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# when a number A is divided by 6, the remainder is 3 and when another

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Joined: 18 Feb 2019
Posts: 718
Location: India
GMAT 1: 460 Q42 V13
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when a number A is divided by 6, the remainder is 3 and when another  [#permalink]

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14 Feb 2020, 19:54
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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
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Joined: 11 Sep 2015
Posts: 4605
GMAT 1: 770 Q49 V46
Re: when a number A is divided by 6, the remainder is 3 and when another  [#permalink]

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15 Feb 2020, 08:53
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Top Contributor
kiran120680 wrote:
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 A² + B² is divided by 12?

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

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
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

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when a number A is divided by 6, the remainder is 3 and when another  [#permalink]

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16 Feb 2020, 12:12
1
kiran120680 wrote:
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

We can write A as $$A = 6i + 3$$ where $$i$$ is any nonegative integer. Similarly $$B = 12j + 9$$ (Note: do not use $$i$$ here as A and B are unrelated). Next we can write $$A^2 + B^2$$ as:

$$(6i + 3)^2 + (12j+ 9)^2$$

Now this next step is important, we are trying to find the remainder after dividing by 12. We can immediately eliminate any term that is a multiple of 12 since those terms will be divisible by 12, and do not affect the remainder. Whatever left that is not divisible by 12 will be part of our remainder.

To demonstrate the first step: $$(12j+ 9)^2$$ will give us three terms but we already know two of them are multiples of 12. The only useful term we will keep is $$9^2 = 81$$.

Thus $$(6i + 3)^2 + (12j+ 9)^2 => 36i^2 + 2 * 6i * 3 + 9 + 81 => 9 + 81 => 90 => 6$$.

90/12 gives us a remainder of 6.

Ans: C
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Re: when a number A is divided by 6, the remainder is 3 and when another  [#permalink]

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16 Feb 2020, 13:54
2
kiran120680 wrote:
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

least value of A is 3
least value of B is 9
3^2+9^2=90
90/12 leaves a remainder of 6
C
Re: when a number A is divided by 6, the remainder is 3 and when another   [#permalink] 16 Feb 2020, 13:54
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