QANT Guide Q 68 - HELP!

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QANT Guide Q 68 - HELP!

by cpay3245 » Fri May 10, 2013 9:43 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 positive integer k such that k+ n is a multiple of 35?

a) 3
b) 4
c) 12
d) 32
e) 35

I do not know how to tackle this question and do not understand the logic in the answer explanation in the textbook. Please help me!

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by srcc25anu » Fri May 10, 2013 9:55 pm
When positive integer n is divided by 5, it leaves a remiander of 1. That means n can be 1,6,11,16,21,26,31 ....
When positive integer n is divided by 7, it leaves a remiander of 3. That means n can be 3,10,17,24,31,38 ....
K + N should be a multiple of 35.
First lets determine what could n stand for. if we see both possibilities above for n, the first common possibility is 31 (if divided by 5 leaves a remainder of 1 and if divided by 7, leaves a remainder of 3). Closest multiple of 35 near 31 is 35 itself. (n=)31 + (k=)4 = multiple of 35
hence K should be 4
Ans B

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by Brent@GMATPrepNow » Fri May 10, 2013 9:56 pm
cpay3245 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
Here's one approach:

There's a nice rule that says, If, when N is divided by D, the remainder is R, then the possible values of N include: R, R+D, R+2D, R+3D,. . .

First we're told that when n is divided by 5, the remainder is 1.
So, possible values of n are 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, etc.

Next we're told that when n is divided by 7, the remainder is 3.
So, possible values of n are 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, etc.

So, we can see that n could equal 31, or 66, or an infinite number of other values.

Important: Since the Least Common Multiple of 7 and 5 is 35, we can conclude that if we list the possible values of n, each value will be 35 greater than the last value.

So, n could equal 31, 66, 101, 136, and so on.

Notice that, if we take any of these values, we need to add 4 to it so that the sum will be a multiple of 35. So, the smallest value of k is 4 such that k+n is a multiple of 35.

Answer = B

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by GMATGuruNY » Sat May 11, 2013 1:43 am
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
For many test-takers, the best approach will be to make a list of values for n.

When positive integer n is divided by the divisor D, the remainder is R.

Given the information above, the smallest possible value of n will be the remainder R.
To determine the other possible values of n, just keep adding multiples of the divisor D.

When positive integer n is divided by 5, the remainder is 1.
Smallest n = 1.
Now add multiples of 5:
1,6,11,16,21,26,31...

When positive integer n is divided by 7, the remainder is 3.
Smallest n = 3.
Now add multiples of 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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by Atekihcan » Sat May 11, 2013 6:51 am
n must be of the form (multiple of 5 + 1)
So, (n + 4) is divisible by 5.

Also, n must be of the form (multiple of 7 + 3)
So, (n + 4) is divisible by 7.

So, (n + 4) is multiple of both 5 and 7, i.e. multiple of 35.
SO, k = 4