A natural number, when divided by 4, 5, 6, or 7, leaves a remainder of 3 in…

When a number leaves the same remainder r on division by several divisors, every such number has the form (LCM of the divisors) × n + r, for a whole number n ≥ 0. The smallest positive number satisfying all the given conditions is therefore LCM of the divisors + r, provided r is smaller than every divisor.

Here the divisors are 4, 5, 6, and 7, and the required remainder is 3. Find the LCM step by step:

1. Write each divisor as a product of primes: 4 = 22, 5 = 5, 6 = 2 × 3, 7 = 7.

2. The LCM takes the highest power of every prime that appears: 22 × 3 × 5 × 7.

3. Multiply these out: 22 × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420.

4. Add the remainder to this LCM: 420 + 3 = 423.

Cross-check by dividing 423 by each divisor independently: 423 ÷ 4 = 105 remainder 3; 423 ÷ 5 = 84 remainder 3; 423 ÷ 6 = 70 remainder 3; 423 ÷ 7 = 60 remainder 3 — every division leaves the required remainder of 3, confirming the value.

Hence, 423 is the smallest natural number that leaves a remainder of 3 when divided by 4, 5, 6, or 7.