Question

What is the remainder when $(5555{)}^{2222}+(2222{)}^{5555}$ is divisible by 7?

• A. 0
• B. 1
• C. 2
• D. 7

Answer 1

The correct option is A 0

Theorem: $a^n-b^n$ is divisible by $a-b$
The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively.
Hence, the problem reduces to finding the remainder when $(4{)}^{2222}+(3{)}^{5555}$ is divided by 7.
Now $(4{)}^{2222}+(3{)}^{5555}=(4^2{)}^{1111}+(3^5{)}^{1111}=(16{)}^{1111}+(243{)}^{1111}$ .
Now $(16{)}^{1111}+(243{)}^{1111}$ is divisible by $16+243$ or it is divisible by 259, which is a multiple of 7.
Hence the remainder when $(5555{)}^{2222}+(2222{)}^{5555}$ is divided by 7 is zero.