What is the remainder when (17^{23} + 23^{23} + 29^{23}) is divided by 23 ?

Q: What is the remainder when (17^{23} + 23^{23} + 29^{23}) is divided by 23 ?

To determine the remainder when $17^{23} + 23^{23} + 29^{23}$ is divided by 23, we can use Fermat's Little Theorem. Fermat's Little Theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$. Let's apply this theorem to each term in the expression: 1. **$17^{23}$ modulo 23:** According to Fermat's Little Theorem, since 23 is a prime number and 17 is not divisible by 23, we have: \[ 17^{22} \equiv 1 \pmod{23} \] Therefore, $17^{23} = 17^{22} \times 17 \equiv 1 \times 17 \equiv 17 \pmod{23}$. 2. **$23^{23}$ modulo 23:** Since 23 is the modulus, any power of 23 will be congruent to 0 modulo 23: \[ 23^{23} \equiv 0 \pmod{23} \] 3. **$29^{23}$ modulo 23:** Again, using Fermat's Little Theorem, since 23 is a prime number and 29 is not divisible by 23, we have: \[ 29^{22} \equiv 1 \pmod{23} \] We need to find $29 \mod 23$, which is 6 (since $29 - 23 = 6$). Therefore: \[ 29^{23} = 29^{22} \times 29 \equiv 1 \times 6 \equiv 6 \pmod{23} \] Now, we sum the results of each term modulo 23: \[ 17^{23} + 23^{23} + 29^{23} \equiv 17 + 0 + 6 \equiv 23 \equiv 0 \pmod{23} \] Thus, the remainder when $17^{23} + 23^{23} + 29^{23}$ is divided by 23 is 0. **Explanation of Incorrect Options:** - **Option 2 (1):** This would imply that the sum $17^{23} + 23^{23} + 29^{23}$ leaves a remainder of 1 when divided by 23, which contradicts our calculation showing the remainder is 0. - **Option 3 (2):** This suggests a remainder of 2, which is incorrect based on our calculation. - **Option 4 (3):** This suggests a remainder of 3, which is also incorrect as per our calculation. Therefore, the correct answer is Option 1, which states that the remainder is 0.

Question

Accepted Answer

To determine the remainder when $17^{23} + 23^{23} + 29^{23}$ is divided by 23, we can use Fermat's Little Theorem. Fermat's Little Theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$.

Let's apply this theorem to each term in the expression:

1. **$17^{23}$ modulo 23:**

According to Fermat's Little Theorem, since 23 is a prime number and 17 is not divisible by 23, we have:
   $$
   17^{22} \equiv 1 \pmod{23}
   $$
   Therefore, $17^{23} = 17^{22} \times 17 \equiv 1 \times 17 \equiv 17 \pmod{23}$.

2. **$23^{23}$ modulo 23:**

Since 23 is the modulus, any power of 23 will be congruent to 0 modulo 23:
   $$
   23^{23} \equiv 0 \pmod{23}
   $$

3. **$29^{23}$ modulo 23:**

Again, using Fermat's Little Theorem, since 23 is a prime number and 29 is not divisible by 23, we have:
   $$
   29^{22} \equiv 1 \pmod{23}
   $$
   We need to find $29 \mod 23$, which is 6 (since $29 - 23 = 6$). Therefore:
   $$
   29^{23} = 29^{22} \times 29 \equiv 1 \times 6 \equiv 6 \pmod{23}
   $$

Now, we sum the results of each term modulo 23:
$$
17^{23} + 23^{23} + 29^{23} \equiv 17 + 0 + 6 \equiv 23 \equiv 0 \pmod{23}
$$

Thus, the remainder when $17^{23} + 23^{23} + 29^{23}$ is divided by 23 is 0.

**Explanation of Incorrect Options:**

- **Option 2 (1):** This would imply that the sum $17^{23} + 23^{23} + 29^{23}$ leaves a remainder of 1 when divided by 23, which contradicts our calculation showing the remainder is 0.

- **Option 3 (2):** This suggests a remainder of 2, which is incorrect based on our calculation.

- **Option 4 (3):** This suggests a remainder of 3, which is also incorrect as per our calculation.

Therefore, the correct answer is Option 1, which states that the remainder is 0.