A natural number leaves a remainder of 3 when divided by 5, and a remainder of 4 when divided by 7. What is the smallest such number?

Numbers leaving remainder 3 on division by 5 are 3, 8, 13, 18, 23, … Test each against division by 7: 3 leaves 3, 8 leaves 1, 13 leaves 6, and 18 leaves 4 because $18=7\times 2+4$. So the smallest number that satisfies both conditions is 18.