Find
a natural number n such that when n is divided by 7, the remainder is 3; when n
is divided by 11, the remainder is 5; and when n is divided by 13, the
remainder is 8.

[Problem revised by Steve Lee, LACC Instructor of
Mathematics.]

**Solution for Problem 8:**

n = 7p + 3 = 11q + 5 = 13r + 8, where p, q, and r are whole
numbers.

Solving for p in terms of r, we get: p = (13r + 5)/7 = r +
(6r + 5)/7.

The smallest value of r
that can make 6r + 5 divisible by 7 is 5.

Therefore (1) r = 5 + 7s, where
s is a whole number.

Similarly, solve for q in
terms of r, we get: (2) q = (13r + 3)/11

Substituting
equation (1) into equation (2), we get

q
= [13(5 + 7s) + 3]/11 = [68 + 91s]/11 = 6 + 8s + (3s + 2)/11.

The
smallest value of s that makes 3s + 2 divisible by11 is 3.

Therefore
s = 3 + 11t, where t is a whole number.

Therefore n = 13r + 8
= 13[5 + 7s] + 8 = 13[5 + 7(3 + 11t)] + 8

= **346 + 13*7*11t**