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Quant Review PS #68

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by tonebeeze » Fri Jan 07, 2011 2:49 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

Quant Review's explanation was a bit convoluted. What is the best approach for these types of remainder questions?
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by anshumishra » Fri Jan 07, 2011 2:58 pm
tonebeeze 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

Quant Review's explanation was a bit convoluted. What is the best approach for these types of remainder questions?
n = 5x + 1 = 7y + 3
5x+1 = {1,6,11,16,21,26,31,...}
7y+3 = {3,10,17,24,31,....}

So one of such number is 31
[By the way: The generic equation of such numbers = 35*z + 31 (get 35 as lcm of 5 and 7, z = any integer)]
n = 31 , so to make n+k a multiple of 35, add 4. (i.e. k=4).

Hence, b
Thanks
Anshu

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by aleph777 » Fri Jan 07, 2011 3:08 pm
I think the quickest solution here is to just pick numbers until you find one that leaves R1 and R3 when divided by 5 and 7, respectively.

So, look for multiples of 5 and add the remainder of 1:

6, 11, 16, 21, 26, 31, 36, 41...

And then for multiples of 7, adding the remainder of 3:

10, 17, 24, 31...

You can stop there, because 31 + 4 = 35, which is a multiple of 35.

Answer: B
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by diaca » Tue Jan 18, 2011 7:40 pm
I got the answer thank you. However I was thinking, has it something to do that the sum of the reminders is 4? So 35-4=31. Could I think the problem in that way?
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by Rahul@gurome » Tue Jan 18, 2011 9:52 pm
diaca wrote:I got the answer thank you. However I was thinking, has it something to do that the sum of the reminders is 4? So 35-4=31. Could I think the problem in that way?
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

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.
Rahul Lakhani
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by OneTwoThreeFour » Wed Jan 26, 2011 8:05 pm
Great Solution! I didn't know you could take the coefficient from the quotient, subtract the remainder, and add it on both sides to make it equal. So much for the tax dollars paid for my education in grade school.

Anyway, thank you so much! I was stuck on this one for quite a well.
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