Find the smallest natural number which when divided by 3 leaves remainder 1, when divided by 4 leaves remainder 2, and when divided by 5 leaves remainder 3.

Let the number be N. Given: N ≡ 1 (mod 3) N ≡ 2 (mod 4) N ≡ 3 (mod 5) Notice the pattern: Remainder is always 2 less than the divisor. So consider N + 2: If N ≡ 1 (mod 3) ⇒ N + 2 ≡ 3 (mod 3) ⇒ N + 2 ≡ 0 (mod 3) If N ≡ 2 (mod 4) ⇒ N + 2 ≡ 4 (mod 4) ⇒ N + 2 ≡ 0 (mod 4) If N ≡ 3 (mod 5) ⇒ N + 2 ≡ 5 (mod 5) ⇒ N + 2 ≡ 0 (mod 5) So N + 2 is divisible by 3, 4 and 5. LCM(3, 4, 5) = 60 Thus N + 2 = 60k, with k a positive integer. Smallest positive N occurs when k = 1: N + 2 = 60 × 1 = 60 N = 60 − 2 = 58 Check: 58 ÷ 3 = 19 remainder 1 58 ÷ 4 = 14 remainder 2 58 ÷ 5 = 11 remainder 3 All conditions satisfied. Answer: 58.