Which is the smallest positive integer that, when divided by 5, leaves a remainder of 2, when divided by 7, leaves a remainder of 3, and when divided by 9, leaves a remainder of 4?
Solution. The double of the sought number, when divided by 5, leaves a remainder of 4; when divided by 7, leaves a remainder of 6; and when divided by 9, leaves a remainder of 8. If we add 1 to the double of the sought number, the resulting number is divisible by 5, 7, and 9. The smallest such positive integer is $5 \cdot 7 \cdot 9=315$, therefore the sought number is $\frac{315-1}{2}=157$.