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VeritasPrepKarishma
sinchicodo
When 15 is divided by the positive integer k, the remainder is 3, for how many different values of k is this true?

A) 1
B) 2
C) 3
D) 4
E) 5

When 15 is divided by k, remainder is 3 i.e. there are 3 balls leftover after grouping. so k must be greater than 3.
It also means that 12 is completely divisible by k.

Factors of 12 are 1, 2, 3, 4, 6, 12

Out of these, k can be 4, 6 and 12.

Answer (C)

why does it mean that "12 is completely divisible by k?"

Since the remainder is 3, it means 3 is extra. If you remove 3 from 15, whatever is left is completely divisible by k.

Check out the video explaining this here: https://youtu.be/A5abKfUBFSc
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sinchicodo
When 15 is divided by the positive integer k, the remainder is 3, for how many different values of k is this true?

A) 1
B) 2
C) 3
D) 4
E) 5

Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

When 15 is divided by the positive integer k, the remainder is 3 --> \(15=kq+3\), where \(k>3\).

From above: \(15=kq+3\) --> \(kq=12\) --> k is a factor of 12 greater than 3, thus k can be 4, 6, or 12.

Answer: C.

Similar question to practice: if-x-y-represents-the-remainder-that-results-when-the-po-169530.html

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199
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VeritasPrepKarishma
sinchicodo
When 15 is divided by the positive integer k, the remainder is 3, for how many different values of k is this true?

A) 1
B) 2
C) 3
D) 4
E) 5

When 15 is divided by k, remainder is 3 i.e. there are 3 balls leftover after grouping. so k must be greater than 3.
It also means that 12 is completely divisible by k.

Factors of 12 are 1, 2, 3, 4, 6, 12

Out of these, k can be 4, 6 and 12.

Answer (C)

Hi Karisma,

Thank you for the posts. I also read your Veritas blog on "grouping marbles" on this topic. However, I still cannot understand why K > 3. Based on grouping marbles analogy, for 15 marbles, I can have a remainder of 3 for:

2 groups: 6 marbles each
6 groups: 2 marbles each
4 groups: 3 marbles each
3 groups 4 marbles each
1 group: 12 marbles

or for Bunuel's formula 15 = KV + 3, where K can be a number for any of the above marble-group combination.

Can your explanation be made even more simpler to understand why K > 3?

Thank you,

TO
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Hi thorinoakenshield,

When dealing with a 'remainder' question, there's a relationship between what you 'divide by' and the 'possible remainders'

If you're supposed to end up with a remainder of 3, then you CANNOT be dividing by 1, 2 or 3 (since dividing by any of those numbers CANNOT yield a reminder of 3):

For example:

17/1 = 17r0

17/2 = 8r1

17/3 = 5r2

To have a remainder of 3, we need to divide by a number that is GREATER than 3....

eg 7/4 = 1r3

When combined with the possibilities that Karishma defined (1,2,3,4,6,12), the ONLY values that fit both restrictions are 4, 6 and 12....

15/4 = 3r3
15/6 = 2r3
15/12 = 1r3

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Hi thorinoakenshield,

When dealing with a 'remainder' question, there's a relationship between what you 'divide by' and the 'possible remainders'

If you're supposed to end up with a remainder of 3, then you CANNOT be dividing by 1, 2 or 3 (since dividing by any of those numbers CANNOT yield a reminder of 3):

For example:

17/1 = 17r0

17/2 = 8r1

17/3 = 5r2

To have a remainder of 3, we need to divide by a number that is GREATER than 3....

eg 7/4 = 1r3

When combined with the possibilities that Karishma defined (1,2,3,4,6,12), the ONLY values that fit both restrictions are 4, 6 and 12....

15/4 = 3r3
15/6 = 2r3
15/12 = 1r3

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

Thank you! It makes sense now.

TO
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Hi there,
I think the question should have constraints such as k<15. Otherwise 15/20 = 3/4 - leaves us the remainder of 3, so K is unlimited.
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Hi there,
I think the question should have constraints such as k<15. Otherwise 15/20 = 3/4 - leaves us the remainder of 3, so K is unlimited.

15 divided by 20 does not give the remainder of 3 as stated in the stem. As far as remainders are concerned, 15/20 is not the same as 3/4, you cannot reduce like this: 15 divided by 20 gives the remainder of 15 but 3 divided by 4 gives the remainder of 3.
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Bunuel
Hero8888
Hi there,
I think the question should have constraints such as k<15. Otherwise 15/20 = 3/4 - leaves us the remainder of 3, so K is unlimited.

15 divided by 20 does not give the remainder of 3 as stated in the stem. As far as remainders are concerned, 15/20 is not the same as 3/4, you cannot reduce like this: 15 divided by 20 gives the remainder of 15 but 3 divided by 4 gives the remainder of 3.

Got it! I didn't know about the fine point tested on GMAT. Thank you!
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15/k = q +3
15=kq+3
12=kq

prime factors for 12 = 2*2*3

Different values that fit: 2*2, 2*3, 4*3 -> 4,6,12

answer C
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Factors of 12 are 1, 2, 3, 4, 6, 12

Out of these, k can be 4, 6 and 12.

Answer (C)
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