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OG Quant Review 2nd Ed. PS Q68

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Junior | Next Rank: 30 Posts
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by futjim » Wed Jul 07, 2010 7:07 pm
Hi all,

Can anyone help me with the following Question?

68) 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 muliple of 35?

- 3
- 4
- 12
- 32
- 35

I don't understand the explaination at the back of the book. Can anyone help?

Thank you in advance.
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by Rahul@gurome » Wed Jul 07, 2010 8:08 pm
dividend = quotient*divisor + remainder
Here, dividend = n, let quotient be q1 in the 1st case and q2 in the 2nd case.
Then, n = 5q1 + 1 and n = 7q2 + 3
Equating, we get 5q1 + 1 = 7q2 + 3, which implies 5q1 = 7q2 + 2

We know that the units digit of multiple of 5 is either 5 or 0, so units digit of 7q2 + 2 should also be 5 or 0
This implies units digit of 7q2 should be 3 or 8.
7q2 can take the values 28, 63, 98,...
So, q2 = 4, 9, 14..
So, q2 = 4 + 5a, for some positive integer a
This implies, n = 7q2 + 3 = 7(4 + 5a) + 3 = 31 + 35a
So, if k is a positive integer, n+k is a multiple of 35 for k = 4, 39...
For k = 4, n + k = 31 + 35a + 4 = 35(a + 1)
Hence, the smallest value is k = 4

The correct answer is (B).
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by kmittal82 » Thu Jul 08, 2010 5:11 am
From the question, we can infer 2 equations:

n = 5a + 1
n = 7b + 3

where a,b are the quotients you get when dividing n by 5 and 7 respectively.

To make something divisible by 35, it should be divisible by both 5 and 7. If we add 4 to both equations above, we can see that in the first case n becomes divisible by 5 and in the second, it becomes divisible by 7 (i.e. 5a + 5 and 7b + 7).

Thus, 4 is the smallest number you can add.
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