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When a number A is divided by 6, the remainder is 3 and when
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17 Apr 2020, 10:53
17 Apr 2020, 10:53
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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?
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6
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When a number A is divided by 6, the remainder is 3 and when
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17 Apr 2020, 11:06
17 Apr 2020, 11:06
1
Carcass 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?
Show: :: OA
6
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: 6
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Re: When a number A is divided by 6, the remainder is 3 and when
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25 May 2020, 10:41
25 May 2020, 10:41
3
For any of these questions with a mystery number getting divided by a given number such that you get a given remainder, I just pick a number that satisfies (usually the lowest number that would satisfy).
So in this case, you can set A = 9, and B = 21. Divide 9 by 6 and you get 1 w/remainder of 3, divide 21 by 12 and you get 1 w/remainder of 9.
Then plug in.
9^2+21^2 = 522
Then do long division. 522 / 12 gives you 43 remainder of 6.
So in this case, you can set A = 9, and B = 21. Divide 9 by 6 and you get 1 w/remainder of 3, divide 21 by 12 and you get 1 w/remainder of 9.
Then plug in.
9^2+21^2 = 522
Then do long division. 522 / 12 gives you 43 remainder of 6.
