n=5a+1 and n=7b+3
find k when (n+k)/35

solution&explanation: the number can be divided by 35 (prime factorization -> 5*7) when it's divided by both 5 and 7. Because n is not divisible neither by 5 (remainder 1) nor by 7 (remainder 3) we conclude that for n to be divided by both 5 and 7 it must be added first the remainder 1 and then 7, as 5 is not divisible by 7 and we need our (n+k) to be divisible by both 5 and 7. So it's (1+7)=8.

scorpion1978 wrote:Hi,
Can some tell me how to solve this problem .. i am half way. It is from the official GMAT Quant review 2nd edition page 70.

When positive integer n divide by 5 the remainder is 1. when n is divided by 7, the remainder is 3. What is the smallest integer K such that K+n is a mutiple of 35.

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

Thank you for your help and support
Best Regards

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.

When positive integer n divide by 5 the remainder is 1. when n is divided by 7, the remainder is 3. What is the smallest integer K such that K+n is a mutiple of 35.

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

let p,q be 2 integers.
Then n can be written as 5p+1 and 7q+3.
5p+1 = 7q+3
5p=7q+2
p=(7q+2)/5
As p is an integer, 7q+2 is multiple of 5.
q=4 makes p=6
Thus n can be 31 (one of the many possible values)
if we add 4 to n, it will become 35 and thus will be divisible by 35
IMO B

Thank you for all your swift replies....got it now.

Question here--please help the non-mathematically-inclined! I had to use a more abstract and primitive approach. Since (k + n) has to be divisible by 5*7, and n is divisible by neither, k must compensate for the missing integer units. The remainder of n/5 is 1, so if n were 4 units larger, it would've been divisible by 5; the remainder of n/7 is 3--so n is lacking 4 units to be divisible by 7. Thus, I reasoned, k has to equal 4 to make its sum with n divisible by both 5 and 7. Was this a legitimate approach, or was I simply lucky in getting the correct answer? [This is important for me to know, since the proper mathematical approach described above would evaporate from my brain in 10 minutes; if I could use my primitive line of thought for such problems, I'd rather do that than attempt to start speaking the same language as the writers of the OG answer explanations.]