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Re: DS: GMATPrep Remainder [#permalink]
03 Feb 2008, 22:43

Expert's post

jimmyjamesdonkey wrote:

I missing you at this part:

m=(5k+2)/4 - m has to be integer. k has to be even and not divisible by 4

Why does k have to be even, and not divisible by 4?

1. m has to be an integer. 2. (5k+2) has to be even and divisible by 4 3. (5k+2) is even when 5k is even. Therefore k is even. 4. if 5k is divisible by 4, (5k+2) will not divisible by 4: 5*4i+2=4*(5i)+2. Therefore, (5k+2) has not to be divisible by 4 => 5k has not to be divisible by 4 5. k is even and indivisible by 4. 6: 2,6,10.... _________________

Re: DS: GMATPrep Remainder [#permalink]
04 Feb 2008, 02:31

1

This post received KUDOS

My approach is :

n=25x+13

multiples of 25 are -> 25,50,75 thus n would be ->38,63,88

Apply 1-> I have 3 values -> Not sufficient Apply 2-> I will have one value ->63, there could be more above 100, thus applying both 1&2 you get n=63 _________________

Re: When the positive integer n is divided by 25, the remainder [#permalink]
25 Jan 2013, 22:58

I used the number pluggin approach until I got this no: 63. When I was about to hit B as the answer, I saw the option A .. Highly unlikely that A couldn't have been there for a reason. .

thought that there must be numbers greater than 100 which will satisfy B. So Chose C :D _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: When the positive integer n is divided by 25, the remainder [#permalink]
06 Oct 2013, 00:55

Bunuel wrote:

When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that n=25q+13, so n could be 13, 38, 63, 88, 113, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> n=20p+3. n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.

(1)+(2) The only value of n which both statements is 63. Sufficient.

Answer: C.

Out of the highlighted values only 63 satisfies we need to find another value isn't it otherwise we can't claim insufficiency? How do we find the other number quickly... 3,23,43,83 and 103 don't satisfy the 1st equation _________________

Re: When the positive integer n is divided by 25, the remainder [#permalink]
06 Oct 2013, 02:24

Expert's post

fozzzy wrote:

Bunuel wrote:

When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that n=25q+13, so n could be 13, 38, 63, 88, 113, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> n=20p+3. n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.

(1)+(2) The only value of n which both statements is 63. Sufficient.

Answer: C.

Out of the highlighted values only 63 satisfies we need to find another value isn't it otherwise we can't claim insufficiency? How do we find the other number quickly... 3,23,43,83 and 103 don't satisfy the 1st equation

Can you please elaborate what you mean? Thank you. _________________

Re: When the positive integer n is divided by 25, the remainder [#permalink]
06 Oct 2013, 02:33

What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113

statement 2

n=20p + 3 => 3,23,43,63,83,103

For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?

don't we have to match the equations here? _________________

Re: When the positive integer n is divided by 25, the remainder [#permalink]
06 Oct 2013, 02:47

1

This post received KUDOS

Expert's post

fozzzy wrote:

What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113

statement 2

n=20p + 3 => 3,23,43,63,83,103

For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?

don't we have to match the equations here?

Well, there cannot be only one value that satisfies both n=25q+13 (13, 38, 63, 88, 113, ...) and n=20p+3 (3, 23, 43, 63, 83, 103, ...).

Next, there is a way to derive general formula for n (of a type n=mx+r, where x is a divisor and r is a remainder) based on above two statements:

Divisor x would be the least common multiple of above two divisors 25 and 20, hence x=100.

Remainder r would be the first common integer in above two patterns, hence r=63.

Therefore general formula based on both statements is n=100m+63. Hence n can be 63, 163, 263, ...