A number leaves a remainder of 3 when divided by 5, and a remainder of 4 when divided by 7. What is the smallest such positive number?
Answer: B
Let the number be N. We have \(N = 5k+3\) and \(N = 7j+4\). We can list the numbers satisfying the first condition: 3, 8, 13, 18, 23, ... Now we check which of these leaves a remainder of 4 when divided by 7. \(3\div 7\) rem 3. \(8\div 7\) rem 1. \(13\div 7\) rem 6. \(18\div 7 = 2 \times 7 + 4\). The remainder is 4. So, 18 is the smallest such number.