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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
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arc601 wrote:
I don't feel this approach is right, but I got the right answer. Brent, please let me know if this approach has any validity.

So, I said:

18Q + x = N = 6Q + y

18Q + x = 6Q + y

Then I plugged in numbers for x and y, for example:

18Q + 9 = 6Q + 3

Then I imagined Q was 1, so we get

27 = 9

It's plain that the two sides aren't equal, but they are multiples of one another.

I used the same thing to look at the other solutions, and saw they weren't multiples of one another.

I read your solution again, and I think I may have made a mistake in writing Q for both sides instead of Q sub 1 and Q sub 2. We are told N is the same, but we're not told Q is the same.

So, I guess I could say

18Q_1 + 9 = 6Q_2 + 3

Given Q_1 and Q_2 could be different values.

The other solutions won't be multiples of each other, and therefore can't be equal whatever Q_1 and Q_2 are. For example, if x = 13 and y = 17, we get

18Q_1 + 13 = 6Q_2 + 7

if both Qs are 1, we get

31 = 13, which are clearly not multiples of each other, and therefore no matter what the Q's are for both sides, the two sides will never be equal.

I found the same to be true for the other answer choices. None of them resulted in multiples.

Was on the right path? Or was getting the right answer just a coincidence?

PS Sorry, if my line of reasoning was confusing. It felt confusing to me writing it.


It's fine to write 18Q_1 + x = 6Q_2 + y
But it might be less confusing to write: 18k + x = 6j + y (where j and k are integers)

Now check the answer choices..
i) x = 9 and y = 3
We get: 18k + 9 = 6j + 3 (it this possible given that k and j are positive integers?)
Rearrange to get: 6j - 18k = 9 - 3
Simplify to get: 6(j - 3k) = 6
Divide both sides by 6 to get: j - 3k = 1
We can see that there are many ways this can be possible. For example, j = 4 and k = 1 is one solution.

Keep going with other statements....
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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
Thanks Brent.
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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
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Please tell me if this is right
N = 18k+ x
N = 6j + y
y-x = 18k - 6j
y-x = 6(3k-j)
This means y-x is a multiple of 6
So A) x=9 y=3 y-x=-6 which is divisible by 6
B) x=16 y=2 y-x=-14 which is not divisible by 6 so eliminate it
C) x=13 y=7 since y<6 eliminate it
Only A is left.
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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
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