21015/099) Solve the system of equations

$a(b+c+d+e+f)=184$

$c(a+b+d+e+f)=301$

$e(a+b+c+d+f)=400$

$b(a+c+d+e+f)=225$

$d(a+b+c+e+f)=225$

$f(a+b+c+d+e)=525$

Solution

We note that all the rest are of the form $n(g-n)$
where $g=a+b+c+d+e+f$ and observe that smaller the value, the smaller n and hence b = d

Factoring, we see that
$a(g-a) = 2^2 * 3 * 7$
$b(g-b) = d(g-d) = 3^2 * 5*2$
$e(g-e) = 2^4 * 5^2$
$f(g-f) = 3 * 5^2 * 7$
and
a < b = d < e < f

Because b and d must be greater than a,

so we have the set for a 2,3,4,7

b = 3,5,

d= 3,5

e= 2,5,10

f = 3,5,15 so on

we can quickly find values that satisfy that and the last four equations: they all work out if g=50, which implies that c=7.

This produces the solution
$a=4, b=d=5, c=7, e=8, f=21$