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.