
| 6    | 2    | 62     |
| 7    | 1    | 71     |
| 8    | 0    | 80     |

So the candidate numbers are: **17, 26, 35, 44, 53, 62, 71, 80**

---

### Step 3: Identify Which of These Are Prime

We now check which of these numbers are **prime**:

- **17**: Prime ✅
- **26**: Not prime (divisible by 2) ❌
- **35**: Not prime (divisible by 5) ❌
- **44**: Not prime (divisible by 2) ❌
- **53**: Prime ✅
- **62**: Not prime (divisible by 2) ❌
- **71**: Prime ✅
- **80**: Not prime (divisible by 2) ❌

---

### Step 4: Count the Prime Numbers

From the list, the prime numbers are:
- 17
- 53
- 71

Thus, there are **three** two-digit prime numbers whose digits sum to 8.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many two-digit prime numbers have digits that add up to 8. Let me start by recalling what two-digit primes are. Two-digit numbers range from 10 to 99, right? And primes are numbers greater than 1 that have no divisors other than 1 and themselves.

First, maybe I should list all two-digit primes and then check which ones have digits summing to 8. But that might take a while. Maybe there's a smarter way. Let me think.

Alternatively, I can think about all two-digit numbers where the sum of the digits is 8 and then check which of those are prime. That might be more efficient. Let me try that approach.

So, a two-digit number can be represented as 10*a + b, where a is the tens digit and b is the units digit. Both a and b are digits, so a is from 1 to 9 (since it's a two-digit number) and b is from 0 to 9. The sum of the digits is a + b = 8. So, I need to find all pairs (a, b) where a is between 1 and 9, b is between 0 and 9, and a + b = 8. Then, for each such number, check if it's prime.

Let me list all possible pairs (a, b):

Starting with a = 1: then b = 8 - 1 = 7. So the number is 17.

a = 2: b = 6. Number is 26.

a = 3: b = 5. Number is 35.

a = 4: b = 4. Number is 44.

a = 5: b = 3. Number is 53.

a = 6: b = 2. Number is 62.

a = 7: b = 1. Number is 71.

a = 8: b = 0. Number is 80.

a = 9: b = -1. Wait, that's not possible because b can't be negative. So a can't be 9 here. So the possible numbers are from a=1 to a=8.

So the numbers are: 17, 26, 35, 44, 53, 62, 71, 80.

Now, I need to check which of these numbers are prime.

Let me recall that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So let's check each number:

1. 17: Let me check if it's prime. The square root of 17 is about 4.12, so I need to check divisibility by primes up to 4. Divided by 2? No, it's odd. Divided by 3? 1+7=8, which isn't divisible by 3. Divided by 5? Doesn't end with 0 or 5. Divided by 7? 7*2=14, 17-14=3, not divisible by 7. So 17 is prime.

2. 26: Even number, ends with 6, so divisible by 2. Not prime.

3. 35: Ends with 5, so divisible by 5. Not prime.

4. 44: Even number, divisible by 2. Not prime.

5. 53: Let's check. Square root of 53 is about 7.28. Check divisibility by 2, 3, 5, 7. 53 is odd, not divisible by 2. 5+3=8, not divisible by 3. Doesn't end with 5, so not divisible by 5. Divided by 7? 7*7=49, 53-49=4, not divisible by 7. So 53 is prime.

6. 62: Even number, ends with 2. Divisible by 2. Not prime.

7. 71: Check if prime. Square root is about 8.43. Check divisibility by 2, 3, 5, 7. 71 is odd, not divisible by 2. 7+1=8, not divisible by 3. Doesn't end with 5. Divided by 7? 7*10=70, 71-70=1, not divisible by 7. Next prime is 11, but 11*6=66, 11*7=77, which is higher. So 71 is prime.

8. 80: Ends with 0, divisible by 10. Not prime.

So from the list, the primes are 17, 53, 71. That's three numbers. Wait, let me check again to make sure I didn't miss any.

Wait, let me check 17, 53, 71. Are there any others?

Wait, let me check 26: 26 is 2*13, not prime. 35 is 5*7. 44 is 4*11. 62 is 2*31. 80 is 8*10. So only 17, 53, 71. So three primes. Therefore, the answer is 3?

Wait, but let me check again. Maybe I missed some numbers? Let me check the list again.

Original numbers generated: 17, 26, 35, 44, 53, 62, 71, 80. That's 8 numbers. Checked all of them. Only 17, 53, 71 are primes. So answer is 3?

Wait, but let me think again. Maybe I made a mistake in generating the numbers? Let me check the pairs again.

a + b = 8, with a from 1-9, b from 0-9.

So:

a=1, b=7: 17

a=2, b=6: 26

a=3, b=5: 35

a=4, b=4: 44

a=5, b=3: 53

a=6, b=2: 62

a=7, b=1: 71

a=8, b=0: 80

a=9, b=-1: invalid. So that's correct. So 8 numbers. So only three primes. So answer is 3?

But wait, let me check if there are other two-digit primes with digits summing to 8. Maybe I missed some numbers?

Wait, for example, is there a two-digit prime where digits add to 8 that I didn't list? Let me think. For example, 17, 53, 71. Are there others?

Wait, let me think of primes with digits adding to 8. Let me think of primes between 10 and 99. Let me list all two-digit primes and check their digit sums. Maybe that's a better approach?

List of two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37, 39 (wait, 39 is not prime),

Wait, no, 39 is divisible by 3. Let me list them properly.

Two-digit primes:

11, 13, 17, 19,

23, 29,

31, 37,

41, 43, 47,

53, 59,

61, 67,

71, 73, 79,

83, 89,

97.

Wait, let me check if I missed any. Let me go step by step:

From 10-19: 11, 13, 17, 19.

From 20-29: 23, 29.

From 30-39: 31, 37.

From 40-49: 41, 43, 47.

From 50-59: 53, 59.

From 60-69: 61, 67.

From 70-79: 71, 73, 79.

From 80-89: 83, 89.

From 90-99: 97.

So that's all two-digit primes. Now, let me check which of these have digits summing to 8.

Let me go through each:

11: 1+1=2. Not 8.

13: 1+3=4. No.

17: 1+7=8. Yes! That's one.

19: 1+9=10. No.

23: 2+3=5. No.

29: 2+9=11. No.

31: 3+1=4. No.

37: 3+7=10. No.

41: 4+1=5. No.

43: 4+3=7. No.

47: 4+7=11. No.

53: 5+3=8. Yes! Second one.

59: 5+9=14. No.

61: 6+1=7. No.

67: 6+7=13. No.

71: 7+1=8. Yes! Third one.

73: 7+3=10. No.

79: 7+9=16. No.

83: 8+3=11. No.

89: 8+9=17. No.

97: 9+7=16. No.

So only 17, 53, 71. So three primes. Therefore, the answer is 3. So that matches my previous result. Therefore, the answer is 3.

But let me check once more to make sure I didn't miss any primes. For example, is there a prime like 89? Wait, 8+9=17. Not 8. How about 73? 7+3=10. No. 19? 1+9=10. No. 13? 4. 17 is 8. 53 is 8. 71 is 8. So yeah, only three.

Alternatively, maybe I can think of another way. For example, the numbers we generated earlier were 17, 26, 35, 44, 53, 62, 71, 80. Then checking which are prime. So 17, 53, 71. So three numbers. So answer is 3.

Therefore, I think the answer is 3. But let me check once again if there's a possibility that I missed a prime. For example, is 89 a prime? Yes, but digits sum to 17. 73 is prime, digits sum to 10. 19 is prime, digits sum to 10. 13 is prime, digits sum to 4. 11 is prime, digits sum to 2. 23 is prime, digits sum to 5. 29 is prime, digits sum to 11. 31 is prime, digits sum to 4. 37 is prime, digits sum to 10. 41 is prime, digits sum to 5. 43 is prime, digits sum to 7. 47 is prime, digits sum to 11. 53 is prime, digits sum to 8. 59 is prime, digits sum to 14. 61 is prime, digits sum to 7. 67 is prime, digits sum to 13. 71 is prime, digits sum to 8. 73 is prime, digits sum to 10. 79 is prime, digits sum to 16. 83 is prime, digits sum to 11. 89 is prime, digits sum to 17. 97 is prime, digits sum to 16.

So yeah, only 17, 53, 71. So three primes. Therefore, the answer is 3.

But just to be thorough, let me check if there are any other numbers with digits summing to 8 that I might have missed. For example, numbers like 80, which is not prime. 80 is 8+0=8, but it's not prime. 71 is prime. 62 is 6+2=8, but 62 is not prime. 53 is prime. 44 is 4+4=8, not prime. 35 is 3+5=8, not prime. 26 is 2+6=8, not prime. 17 is prime. So yeah, only those three.

Therefore, I think the answer is 3.

**Final Answer**
The number of two-digit primes with digits summing to 8 is \boxed{3}.

To determine how many two-digit prime numbers have digits that sum to 8, we can approach the problem in a structured way.

---

### Step 1: Understand the Structure of Two-Digit Numbers

A two-digit number can be written as $10a + b$, where:
- $a$ is the tens digit (from 1 to 9),
- $b$ is the units digit (from 0 to 9).

We are looking for numbers where the **sum of the digits** is 8, i.e., $a + b = 8$.

---

### Step 2: Generate All Valid Two-Digit Numbers with Digit Sum 8

We iterate over all possible values of $a$ from 1 to 8 (since $a = 9$ would require $b = -1$, which is invalid), and compute $b = 8 - a$:

| $a$ | $b$ | Number |
|------|------|--------|
| 1    | 7    | 17     |
| 2    | 6    | 26     |
| 3    | 5    | 35     |
| 4    | 4    | 44     |
| 5    | 3    | 53     |
| 6    | 2    | 62     |
| 7    | 1    | 71     |
| 8    | 0    | 80     |

So the candidate numbers are: **17, 26, 35, 44, 53, 62, 71, 80**

---

### Step 3: Identify Which of These Are Prime

We now check which of these numbers are **prime**:

- **17**: Prime ✅
- **26**: Not prime (divisible by 2) ❌
- **35**: Not prime (divisible by 5) ❌
- **44**: Not prime (divisible by 2) ❌
- **53**: Prime ✅
- **62**: Not prime (divisible by 2) ❌
- **71**: Prime ✅
- **80**: Not prime (divisible by 2) ❌

---

### Step 4: Count the Prime Numbers

From the list, the prime numbers are:
- 17
- 53
- 71

Thus, there are **three** two-digit prime numbers whose digits sum to 8.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
First, we list all the two-digit numbers with a digits sum of 8:

17, 26, 35, 44, 53, 62, 71, 80

Clearly, 26, 44, 62, and 80 are not prime.  35 is not prime, but 17, 53, and 71 are (we can verify this by dividing each number by all the primes less than its square root (why is this?)).  Hence, the answer is $\boxed{3}$ two-digit primes.


Explanation:
Okay, so I need to figure out how many integers satisfy the inequality |x - 2| ≤ 5.6. Hmm, let me start by recalling what absolute value inequalities mean. The expression |x - 2| ≤ 5.6 means that the distance between x and 2 is less than or equal to 5.6. So, I can rewrite this inequality without the absolute value by considering the definition of absolute value.

Right, if |a| ≤ b, then -b ≤ a ≤ b. Applying that here, I can write:

-5.6 ≤ x - 2 ≤ 5.6

Now, I need to solve for x. To do that, I should add 2 to all three parts of the inequality. Let me do that step by step.

Starting with the left part: -5.6 + 2. Let me calculate that. Hmm, -5.6 + 2 is the same as -(5.6 - 2) which is -3.6? Wait, no. Wait, adding 2 to -5.6. Let me think. If I have -5.6 and add 2, that's like moving 2 units to the right on the number line. So, -5.6 + 2 = -3.6. Yeah, that's right.

Then the middle part is x - 2 + 2, which simplifies to x.

And the right part is 5.6 + 2, which is 7.6.

So putting it all together, the inequality becomes:

-3.6 ≤ x ≤ 7.6

So, x has to be greater than or equal to -3.6 and less than or equal to 7.6. Now, the question is asking for the number of integers in this solution set.

Alright, so I need to find all integers x such that -3.6 ≤ x ≤ 7.6. Let me think about how to approach this.

First, let me note that integers are whole numbers, so they don't include fractions or decimals. So, I need to find all integers between -3.6 and 7.6, including the endpoints if they are integers.

But wait, the endpoints here are -3.6 and 7.6. Are these integers? No, they are not. So, the integers that satisfy the inequality will be the integers greater than or equal to the next integer greater than -3.6, and less than or equal to the integer less than 7.6.

Let me clarify. Since x has to be greater than or equal to -3.6, the smallest integer that satisfies this is -3, because -3 is greater than -3.6. Similarly, the largest integer that satisfies x ≤ 7.6 is 7, since 7 is less than 7.6.

Therefore, the integers in the solution set are from -3 up to 7, inclusive. Now, I need to count how many integers are in this range.

To count the integers from -3 to 7 inclusive, I can use the formula: number of integers = (last - first) + 1.

Let me check that formula. For example, from 1 to 3 inclusive, that's 3 - 1 + 1 = 3 numbers, which is correct. So applying that here:

Last integer is 7, first integer is -3. So, 7 - (-3) + 1 = 7 + 3 + 1 = 11. Wait, 7 - (-3) is 10, then +1 is 11. So, 11 integers? Let me verify by listing them.

Starting from -3: -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. Let me count them:

1. -3
2. -2
3. -1
4. 0
5. 1
6. 2
7. 3
8. 4
9. 5
10. 6
11. 7

Yes, that's 11 integers. So, the answer should be 11.

But wait, let me make sure I didn't make a mistake in determining the range. Let me check the original inequality again.

Original inequality: |x - 2| ≤ 5.6. So solving that gives -5.6 ≤ x - 2 ≤ 5.6, then adding 2 gives -3.6 ≤ x ≤ 7.6. So, x is between -3.6 and 7.6.

So, integers greater than or equal to -3.6. The integers greater than or equal to -3.6 are -3, -2, ..., up to 7.6. So, the upper bound is 7.6, so integers less than or equal to 7.6 are up to 7.

But let me check if maybe there's an integer that's just below 7.6, which is 7, and above -3.6, which is -3. So, yes, the integers from -3 to 7 inclusive.

But let me check if maybe there's a mistake here. For example, if the upper bound was 7.6, does that include 7.6? Well, the inequality is x ≤ 7.6, so 7.6 is not an integer, so the largest integer less than or equal to 7.6 is 7. Similarly, the smallest integer greater than or equal to -3.6 is -3.

But let me check if maybe there's an integer that's less than -3.6? For example, -4. Is -4 less than -3.6? Yes, because -4 is to the left of -3.6 on the number line. But since the inequality is x ≥ -3.6, -4 is not included. So, the first integer is -3.

Similarly, 7.6 is not an integer, so 8 is not included. So, the last integer is 7.

Therefore, the integers are from -3 to 7 inclusive.

But let me count again. From -3 to 7 inclusive. How many numbers is that?

One way to calculate is to take the difference between the upper and lower bounds and add 1.

So, upper bound is 7, lower bound is -3.

So, 7 - (-3) = 10. Then add 1, so 11 numbers.

Alternatively, think of how many numbers from -3 to 7.

Starting from 0, the numbers from -3 to 3 would be 7 numbers: -3, -2, -1, 0, 1, 2, 3. Then from 4 to 7 is 4 numbers: 4, 5, 6, 7. So total is 7 + 4 = 11.

Yes, that matches.

Alternatively, think of the number of integers between two numbers. If you have a range from a to b inclusive, the number of integers is floor(b) - ceil(a) + 1.

In this case, a is -3.6, so ceil(a) is -3. And b is 7.6, so floor(b) is 7. Then 7 - (-3) + 1 = 11.

So, that seems consistent.

But let me check if there's a possibility that maybe the original inequality includes some other integers? For example, maybe I made a mistake in solving the inequality? Let me check again.

Original inequality: |x - 2| ≤ 5.6.

This is equivalent to -5.6 ≤ x - 2 ≤ 5.6.

Adding 2 to all parts:

Left side: -5.6 + 2 = -3.6

Right side: 5.6 + 2 = 7.6

So, yes, the solution set is x ∈ [-3.6, 7.6].

Therefore, integers in this interval are from -3 to 7 inclusive.

So, 11 integers.

But let me check with specific numbers. Let me take x = -3.

Compute | -3 - 2 | = | -5 | = 5 ≤ 5.6. Yes, that's true.

x = -4: | -4 - 2 | = | -6 | = 6, which is 6 > 5.6. So, not included.

x = 7: |7 - 2| = |5| = 5 ≤ 5.6. True.

x = 8: |8 - 2| = 6 > 5.6. Not included.

So, that seems correct.

Therefore, the integers are from -3 to 7 inclusive.

So, the count is 11.

But just to be thorough, let me list them again:

Starting at -3:

-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.

That's 11 numbers.

Yes, that seems correct.

Therefore, the number of integers in the solution set is 11.

I think that's the answer.

**Final Answer**
The number of integers in the solution set is \boxed{11}.

To determine how many integers satisfy the inequality $ |x - 2| \leq 5.6 $, we begin by interpreting the absolute value inequality.

---

### Step 1: Rewrite the Inequality

The inequality $ |x - 2| \leq 5.6 $ means that the distance between $ x $ and 2 is at most 5.6. Using the definition of absolute value, we can rewrite this as:

$$
-5.6 \leq x - 2 \leq 5.6
$$

---

### Step 2: Solve for $ x $

To isolate $ x $, add 2 to all parts of the inequality:

$$
-5.6 + 2 \leq x \leq 5.6 + 2
$$

$$
-3.6 \leq x \leq 7.6
$$

So, the solution set is all real numbers $ x $ such that $ -3.6 \leq x \leq 7.6 $.

---

### Step 3: Identify Integer Solutions

We are asked to find how many **integers** lie in this interval. Since the endpoints are not integers, we consider the **smallest integer greater than or equal to -3.6**, which is **-3**, and the **largest integer less than or equal to 7.6**, which is **7**.

Thus, the integers in the solution set are:

$$
-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7
$$

---

### Step 4: Count the Integers

To count the number of integers from -3 to 7 inclusive, we use the formula:

$$
\text{Number of integers} = \text{Last} - \text{First} + 1
$$

$$
7 - (-3) + 1 = 7 + 3 + 1 = 11
$$

---

### Final Answer

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many integers satisfy the inequality |x - 2| ≤ 5.6. Hmm, let me start by recalling what absolute value inequalities mean. The expression |x - 2| ≤ 5.6 means that the distance between x and 2 is less than or equal to 5.6. So, I can rewrite this inequality without the absolute value by considering the definition of absolute value.

Right, if |a| ≤ b, then -b ≤ a ≤ b. Applying that here, I can write:

-5.6 ≤ x - 2 ≤ 5.6

Now, I need to solve for x. To do that, I should add 2 to all three parts of the inequality. Let me do that step by step.

Starting with the left part: -5.6 + 2. Let me calculate that. Hmm, -5.6 + 2 is the same as -(5.6 - 2) which is -3.6? Wait, no. Wait, adding 2 to -5.6. Let me think. If I have -5.6 and add 2, that's like moving 2 units to the right on the number line. So, -5.6 + 2 = -3.6. Yeah, that's right.

Then the middle part is x - 2 + 2, which simplifies to x.

And the right part is 5.6 + 2, which is 7.6.

So putting it all together, the inequality becomes:

-3.6 ≤ x ≤ 7.6

So, x has to be greater than or equal to -3.6 and less than or equal to 7.6. Now, the question is asking for the number of integers in this solution set.

Alright, so I need to find all integers x such that -3.6 ≤ x ≤ 7.6. Let me think about how to approach this.

First, let me note that integers are whole numbers, so they don't include fractions or decimals. So, I need to find all integers between -3.6 and 7.6, including the endpoints if they are integers.

But wait, the endpoints here are -3.6 and 7.6. Are these integers? No, they are not. So, the integers that satisfy the inequality will be the integers greater than or equal to the next integer greater than -3.6, and less than or equal to the integer less than 7.6.

Let me clarify. Since x has to be greater than or equal to -3.6, the smallest integer that satisfies this is -3, because -3 is greater than -3.6. Similarly, the largest integer that satisfies x ≤ 7.6 is 7, since 7 is less than 7.6.

Therefore, the integers in the solution set are from -3 up to 7, inclusive. Now, I need to count how many integers are in this range.

To count the integers from -3 to 7 inclusive, I can use the formula: number of integers = (last - first) + 1.

Let me check that formula. For example, from 1 to 3 inclusive, that's 3 - 1 + 1 = 3 numbers, which is correct. So applying that here:

Last integer is 7, first integer is -3. So, 7 - (-3) + 1 = 7 + 3 + 1 = 11. Wait, 7 - (-3) is 10, then +1 is 11. So, 11 integers? Let me verify by listing them.

Starting from -3: -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. Let me count them:

1. -3
2. -2
3. -1
4. 0
5. 1
6. 2
7. 3
8. 4
9. 5
10. 6
11. 7

Yes, that's 11 integers. So, the answer should be 11.

But wait, let me make sure I didn't make a mistake in determining the range. Let me check the original inequality again.

Original inequality: |x - 2| ≤ 5.6. So solving that gives -5.6 ≤ x - 2 ≤ 5.6, then adding 2 gives -3.6 ≤ x ≤ 7.6. So, x is between -3.6 and 7.6.

So, integers greater than or equal to -3.6. The integers greater than or equal to -3.6 are -3, -2, ..., up to 7.6. So, the upper bound is 7.6, so integers less than or equal to 7.6 are up to 7.

But let me check if maybe there's an integer that's just below 7.6, which is 7, and above -3.6, which is -3. So, yes, the integers from -3 to 7 inclusive.

But let me check if maybe there's a mistake here. For example, if the upper bound was 7.6, does that include 7.6? Well, the inequality is x ≤ 7.6, so 7.6 is not an integer, so the largest integer less than or equal to 7.6 is 7. Similarly, the smallest integer greater than or equal to -3.6 is -3.

But let me check if maybe there's an integer that's less than -3.6? For example, -4. Is -4 less than -3.6? Yes, because -4 is to the left of -3.6 on the number line. But since the inequality is x ≥ -3.6, -4 is not included. So, the first integer is -3.

Similarly, 7.6 is not an integer, so 8 is not included. So, the last integer is 7.

Therefore, the integers are from -3 to 7 inclusive.

But let me count again. From -3 to 7 inclusive. How many numbers is that?

One way to calculate is to take the difference between the upper and lower bounds and add 1.

So, upper bound is 7, lower bound is -3.

So, 7 - (-3) = 10. Then add 1, so 11 numbers.

Alternatively, think of how many numbers from -3 to 7.

Starting from 0, the numbers from -3 to 3 would be 7 numbers: -3, -2, -1, 0, 1, 2, 3. Then from 4 to 7 is 4 numbers: 4, 5, 6, 7. So total is 7 + 4 = 11.

Yes, that matches.

Alternatively, think of the number of integers between two numbers. If you have a range from a to b inclusive, the number of integers is floor(b) - ceil(a) + 1.

In this case, a is -3.6, so ceil(a) is -3. And b is 7.6, so floor(b) is 7. Then 7 - (-3) + 1 = 11.

So, that seems consistent.

But let me check if there's a possibility that maybe the original inequality includes some other integers? For example, maybe I made a mistake in solving the inequality? Let me check again.

Original inequality: |x - 2| ≤ 5.6.

This is equivalent to -5.6 ≤ x - 2 ≤ 5.6.

Adding 2 to all parts:

Left side: -5.6 + 2 = -3.6

Right side: 5.6 + 2 = 7.6

So, yes, the solution set is x ∈ [-3.6, 7.6].

Therefore, integers in this interval are from -3 to 7 inclusive.

So, 11 integers.

But let me check with specific numbers. Let me take x = -3.

Compute | -3 - 2 | = | -5 | = 5 ≤ 5.6. Yes, that's true.

x = -4: | -4 - 2 | = | -6 | = 6, which is 6 > 5.6. So, not included.

x = 7: |7 - 2| = |5| = 5 ≤ 5.6. True.

x = 8: |8 - 2| = 6 > 5.6. Not included.

So, that seems correct.

Therefore, the integers are from -3 to 7 inclusive.

So, the count is 11.

But just to be thorough, let me list them again:

Starting at -3:

-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.

That's 11 numbers.

Yes, that seems correct.

Therefore, the number of integers in the solution set is 11.

I think that's the answer.

**Final Answer**
The number of integers in the solution set is \boxed{11}.

To determine how many integers satisfy the inequality $ |x - 2| \leq 5.6 $, we begin by interpreting the absolute value inequality.

---

### Step 1: Rewrite the Inequality

The inequality $ |x - 2| \leq 5.6 $ means that the distance between $ x $ and 2 is at most 5.6. Using the definition of absolute value, we can rewrite this as:

$$
-5.6 \leq x - 2 \leq 5.6
$$

---

### Step 2: Solve for $ x $

To isolate $ x $, add 2 to all parts of the inequality:

$$
-5.6 + 2 \leq x \leq 5.6 + 2
$$

$$
-3.6 \leq x \leq 7.6
$$

So, the solution set is all real numbers $ x $ such that $ -3.6 \leq x \leq 7.6 $.

---

### Step 3: Identify Integer Solutions

We are asked to find how many **integers** lie in this interval. Since the endpoints are not integers, we consider the **smallest integer greater than or equal to -3.6**, which is **-3**, and the **largest integer less than or equal to 7.6**, which is **7**.

Thus, the integers in the solution set are:

$$
-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7
$$

---

### Step 4: Count the Integers

To count the number of integers from -3 to 7 inclusive, we use the formula:

$$
\text{Number of integers} = \text{Last} - \text{First} + 1
$$

$$
7 - (-3) + 1 = 7 + 3 + 1 = 11
$$

---

### Final Answer

$$
\boxed{11}
$$

Solution: Getting rid of the absolute value, we have $-5.6 \le x-2 \le 5.6$, or $-3.6 \le x \le 7.6$. Thus, $x$ can be any integer from -3 to 7, inclusive. There are $7-(-3)+1=\boxed{11}$ integers in this range.


Explanation:
Okay, so I need to find sin T in triangle RST. They told me that sin R is 2/5. Let me try to visualize the triangle first. The Asymptote code shows that T is at (0,0), S is at (2,0), and R is at (2, sqrt(21)). There's a right angle mark at R, S, T. Wait, hold on, right angle mark is usually at the vertex where the right angle is. The Asymptote code says rightanglemark(R,S,T,10), so that means the right angle is at S? Because the right angle mark is between R, S, T. So the right angle is at S? Let me confirm.

In the Asymptote code, they draw R connected to S connected to T connected to R. Then the right angle mark is at R, S, T. Wait, the rightanglemark is usually drawn between three points, with the right angle at the middle one. So if it's rightanglemark(R,S,T,10), then the right angle is at S. So triangle RST is right-angled at S. Therefore, angle S is 90 degrees. So the triangle has a right angle at S, with legs RS and ST, and hypotenuse RT.

Wait, but the Asymptote code labels the side RT as 5. Let me check. The Asymptote code says label("$5$",(R+T)/2,NW);. The midpoint between R and T is ( (2 + 0)/2, (sqrt(21) + 0)/2 ) which is (1, sqrt(21)/2). So they label the segment RT as 5. Therefore, RT is the hypotenuse of the right triangle, and its length is 5. So triangle RST is right-angled at S, with hypotenuse RT = 5. Then RS and ST are the legs.

But they also mention sin R = 2/5. Let me recall that in a right-angled triangle, the sine of an angle is opposite over hypotenuse. But angle R is at vertex R. Wait, if the triangle is right-angled at S, then angles at R and T are acute angles. So angle R is at point R, which is connected to S and T. Let me try to sketch mentally.

Point S is at (2,0), T is at (0,0), R is at (2, sqrt(21)). So RS is vertical from (2,0) to (2, sqrt(21)), and ST is horizontal from (2,0) to (0,0). So the right angle is at S, between RS and ST. Therefore, angle at S is 90 degrees. Then angle R is at point R, between RS and RT. Similarly, angle T is at point T, between ST and RT.

So, given that, sin R is 2/5. Let me recall that in triangle RST, angle R is opposite to side ST. Wait, in a triangle, the sine of an angle is opposite over hypotenuse. Wait, but since it's a right-angled triangle, hypotenuse is RT. So for angle R, the opposite side is ST, and the hypotenuse is RT. Therefore, sin R = opposite / hypotenuse = ST / RT = 2/5. But they told us that RT is 5, so ST / 5 = 2/5, so ST = 2. Therefore, ST is 2 units long. Then, since the triangle is right-angled at S, we can find RS using Pythagoras. RT^2 = RS^2 + ST^2. So 5^2 = RS^2 + 2^2 => 25 = RS^2 + 4 => RS^2 = 21 => RS = sqrt(21). Which matches the coordinates given in the Asymptote code, since R is at (2, sqrt(21)), so RS is vertical from (2,0) to (2, sqrt(21)), so length sqrt(21). So that checks out.

Now, the question is asking for sin T. So angle T is at point T, which is the right angle? Wait no, angle T is at point T, which is connected to S and R. Since the right angle is at S, angle T is one of the acute angles. So in triangle RST, angle T is opposite to side RS. Wait, let me think again. In triangle RST, angle T is at point T. The sides opposite to angles are as follows: side opposite angle R is ST, side opposite angle S is RT, and side opposite angle T is RS. Therefore, sin T would be opposite over hypotenuse. The hypotenuse is RT, which is 5. The side opposite angle T is RS, which is sqrt(21). Therefore, sin T = RS / RT = sqrt(21)/5. But let me check if that's correct.

Alternatively, maybe I can use the coordinates to compute it. Since coordinates are given, maybe I can compute the lengths and then compute the sine. Let me confirm.

Coordinates:

T is at (0,0), S is at (2,0), R is at (2, sqrt(21)).

So, angle T is at (0,0). Let me find the sides adjacent and opposite to angle T. In triangle RST, angle T is between sides ST and RT. So, to find sin T, we can think of the opposite side over hypotenuse. Wait, but in a right-angled triangle, the hypotenuse is opposite the right angle. Since the right angle is at S, hypotenuse is RT. Therefore, for angle T, which is at (0,0), the sides are:

- Opposite side: RS, which is sqrt(21)

- Adjacent side: ST, which is 2

- Hypotenuse: RT, which is 5

Therefore, sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5. Alternatively, if I think of angle T, then in terms of the triangle, sin T = opposite / hypotenuse. So that would be RS / RT. Which is sqrt(21)/5. So that would be the answer. But let me check if that's correct.

Alternatively, maybe I can compute it using coordinates. Let me compute the coordinates. The coordinates of T are (0,0), S is (2,0), R is (2, sqrt(21)). So vector from T to S is (2,0), and vector from T to R is (2, sqrt(21)). So angle at T is between vectors TS and TR. To find sin T, maybe I can compute the angle between those two vectors.

Alternatively, since we know the lengths of the sides, we can use trigonometry. Since we have a right-angled triangle, we can use the sides. Since angle T is adjacent to side ST (length 2) and opposite to side RS (length sqrt(21)), so sin T = opposite / hypotenuse = sqrt(21)/5. Therefore, the answer is sqrt(21)/5. But let me check if that's the case.

Alternatively, maybe I can use the fact that in a right-angled triangle, the two non-right angles are complementary. So angle R + angle T = 90 degrees. Therefore, sin T = sin(90 - R) = cos R. So if I can find cos R, then that would be sin T. Since sin R is 2/5, then cos R can be found using sin^2 R + cos^2 R = 1. Therefore, cos R = sqrt(1 - (2/5)^2) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5. Therefore, sin T = cos R = sqrt(21)/5. So that's another way to get the same answer. Therefore, sin T is sqrt(21)/5.

But let me check if that's consistent with the coordinates. Let me compute the coordinates. The coordinates of the triangle are T(0,0), S(2,0), R(2, sqrt(21)). So angle at T. Let me compute the lengths of the sides adjacent to angle T. The sides adjacent to angle T are ST and RT? Wait, no. In angle T, the sides are ST and RT. Wait, but in a triangle, the sides adjacent to angle T are the two sides that form the angle, which are ST and RT. However, in the right-angled triangle, the hypotenuse is RT, and the other sides are ST and RS. Wait, maybe I need to think again.

Wait, angle T is at point T. The sides forming angle T are ST and RT. The side opposite angle T is RS. So, in terms of trigonometry, sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5. Which is the same as before. Alternatively, if I use coordinates, I can compute the angle. Let me compute the coordinates.

Vector TS is from T to S: (2,0) - (0,0) = (2,0). Vector TR is from T to R: (2, sqrt(21)) - (0,0) = (2, sqrt(21)). The angle between vectors TS and TR is angle T. To find sin T, we can use the cross product formula. The magnitude of the cross product of vectors TS and TR is |TS x TR| = |(2)(sqrt(21)) - (0)(2)| = 2*sqrt(21). The magnitude of TS is sqrt(2^2 + 0^2) = 2. The magnitude of TR is sqrt(2^2 + (sqrt(21))^2) = sqrt(4 + 21) = sqrt(25) = 5. Then, the sine of the angle between them is |TS x TR| / (|TS| |TR|) = 2*sqrt(21) / (2*5) = sqrt(21)/5. Which matches our previous results. Therefore, sin T is sqrt(21)/5.

Therefore, the answer is sqrt(21)/5, which can be written as \boxed{\dfrac{\sqrt{21}}{5}}.

But let me check again if I didn't make any mistake. The problem says sin R is 2/5. Let me confirm that with the coordinates. In triangle RST, angle R is at point R. The sides adjacent to angle R are RS and RT. Wait, no. In angle R, the sides are RS and RT. Wait, but in the right-angled triangle, angle R is not the right angle. The right angle is at S. So angle R is at point R, between RS and RT. Therefore, the sides adjacent to angle R are RS and RT? Wait, no. Wait, angle R is between sides RS and RT? Wait, RS is from R to S, and RT is from R to T. So yes, angle R is between RS and RT. Therefore, in terms of the triangle, the sides adjacent to angle R are RS and RT, but the side opposite angle R is ST. Therefore, sin R = opposite / hypotenuse = ST / RT. Which is 2 / 5, as given. Which matches the problem statement. Therefore, that's correct. Therefore, sin T is sqrt(21)/5. So that's the answer.

Alternatively, since angle R and angle T are complementary, as the triangle is right-angled at S, so angle R + angle T = 90 degrees. Therefore, sin T = sin(90 - R) = cos R. And since sin R = 2/5, then cos R = sqrt(1 - (2/5)^2) = sqrt(21)/5. Therefore, sin T = sqrt(21)/5. So that's another way to get the answer. Therefore, regardless of the method, the answer is sqrt(21)/5.

Therefore, I think that's the correct answer. Let me check once more with the coordinates. Coordinates of T(0,0), S(2,0), R(2, sqrt(21)). So, triangle with right angle at S. So, coordinates are such that ST is horizontal from (0,0) to (2,0), RS is vertical from (2,0) to (2, sqrt(21)), and RT is the hypotenuse from (2, sqrt(21)) to (0,0). So, the length of RT is 5, as given. So, coordinates are consistent with the given information. Therefore, sin T is sqrt(21)/5. Therefore, I think that's the answer.

**Final Answer**
\boxed{\dfrac{\sqrt{21}}{5}}

To solve for $\sin T$ in triangle $RST$, we are given that $\sin R = \frac{2}{5}$, and the triangle is right-angled at $S$. Let's proceed step by step.

---

### Step 1: Understand the Triangle

From the Asymptote code and the description:

- $T = (0, 0)$
- $S = (2, 0)$
- $R = (2, \sqrt{21})$

This places the triangle in the coordinate plane with:

- $S$ at the origin,
- $T$ to the right of $S$,
- $R$ directly above $S$.

The right angle is at $S$, so triangle $RST$ is a **right triangle** with the right angle at $S$. Therefore:

- $RT$ is the **hypotenuse**,
- $RS$ and $ST$ are the **legs**.

---

### Step 2: Use Given Information

We are told that:

$$
\sin R = \frac{2}{5}
$$

In triangle $RST$, angle $R$ is at vertex $R$, and the side opposite to angle $R$ is $ST$, while the hypotenuse is $RT$. So:

$$
\sin R = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{ST}{RT}
$$

We are also told that the length of $RT = 5$, so:

$$
\frac{ST}{5} = \frac{2}{5} \Rightarrow ST = 2
$$

---

### Step 3: Use the Pythagorean Theorem

Now that we know two sides of the triangle, we can find the third using the Pythagorean Theorem:

$$
RT^2 = RS^2 + ST^2
$$

$$
5^2 = RS^2 + 2^2 \Rightarrow 25 = RS^2 + 4 \Rightarrow RS^2 = 21 \Rightarrow RS = \sqrt{21}
$$

---

### Step 4: Find $\sin T$

Angle $T$ is at vertex $T$, and the side opposite to angle $T$ is $RS$, while the hypotenuse is still $RT = 5$. So:

$$
\sin T = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{RS}{RT} = \frac{\sqrt{21}}{5}
$$

---

### Step 5: Confirm Using Complementary Angles

Since the triangle is right-angled at $S$, angles $R$ and $T$ are complementary:

$$
\angle R + \angle T = 90^\circ \Rightarrow \sin T = \cos R
$$

We can compute $\cos R$ using the identity:

$$
\sin^2 R + \cos^2 R = 1 \Rightarrow \cos R = \sqrt{1 - \sin^2 R} = \sqrt{1 - \left(\frac{2}{5}\right)^2} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}
$$

So again, we find:

$$
\sin T = \frac{\sqrt{21}}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{\sqrt{21}}{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find sin T in triangle RST. They told me that sin R is 2/5. Let me try to visualize the triangle first. The Asymptote code shows that T is at (0,0), S is at (2,0), and R is at (2, sqrt(21)). There's a right angle mark at R, S, T. Wait, hold on, right angle mark is usually at the vertex where the right angle is. The Asymptote code says rightanglemark(R,S,T,10), so that means the right angle is at S? Because the right angle mark is between R, S, T. So the right angle is at S? Let me confirm.

In the Asymptote code, they draw R connected to S connected to T connected to R. Then the right angle mark is at R, S, T. Wait, the rightanglemark is usually drawn between three points, with the right angle at the middle one. So if it's rightanglemark(R,S,T,10), then the right angle is at S. So triangle RST is right-angled at S. Therefore, angle S is 90 degrees. So the triangle has a right angle at S, with legs RS and ST, and hypotenuse RT.

Wait, but the Asymptote code labels the side RT as 5. Let me check. The Asymptote code says label("$5$",(R+T)/2,NW);. The midpoint between R and T is ( (2 + 0)/2, (sqrt(21) + 0)/2 ) which is (1, sqrt(21)/2). So they label the segment RT as 5. Therefore, RT is the hypotenuse of the right triangle, and its length is 5. So triangle RST is right-angled at S, with hypotenuse RT = 5. Then RS and ST are the legs.

But they also mention sin R = 2/5. Let me recall that in a right-angled triangle, the sine of an angle is opposite over hypotenuse. But angle R is at vertex R. Wait, if the triangle is right-angled at S, then angles at R and T are acute angles. So angle R is at point R, which is connected to S and T. Let me try to sketch mentally.

Point S is at (2,0), T is at (0,0), R is at (2, sqrt(21)). So RS is vertical from (2,0) to (2, sqrt(21)), and ST is horizontal from (2,0) to (0,0). So the right angle is at S, between RS and ST. Therefore, angle at S is 90 degrees. Then angle R is at point R, between RS and RT. Similarly, angle T is at point T, between ST and RT.

So, given that, sin R is 2/5. Let me recall that in triangle RST, angle R is opposite to side ST. Wait, in a triangle, the sine of an angle is opposite over hypotenuse. Wait, but since it's a right-angled triangle, hypotenuse is RT. So for angle R, the opposite side is ST, and the hypotenuse is RT. Therefore, sin R = opposite / hypotenuse = ST / RT = 2/5. But they told us that RT is 5, so ST / 5 = 2/5, so ST = 2. Therefore, ST is 2 units long. Then, since the triangle is right-angled at S, we can find RS using Pythagoras. RT^2 = RS^2 + ST^2. So 5^2 = RS^2 + 2^2 => 25 = RS^2 + 4 => RS^2 = 21 => RS = sqrt(21). Which matches the coordinates given in the Asymptote code, since R is at (2, sqrt(21)), so RS is vertical from (2,0) to (2, sqrt(21)), so length sqrt(21). So that checks out.

Now, the question is asking for sin T. So angle T is at point T, which is the right angle? Wait no, angle T is at point T, which is connected to S and R. Since the right angle is at S, angle T is one of the acute angles. So in triangle RST, angle T is opposite to side RS. Wait, let me think again. In triangle RST, angle T is at point T. The sides opposite to angles are as follows: side opposite angle R is ST, side opposite angle S is RT, and side opposite angle T is RS. Therefore, sin T would be opposite over hypotenuse. The hypotenuse is RT, which is 5. The side opposite angle T is RS, which is sqrt(21). Therefore, sin T = RS / RT = sqrt(21)/5. But let me check if that's correct.

Alternatively, maybe I can use the coordinates to compute it. Since coordinates are given, maybe I can compute the lengths and then compute the sine. Let me confirm.

Coordinates:

T is at (0,0), S is at (2,0), R is at (2, sqrt(21)).

So, angle T is at (0,0). Let me find the sides adjacent and opposite to angle T. In triangle RST, angle T is between sides ST and RT. So, to find sin T, we can think of the opposite side over hypotenuse. Wait, but in a right-angled triangle, the hypotenuse is opposite the right angle. Since the right angle is at S, hypotenuse is RT. Therefore, for angle T, which is at (0,0), the sides are:

- Opposite side: RS, which is sqrt(21)

- Adjacent side: ST, which is 2

- Hypotenuse: RT, which is 5

Therefore, sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5. Alternatively, if I think of angle T, then in terms of the triangle, sin T = opposite / hypotenuse. So that would be RS / RT. Which is sqrt(21)/5. So that would be the answer. But let me check if that's correct.

Alternatively, maybe I can compute it using coordinates. Let me compute the coordinates. The coordinates of T are (0,0), S is (2,0), R is (2, sqrt(21)). So vector from T to S is (2,0), and vector from T to R is (2, sqrt(21)). So angle at T is between vectors TS and TR. To find sin T, maybe I can compute the angle between those two vectors.

Alternatively, since we know the lengths of the sides, we can use trigonometry. Since we have a right-angled triangle, we can use the sides. Since angle T is adjacent to side ST (length 2) and opposite to side RS (length sqrt(21)), so sin T = opposite / hypotenuse = sqrt(21)/5. Therefore, the answer is sqrt(21)/5. But let me check if that's the case.

Alternatively, maybe I can use the fact that in a right-angled triangle, the two non-right angles are complementary. So angle R + angle T = 90 degrees. Therefore, sin T = sin(90 - R) = cos R. So if I can find cos R, then that would be sin T. Since sin R is 2/5, then cos R can be found using sin^2 R + cos^2 R = 1. Therefore, cos R = sqrt(1 - (2/5)^2) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5. Therefore, sin T = cos R = sqrt(21)/5. So that's another way to get the same answer. Therefore, sin T is sqrt(21)/5.

But let me check if that's consistent with the coordinates. Let me compute the coordinates. The coordinates of the triangle are T(0,0), S(2,0), R(2, sqrt(21)). So angle at T. Let me compute the lengths of the sides adjacent to angle T. The sides adjacent to angle T are ST and RT? Wait, no. In angle T, the sides are ST and RT. Wait, but in a triangle, the sides adjacent to angle T are the two sides that form the angle, which are ST and RT. However, in the right-angled triangle, the hypotenuse is RT, and the other sides are ST and RS. Wait, maybe I need to think again.

Wait, angle T is at point T. The sides forming angle T are ST and RT. The side opposite angle T is RS. So, in terms of trigonometry, sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5. Which is the same as before. Alternatively, if I use coordinates, I can compute the angle. Let me compute the coordinates.

Vector TS is from T to S: (2,0) - (0,0) = (2,0). Vector TR is from T to R: (2, sqrt(21)) - (0,0) = (2, sqrt(21)). The angle between vectors TS and TR is angle T. To find sin T, we can use the cross product formula. The magnitude of the cross product of vectors TS and TR is |TS x TR| = |(2)(sqrt(21)) - (0)(2)| = 2*sqrt(21). The magnitude of TS is sqrt(2^2 + 0^2) = 2. The magnitude of TR is sqrt(2^2 + (sqrt(21))^2) = sqrt(4 + 21) = sqrt(25) = 5. Then, the sine of the angle between them is |TS x TR| / (|TS| |TR|) = 2*sqrt(21) / (2*5) = sqrt(21)/5. Which matches our previous results. Therefore, sin T is sqrt(21)/5.

Therefore, the answer is sqrt(21)/5, which can be written as \boxed{\dfrac{\sqrt{21}}{5}}.

But let me check again if I didn't make any mistake. The problem says sin R is 2/5. Let me confirm that with the coordinates. In triangle RST, angle R is at point R. The sides adjacent to angle R are RS and RT. Wait, no. In angle R, the sides are RS and RT. Wait, but in the right-angled triangle, angle R is not the right angle. The right angle is at S. So angle R is at point R, between RS and RT. Therefore, the sides adjacent to angle R are RS and RT? Wait, no. Wait, angle R is between sides RS and RT? Wait, RS is from R to S, and RT is from R to T. So yes, angle R is between RS and RT. Therefore, in terms of the triangle, the sides adjacent to angle R are RS and RT, but the side opposite angle R is ST. Therefore, sin R = opposite / hypotenuse = ST / RT. Which is 2 / 5, as given. Which matches the problem statement. Therefore, that's correct. Therefore, sin T is sqrt(21)/5. So that's the answer.

Alternatively, since angle R and angle T are complementary, as the triangle is right-angled at S, so angle R + angle T = 90 degrees. Therefore, sin T = sin(90 - R) = cos R. And since sin R = 2/5, then cos R = sqrt(1 - (2/5)^2) = sqrt(21)/5. Therefore, sin T = sqrt(21)/5. So that's another way to get the answer. Therefore, regardless of the method, the answer is sqrt(21)/5.

Therefore, I think that's the correct answer. Let me check once more with the coordinates. Coordinates of T(0,0), S(2,0), R(2, sqrt(21)). So, triangle with right angle at S. So, coordinates are such that ST is horizontal from (0,0) to (2,0), RS is vertical from (2,0) to (2, sqrt(21)), and RT is the hypotenuse from (2, sqrt(21)) to (0,0). So, the length of RT is 5, as given. So, coordinates are consistent with the given information. Therefore, sin T is sqrt(21)/5. Therefore, I think that's the answer.

**Final Answer**
\boxed{\dfrac{\sqrt{21}}{5}}

To solve for $\sin T$ in triangle $RST$, we are given that $\sin R = \frac{2}{5}$, and the triangle is right-angled at $S$. Let's proceed step by step.

---

### Step 1: Understand the Triangle

From the Asymptote code and the description:

- $T = (0, 0)$
- $S = (2, 0)$
- $R = (2, \sqrt{21})$

This places the triangle in the coordinate plane with:

- $S$ at the origin,
- $T$ to the right of $S$,
- $R$ directly above $S$.

The right angle is at $S$, so triangle $RST$ is a **right triangle** with the right angle at $S$. Therefore:

- $RT$ is the **hypotenuse**,
- $RS$ and $ST$ are the **legs**.

---

### Step 2: Use Given Information

We are told that:

$$
\sin R = \frac{2}{5}
$$

In triangle $RST$, angle $R$ is at vertex $R$, and the side opposite to angle $R$ is $ST$, while the hypotenuse is $RT$. So:

$$
\sin R = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{ST}{RT}
$$

We are also told that the length of $RT = 5$, so:

$$
\frac{ST}{5} = \frac{2}{5} \Rightarrow ST = 2
$$

---

### Step 3: Use the Pythagorean Theorem

Now that we know two sides of the triangle, we can find the third using the Pythagorean Theorem:

$$
RT^2 = RS^2 + ST^2
$$

$$
5^2 = RS^2 + 2^2 \Rightarrow 25 = RS^2 + 4 \Rightarrow RS^2 = 21 \Rightarrow RS = \sqrt{21}
$$

---

### Step 4: Find $\sin T$

Angle $T$ is at vertex $T$, and the side opposite to angle $T$ is $RS$, while the hypotenuse is still $RT = 5$. So:

$$
\sin T = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{RS}{RT} = \frac{\sqrt{21}}{5}
$$

---

### Step 5: Confirm Using Complementary Angles

Since the triangle is right-angled at $S$, angles $R$ and $T$ are complementary:

$$
\angle R + \angle T = 90^\circ \Rightarrow \sin T = \cos R
$$

We can compute $\cos R$ using the identity:

$$
\sin^2 R + \cos^2 R = 1 \Rightarrow \cos R = \sqrt{1 - \sin^2 R} = \sqrt{1 - \left(\frac{2}{5}\right)^2} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}
$$

So again, we find:

$$
\sin T = \frac{\sqrt{21}}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{\sqrt{21}}{5}}
$$

Solution:
Because $\triangle RST$ is a right triangle, $\sin R = \frac{ST}{RT}$. So $\sin R = \frac{2}{5} = \frac{ST}{5}$. Then $ST=2$.

We know that $\sin T = \frac{RS}{RT}$. By the Pythagorean Theorem, $RS = \sqrt{RT^2 - ST^2} = \sqrt{25-4} = \sqrt{21}$. Then $\sin T = \boxed{\frac{\sqrt{21}}{5}}$.