A heap of stones can be made up into group of 21. When made up into groups of…

Solution:

1. Find the common pattern for numbers that leave remainder 3 when divided by 16, 20, 25 and 45.

LCM(16, 20, 25, 45) = 3600 (since 16 = 2^4, 20 = 2^2·5, 25 = 5^2, 45 = 3^2·5, so LCM = 2^4·3^2·5^2 = 3600).

Therefore any number that leaves remainder 3 for each of those divisors has the form 3600k + 3 for some integer k ≥ 0.

2. Impose the additional condition that the number is divisible by 21.

We need 3600k + 3 to be divisible by 21. Reduce modulo 21:

3600 ≡ 9 (mod 21), so 3600k + 3 ≡ 9k + 3 (mod 21).

Solve 9k + 3 ≡ 0 (mod 21) ⇒ 9k ≡ 18 (mod 21). Divide both sides by gcd(9,21)=3 to get 3k ≡ 6 (mod 7).

Compute the inverse of 3 modulo 7 (which is 5), so k ≡ 5×6 ≡ 30 ≡ 2 (mod 7). The smallest nonnegative k satisfying this is k = 2.

3. Therefore the least number is 3600×2 + 3 = 7203.

Check: 7203 leaves remainder 3 when divided by 16, 20, 25 and 45, and 7203 ÷ 21 = 343, so it is divisible by 21.