There are 2 ways of doing this:

1) Hard way (Using algebra)

n= 5p + 1
n= 7q + 3

n+k = 5p + (1+k)
n+k = 7q + (3+k)

if k = 4
n+4 = 5p + 5 = 5(p+1)
n+4 = 7q + 7 = 7(q+1)

Thus, 4 is the smallest number which gives us no remainder for each case, since any number divisible by 35 must be divisible by both 7 and 5.

2) Easy way (picking numbers)

31 is a good number which satisfies this criteria, and you can see adding 4 to 31 will make it divisible by 35

Hence, (B)

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[/quote]

I tried as follows:

K+n may be one of the 35, 70, 105, 140....etc

last digit of n may be = 1 or 6

here the k+n= 105 satisfying the above condition , considering the k=4 (smallest +ve integer)

ans: B

kourousafama wrote:Hi,

Does anybody have a clear and simple answer for the following problem? The answer given is the OG quant review seems to me very complicated.

Many thanks,

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

The simplest way in my opinion is as follows -

n = 5p + 1 ==> This means that for n to be exactly divisible by 5, we must add 4 or subtract 1
n = 7q + 3 ==> This means that for n to be exactly divisible by 7, we must add 4 or subtract 3

The question asks us what no. "k" should be added to make the no. k+n divisible by 35. If a no. has to be divisible by 35, clearly it must be divisible by both 5 and 7. Looking at the above two equations for n, it is obvious that the least value that must be added to make the no. divisible by both 5 and 7 is 4. Hence B is the answer.

This method works for me. Hopefully it will work for you too!!

Cheers!