BTGModeratorVI wrote: ↑

Sun Jul 26, 2020 6:44 am

When a number A is divided by 6, the remainder is 3 and when another number B is divided by 12, the remainder is 9. What is the remainder when A^2+B^2 is divided by 12?

A. 4
B. 5
C. 6
D. 10
E. Cannot be determined

Answer: C
Source: Jamboree

When a number A is divided by 6, the remainder is 3
In other words, A is 3 greater than some of multiple of 6.
In other words, A = 6k + 3, for some integer k

When B is divided by 12, the remainder is 9.
Another words, B is 9 greater than some multiple of 12
In other words, B = 12j + 9, for some integer j

What is the remainder when A² + B² is divided by 12?
We have: A² + B² = (6k + 3)² + (12j + 9)²
Expand and simplify: A² + B² = (36k² + 36k + 9) + (144j² + 216j + 81)
Simplify: A² + B² = 36k² + 36k + 144j² + 216j + 90
Rewrite 90 as follows to get: A² + B² = 36k² + 36k + 144j² + 216j + 84 + 6
Factor out at 12 from the first five terms to get: A² + B² = 12(3k² + 3k + 12j² + 18j + 7) + 6

We can now see that A² + B² is 6 greater than some multiple of 12.
So, when we divide A² + B² by 12, the remainder will be 6

Answer: C

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