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Answer: there are 2 possible values for $N$ ($90$ or $405$)


Factors come in pairs with product $N$:
So $f\times45f=N\Rightarrow N=45f^2$.
 
$f$ must be prime because factors of $f$ are also factors of $N$

Factors of $45$ are also factors of $N$ so $3$ and $5$ are factors of $N$

$f\le3$ so $f=2$ or $f=3$

So $N=45\times2^2=90$ or $N= 45\times 3^2 = 405$
 
 
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.