Question

What is the remainder when $7^{13}-{41}^{13}+{75}^{13}-{42}^{13}$ is divided by 66?

• A. 2
• B. 64
• C. 1
• D. 0

Answer 1

Answer: (D)

0

Here 66 cab be written as the product of prime numbers
66 = $2\times 3\times 11$
reduce the given expression in the power of (11-1)
then $7^3-{41}^3+{75}^3-{42}^3$
now 75 and 42 are divisible by 3, so removing these terms.
the expression becomes $7^3-{41}^3$
Now reduce the expression in power of (3-1)
the expression becomes 7-41 = -36 which is divisible by 2.
so the remainder is 0.