M3. Three positive integers have sum 25 and product 360 . Find all possible triples of these integers.
Solution
Let the integers be $a, b, c$. Then $a+b+c=25$ and $a b c=360=2^{3} \times 3^{2} \times 5$. We must split this factorised product up into parts so that the sum of the resulting numbers is 25 .
It is possible, of course, to make a list of all the different ways of doing this, but there are 32 possibilities and ensuring that no alternatives are missing is difficult. Such an approach depends on the list being exhaustive and so omissions result in it being invalid. However, there are several methods which avoid making such a list.

Method 1 :
This focuses on the fact that one of the three numbers is divisible by 5 . Without loss of generality, take this to be $a$. It is either $5,10,15$ or 20 , since 25 is too large to obtain the required sum.
If $a=5$, then $b+c=20$ and $b c=72$. It is possible to list all the factorisations of 72 and show that none of them produce a sum of 20 . However, a better approach is to consider the quadratic $x^{2}-20 x+72=0$. The sum of its roots is 20 and the product is 72 , so the roots will be $b$ and $c$. However, the discriminant of this quadratic is negative, so there are no real roots.
This approach is useful in checking all such cases.
If $a=10$, the quadratic is $x^{2}-15 x+36=0$, which has roots 3 and 12 . Hence there is a solution $10,3,12$.
If $a=15$, the quadratic is $x^{2}-10 x+24=0$, which has roots 4 and 6 , and so there is a solution $15,4,6$.
If $a=20$, the quadratic is $x^{2}-15 x+18=0$, and again the discriminant is negative and there are no roots.
Hence there are exactly two triples of numbers, namely $4,6,15$ and $3,10,12$.
Method 2:
This focuses on the largest number of the three. Since their sum is 25 , this is greater than 8 and smaller than 24 . Since it is also a factor of 360 , the only possible values are 9,10 , 12, 15, 18 and 20. The approach in Method 1 can now be used to eliminate 9,18 and 20. The values 10 and 12 result in the same triple.

Method 3:
This focuses on the $2 \mathrm{~s}$ in the factorisation of 360 . Since $a+b+c=25$, they cannot be all even, nor can two be odd and one even, and since $a b c=360$, they cannot be all odd. It follows that two are even and one is odd. The factor of 8 is split between the two even numbers.
Without loss of generality, let $a=2 a_{1}$ and $b=4 b_{1}$, with $a_{1}, b_{1}, c$ formed from the factors 3,3 and 5 . Now $b_{1}$ is not a multiple of 5 or 9 , since the sum would be greater than 25 , so it is either 1 or 3 .
If $b_{1}=1$, then $b=4$ and $2 a_{1}+c=21$. Using the factors 3,3 and 5 , the only possibility is $a_{1}=3, c=15$ and so $a=6$.
If $b_{1}=3$, then $b=12$ and $2 a_{1}+c=13$. Using the factors 3 and 5 , the only possibility is $a_{1}=5, c=3$ and so $a=10$.
Hence there are two triples of numbers, namely $4,6,15$ and $3,10,12$.