Remainders,Divisibility(DS)

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Remainders,Divisibility(DS)

by watashiwasakshi » Thu Jul 07, 2016 12:16 am
If r is the remainder when integer n is divided by 7, what is the value of r?
(a) When n is divided by 21, the remainder is an odd number
(b) When n is divided by 28, the remainder is 3

And
Is n/18 an integer?
(a) 5n/18 is an integer
(b) 3n/18 is an integer
Source: — Data Sufficiency |

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by GMATGuruNY » Thu Jul 07, 2016 2:20 am
Moving forward, please post each problem in a separate thread.
Is n/18 an integer?

a) 5n/18 is an integer.
b) 3n/18 is an integer.
Statement 1: 5n/18 is an integer
If 5n/18 = 1, then n=18/5.
In this case, n/18 is not an integer.
If 5n/18 = 5, then n=18.
In this case, n/18 is an integer.
INSUFFICIENT.

Statement 2: 3n/18 is an integer
If 3n/18 = 1, then n=18/3=6.
In this case, n/18 is not an integer.
If 3n/18 = 3, then n=18.
In this case, n/18 is an integer.
INSUFFICIENT.

Statements combined:
5n/18 - 3n/18 = 2n/18.
Since both 5n/18 and 3n/18 are integers, we get:
2n/18 = 5n/18 - 3n/18 = integer - integer = integer.

3n/18 - 2n/18 = n/18.
Since both 3n/18 and 2n/18 are integers, we get:
n/18 = 3n/18 - 2n/18 = integer - integer = integer.
SUFFICIENT.

The correct answer is C.
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by GMATGuruNY » Thu Jul 07, 2016 2:30 am
watashiwasakshi wrote:If r is the remainder when integer n is divided by 7, what is the value of r?
(a) When n is divided by 21, the remainder is an odd number
(b) When n is divided by 28, the remainder is 3

Statement 1: When n is divided by 21, the remainder is an odd number

It's possible that n=22, since dividing 22 by 21 yields an odd remainder of 1.
If we divide n=22 by 7, we get:
22/7 = 3 R1.
In this case, r=1.

It's possible that n=24, since dividing 24 by 21 yields an odd remainder of 3.
If we divide n=24 by 7, we get:
24/7 = 3 R3.
In this case, r=3.

Since r can be different values, INSUFFICIENT.

Statement 2: When n is divided by 28, the remainder is 3.
In other words:
n is 3 more than a MULTIPLE OF 28.
Since any multiple of 28 is also a multiple of 7, the following is implied:
n is 3 more than a MULTIPLE OF 7.
Thus, dividing n by 7 will yield a remainder of 3, with the result that r=3.
SUFFICIENT.

The correct answer is B.
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by Brent@GMATPrepNow » Thu Jul 07, 2016 4:42 am
watashiwasakshi wrote: Is n/18 an integer?
(a) 5n/18 is an integer
(b) 3n/18 is an integer
One small point: Many students answer this question incorrectly, because they assume that n is an INTEGER.
If n were, indeed, an integer, then the correct answer would be A.
However, since we can't assume that n is an integer, the correct answer is C (for the reasons stated in the above posts).

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by Matt@VeritasPrep » Fri Jul 08, 2016 1:59 pm
For the first problem:

S1:

n = 21*(some integer) + (some odd number)
n = 21*k + odd

When we divide this by 7, we get 3k + (odd/7), so we need to know the remainder of the odd number when divided by 7. We obviously don't know that; INSUFFICIENT

S2:

n = 28*(some integer) + 3

Dividing this by 7, we get 4*(some integer) + (3/7). Since 3 / 7 has remainder 3, that's our answer! SUFFICIENT

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by Matt@VeritasPrep » Fri Jul 08, 2016 2:11 pm
For the second problem:

S1:
5n/18 = m
5n = 18m

If n and m are both integers, then we know that n is a multiple of 18 and m is a multiple of 5. But we don't know that they're both integers! We could have n = 18/5 and m = 1, or whatever; NOT SUFFICIENT

S2:
Same idea as S1. We'll have 3n = 18k, or n = 6k. NOT SUFFICIENT

S1+S2:

5n/18 = integer
3n/18 = integer

(5n/18) - (3n/18) - (3n/18) = integer - integer - integer = integer

so

(-n)/18 = integer

-n = 18 * integer

Since -n is the product of two integers, it itself is an integer. Since -n = a multiple of 18, n must be a multiple of 18.