You may also like

A Biggy

Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.

Why 24?

Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.

Factoring a Million

In how many ways can the number 1 000 000 be expressed as the product of three positive integers?

Three Primes

Age 14 to 16 Short
Challenge Level

Answer: 2, 5 and 7

Searching for possibilities
Product is $5\times$sum, so product is a multiple of $5$
$\therefore$ one of the numbers is $5$
$5\times\underline{     }\times\underline{     }=5\times(5+\underline{     }+\underline{     })$
$\Rightarrow \underline{     }\times\underline{     }=5+\underline{     }+\underline{     }$
$3\times7=21$   $5+3+7=15$   numbers too big
$3\times5=15$   $5+3+5=13$   numbers too small
$5\times5=25$   $5+5+5=15$   numbers too big
$2\times7=14$   $5+2+7=14$   yes
All reasonable possibilities tested

Using algebra
Let $p$, $q$ and $r$ be three prime numbers such that $pqr=5(p+q+r)$. Then one of the prime numbers must be $5$, say $r$.

This implies that $5pq=5(p+q+5)\Rightarrow pq=p+q+5\Rightarrow pq-p-q+1=6\Rightarrow (p-1)(q-1)=6$.

Therefore either $p-1=1$ and so $q-1=6$ i.e. $(p,q)=(2,7)$ (or vice versa) or $p-1=2$ and so $q-1=3$ i.e. $(p,q)=(3,4)$ (or vice versa). But $4$ is not prime, so the only triple of primes which satisfies the condition is $(2,5,7)$.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.