Quant Review PS #68

Question

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?

anshumishra · Answer

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

(Every mistake is a lesson learned )

aleph777 · Answer

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

diaca · Answer

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?

Rahul@gurome · Answer

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
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

OneTwoThreeFour · Answer

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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