Question

The remainder when ($4^{87}$ + $6^{87}$) is divided by 25, is

• A. Zero
• B. 10
• C. 20
• D. 24

Answer 1

Answer: (C)

20

Answer 2

To find the remainder of ($4^{87}$ + $6^{87}$) when divided by 25, we use modular arithmetic.

We calculate $4^{87} \pmod{25}$: $4^5 = 1024 \equiv -1 \pmod{25}$. Since $87 = 17 \times 5 + 2$, $4^{87} = (4^5)^{17} \cdot 4^2 \equiv (-1)^{17} \cdot 16 \equiv -16 \equiv 9 \pmod{25}$.

We calculate $6^{87} \pmod{25}$: $6^5 = 7776 \equiv 1 \pmod{25}$. Since $87 = 17 \times 5 + 2$, $6^{87} = (6^5)^{17} \cdot 6^2 \equiv 1^{17} \cdot 36 \equiv 36 \equiv 11 \pmod{25}$.

Adding the remainders: $4^{87} + 6^{87} \equiv 9 + 11 \equiv 20 \pmod{25}$.