Let the smallest prime factor of N be p, whence the second largest factor, and the highest factor tina wrote down, is ${N}\over{p}$.

Now we have $\frac{N}{p}= 45p$. So ${N}= 45 p^2$.

Since N is a multiple of $45$, it has prime factors of $3$ and $5$ and, because p is the smallest prime factor of N, we can conclude that p can be only $2$ or $3$.

Hence either $N = 45\times 2^2 = 180$ or $N = 45\times 3^2 = 405$

*This problem is taken from the UKMT Mathematical Challenges.*

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