Prove that $43^{101}+23^{101}$ is divisible by $66$ using modular arithmetic

Question

Prove that $43^{101}+23^{101}$ is divisible by 66. I was able to find it to be divisible by 22 but not by 66. Please help

It seems to me that none of the answers in the "duplicate" question use modular arithmetic. If that is important to you and you are not satisfied with the answer already given, I suggest you edit your question to clarify what you really need and say more about what you already know. (For example, how did you show the number is divisible by 22?) — David K, May 16 '17 at 16:01

score 4 · Answer 1 · edited May 16 '17 at 22:56

4

$43\equiv -23 \pmod{66}$

$43^{101}\equiv -(23^{101})\pmod{66}$

$43^{101}+23^{101}\equiv 0 \pmod{66}$

edited May 16 '17 at 22:56

amWhy

answered May 16 '17 at 15:50

CY Aries

1 Answers1