When positive integer A is divided by 6, the remainder

BTGModeratorVI wrote: ↑

Mon Jun 08, 2020 11:55 am

When positive integer A is divided by 6, the remainder is 3 and when positive integer 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

Given the information, say A = 6 + 3 = 9 and B = 12 + 9 = 21

So, A^2 + B^2 = 9^2 + 21^2 = 81 + 441 = 522

So, the remainder when A^2 + B^2 is divided by 12 = the remainder when 522 is divided by 12 = 6

The correct answer: C

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Mon Jun 08, 2020 11:55 am

When positive integer A is divided by 6, the remainder is 3 and when positive integer 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

BTGModeratorVI wrote: ↑

Mon Jun 08, 2020 11:55 am

When positive integer A is divided by 6, the remainder is 3 and when positive integer 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

Solution:

We can write A = 6a + 3 and B = 12b + 9 for some integers a and b. Therefore,

A^2 + B^2 = (6a + 3)^2 + (12b + 9)^2 = 36a^2 + 36a + 9 + 144b^2 + 216b + 81

We see that all the terms except the constant terms are multiples of 12 regardless what a and b are. The two constant terms add up to 9 + 81 = 90, which has a remainder of 6 when divided by 12.

Answer: C