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When a number A is divided by 6, the remainder is 3 and

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by BTGModeratorVI » Fri May 29, 2020 6:38 am

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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
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Re: When a number A is divided by 6, the remainder is 3 and

by Brent@GMATPrepNow » Wed Jun 03, 2020 4:55 am
BTGModeratorVI wrote: ↑
Fri May 29, 2020 6:38 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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Re: When a number A is divided by 6, the remainder is 3 and

by quichopipe » Wed Jun 03, 2020 12:39 pm
\(\frac{A}{6}=\ Q\ +\ \frac{3}{6}\ and\ \frac{B}{12}\ =\ Q\ +\frac{9}{12}\)

Therefore...

A = 6Q + 3
B = 12Q + 9

We want to know that is the remainder when (A^2 + B^2) is divided by 12. Well, we know what A and B is so let's plug in

Let's expand the numerator and denominator
\(\frac{\left(36Q^2\ +\ 36Q\ +9\right)\ +\ \left(144Q^2\ +\ 216Q\ +\ 81\right)}{12}\)

From here, we notice that the only two elements that ARE NOT divisible by 12 are 9 and 81, so we know our remainder must come from there. Let's add those two elements:
\(\frac{9}{12}+\frac{81}{12}\ =\ 7\left(\frac{6}{12}\right)\)

Our remainder is 6.

Answer choice C.
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Re: When a number A is divided by 6, the remainder is 3 and

by Scott@TargetTestPrep » Sat Apr 03, 2021 10:00 am
BTGModeratorVI wrote: ↑
Fri May 29, 2020 6:38 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
Solution:

We can let A be 6m + 3 for some integer m and B be 12n + 9 for some integer n. Therefore,

A^2 + B^2 = (6m + 3)^2 + (12n + 9)^2 = 36m^2 + 36m + 9 + 144n^2 + 216n + 81 = 36m^2 + 36m + 144n^2 + 216n + 90

We see that all the terms of the final expression are a multiple of 12 except the last term. Since the last term is 90 and 90/12 = 7 R 6, the remainder is 6.


Answer: C

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