In modular arithmetic, what is the remainder when 15^23 + 23^23 is divided by 19?

Introduction:
This question tests your ability to work with large powers using modular arithmetic. Direct calculation of 15^23 and 23^23 is impossible by hand, so you must use properties of remainders and patterns in powers to find the remainder when divided by 19.

Given Data / Assumptions:

• We need the value of (15^23 + 23^23) modulo 19.
• 19 is a prime number.
• We are allowed to use properties of modular arithmetic, such as cyclicity of powers.

Concept / Approach:
When we work modulo a number, we can replace any base with its remainder modulo that number. Also, powers of a given base often repeat in cycles. Since 19 is prime, Fermat type arguments and modular patterns are especially convenient.

Step-by-Step Solution:
Step 1: Reduce the bases modulo 19. 15 mod 19 = 15. 23 mod 19 = 23 - 19 = 4. So we need (15^23 + 4^23) mod 19.
Step 2: Find 15^2 mod 19. 15^2 = 225. 225 mod 19 = 225 - 19 * 11 = 225 - 209 = 16. So 15^2 ≡ 16 (mod 19). That implies: 15^4 ≡ 16^2 = 256 ≡ 256 - 19 * 13 = 256 - 247 = 9 (mod 19). Continuing this way, or by using a calculator for residues only, one finds that: 15^23 ≡ 2 (mod 19). Step 3: Work with 4^23 modulo 19. Observe that 4^2 = 16 and 4^3 = 64 ≡ 64 - 57 = 7 (mod 19). Powers of 4 modulo 19 repeat in a cycle of length 9. After working through this pattern, we find: 4^23 ≡ 17 (mod 19). Step 4: Add the remainders. 15^23 + 23^23 ≡ 2 + 17 = 19 ≡ 0 (mod 19). Thus the required remainder is 0.

Verification / Alternative Check:
We can check consistency: if the sum of two numbers is exactly divisible by 19, their remainders must add up to a multiple of 19. Here, 2 + 17 = 19, which is 19 * 1, confirming that the remainder is 0.

Why Other Options Are Wrong:
4, 3, 1, and 2 are all non-zero remainders. Our detailed remainder calculations show that the expression is perfectly divisible by 19. Therefore these options do not match the correct modular result.

Common Pitfalls:
Common mistakes include trying to compute the huge powers directly, ignoring modular reductions, or making arithmetic errors while reducing powers step by step. Always reduce the base first and use patterns or cycles in exponents to manage large powers efficiently.

Final Answer:
The expression 15^23 + 23^23 leaves a remainder of 0 when divided by 19.