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When positive integer N is divided by 18, the remainder is x [#permalink]
05 Jun 2018, 07:54

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Expert's post

00:00

Question Stats:

54% (01:42) correct
45% (01:57) wrong based on 33 sessions

When positive integer N is divided by 18, the remainder is x. When N is divided by 6, the remainder is y. Which of the following are possible values of x and y?

i) x = 9 and y = 3 ii) x = 16 and y = 2 iii) x = 13 and y = 7

A) i only B) i and ii only C) i and iii only D) ii and iii only E) i,ii and iii

Re: When positive integer N is divided by 18, the remainder is x [#permalink]
07 Jun 2018, 06:47

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This post received KUDOS

Expert's post

GreenlightTestPrep wrote:

When positive integer N is divided by 18, the remainder is x. When N is divided by 6, the remainder is y. Which of the following are possible values of x and y?

i) x = 9 and y = 3 ii) x = 16 and y = 2 iii) x = 13 and y = 7

A) i only B) i and ii only C) i and iii only D) ii and iii only E) i,ii and iii

Let's examine each statement separately...

i) x = 9 and y = 3 Let's come up with a value of N that satisfies this condition. How about N = 9? 9 divided by 18 = 0 with remainder 9 (i.e., x = 9) ...and 9 divided by 6 = 1 with remainder 3 (i.e., y = 3) Perfect, statement i is TRUE Check the answer choices.....ELIMINATE D

ii) x = 16 and y = 2 Can you come up with a value of N that satisfies this condition? How about N = 16? 16 divided by 18 = 0 with remainder 16 (i.e., x = 16, which WORKS) However, 16 divided by 6 = 2 with remainder 4 (i.e., y = 4. NO GOOD)

How about N = 34? 34 divided by 18 = 1 with remainder 16 (i.e., x = 16, which WORKS) However, 34 divided by 6 = 5 with remainder 4 (i.e., y = 4. NO GOOD)

We can keep testing N-values until we convince ourselves that there are no values of N that makes those values (x = 16 and y = 2) possible. So, statement ii is FALSE Check the answer choices.....ELIMINATE B and E

HOWEVER, if you need more convincing that statement ii is FALSE, we can make the following observations: When positive integer N is divided by 18, the remainder is x: so, we can say that N = 18k + x for some integer k When positive integer N is divided by 6, the remainder is y: so, we can say that N = 6j + y for some integer j We can combine the two equations to get: 18k + x = 6j + y Isolate x to get: x = y + 6j - 18k Factor right side to get: x = y + 6(j - 3k) Rewrite as: x = y + some multiple of 6 This means we'll never have the case where x = 16 and y = 2, because 16 CANNOT be written as 2 + some multiple of 6 So, statement ii is FALSE

iii) x = 13 and y = 7 We must be careful with this one. While it is true that 13 CAN be written as 7 + some multiple of 6, we must also consider the following property of remainders: When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

Based on the above property, when we divide N by 6, the remainder can be 5, 4, 3, 2, 1, or 0 So, the remainder CANNOT be 7 In other words, y CANNOT equal 7 So, statement iii is FALSE ELIMINATE C

Answer: A

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Re: When positive integer N is divided by 18, the remainder is x [#permalink]
15 Sep 2019, 12:51

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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.

Re: When positive integer N is divided by 18, the remainder is x [#permalink]
16 Sep 2019, 06:54

1

This post received KUDOS

Expert's post

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]
24 Mar 2020, 04:09

1

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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.

greprepclubot

Re: When positive integer N is divided by 18, the remainder is x
[#permalink]
24 Mar 2020, 04:09