\div 7 = 17 \text{ remainder } 1 \Rightarrow \text{Yes}
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- So $k = 7m + 2$. Then $x = 60k = 60(7m + 2) = 420m + 120$. We seek the smallest three-digit such $x$:
- x = 420m + 120 \geq 100 \Rightarrow m \geq 0
- Try $m = 0$: $x = 120$ (valid, three-digit). Check $120 \mod 7 = 1$?
- Thus, the smallest such number is $\boxed{120}$.
- Question: What is the largest integer that must divide the product of any four consecutive odd integers?
- Solution: Let the four consecutive odd integers be $n, n+2, n+4, n+6$, where $n$ is odd. Their product is:
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