The remainder when $(15^{23} +23^{23})$ is divided by 19, is:

0

The correct answer is Option (4) → 0

We find the remainder of

$15^{23} + 23^{23} \quad \text{when divided by } 19$.

Since 19 is prime, use Fermat’s Little Theorem:

$a^{18} \equiv 1 \pmod{19}$

So,

$a^{23} \equiv a^{5} \pmod{19}$

Step 1: Reduce bases modulo 19

• $15 \equiv -4 \pmod{19}$
• $23 \equiv 4 \pmod{19}$

Step 2: Compute powers

$(-4)^{23} \equiv -4^{5} \pmod{19}$

$4^{5} = 1024 \equiv 17 \pmod{19}$

So,

$(-4)^{23} \equiv -17 \equiv 2 \pmod{19}$

$4^{23} \equiv 4^{5} \equiv 17 \pmod{19}$

Step 3: Add results

$2 + 17 = 19 \equiv 0 \pmod{19}$