Quant Review- Divisibility

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Quant Review- Divisibility

by DCS80 » Tue Mar 05, 2013 7:29 pm
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest possible integer k such that k+n is a multiple of 35?

A)3
B) 4
C) 12
D) 32
E) 35

answer later.....(quant review wording for answer is very unclear for this problem)
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by Anurag@Gurome » Tue Mar 05, 2013 7:32 pm
DCS80 wrote:When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest possible integer k such that k+n is a multiple of 35?

A)3
B) 4
C) 12
D) 32
E) 35

answer later.....(quant review wording for answer is very unclear for this problem)
n = 5*a + 1.
n = 7*b + 3.
Here, a and b are integers.
Note that the difference between divider and remainder (5 - 1 and 7 - 3) is 4 in both the case.
So add 4 on both sides of each of the 2 equations.
So, we get n + 4 = 5*a + 5 = 5*(a + 1).
n+4 = 7*b + 7 = 7*(b + 1).
This means n+4 is a multiple of both 5 and 7.
Since 5 and 7 are co-prime, n + 4 has to be a multiple of 5 * 7 = 35 as well.
So the smallest possible value of k is 4.

The correct answer is B.
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by GMATGuruNY » Tue Mar 05, 2013 9:39 pm
kannans3 wrote: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k+n is a multiple of 35?

(A) 3
(B) 4
(C) 12
(D) 32
(E) 35
When positive integer n is divided by 5, the remainder is 1.
The smallest possible value of n that satisfies the statement above is the given remainder of 1.
To determine the other possible values of n, just keep adding multiples of the divisor 5:
1,6,11,16,21,26,31...

When positive integer n is divided by 7, the remainder is 3.
The smallest possible value of n that satisfies the statement above is the given remainder of 3.
To determine the other possible values of n, just keep adding multiples of the divisor 7:
3,10,17,24,31...

The smallest value included in both lists is n=31.

Now we can plug in the answers, which represent the smallest possible value of k.
When the correct answer is added to n=31, the sum will be a multiple of 35.
Since we need the smallest possible value of k, we should start with the smallest answer choice.

Answer choice A: k=3
n+k = 31+3 = 34. Not a multiple of 35.
Eliminate A.

Answer choice B: k=4
n+k = 31+4 = 35. Success!

The correct answer is B.

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