Find the smallest number that leaves a remainder of 4 on division by 5, 5 on…

Solution: Each required remainder is one less than its divisor, so adding 1 to the unknown number makes it divisible by each divisor.

Therefore N + 1 must be a common multiple of 5, 6, 7, 8 and 9.

• Compute the least common multiple (LCM) of 5, 6, 7, 8 and 9.

• Prime-power factors: 8 supplies 2^3, 9 supplies 3^2, and include 5 and 7. So LCM = 2^3 * 3^2 * 5 * 7 = 2520.

• The smallest N is then 2520 - 1 = 2519.

Thus the smallest number satisfying all the given remainders is 2519.

Note: All solutions have the form N = k*2520 - 1 for integer k ≥ 1. For example, k = 2 gives 5039, which also meets the conditions but is larger than the smallest.