Let $a,b,c,d,e,f\in\mathbb{Z^+}$ where $a+b=c+d=e+f=101$. Prove that there does not exist ${ace\over bdf}={m\over n}$ where $m,n\in\mathbb{Z^+}$ and $m+n\lt101$.
I tried to attempt this question but it was hard. I tried to say that $a=101-b, c=101-d, e=101-f$ and when substituting I got this: $$101^3-101^2 (b+d+f)-101(bd+df+fb)+bdf\over bdf$$ I tried to find the gcd between the numerator and denominator but couldn't. Is it also correct to assume that $a>b,c>d,e>f$? Any help would be appreciated.
There are sign mistakes in the calculation the OP made, but it is leading to a solution:
Assume that such $m,n$ with $m+n < 101$ exist, then
$$\frac{m}n = {101^3-101^2 (b+d+f)+101(bd+df+fb)-bdf\over bdf}$$
$$\frac{m}n +1 = {101^3-101^2 (b+d+f)+101(bd+df+fb)\over bdf}$$
Since $0 < b,d,f < 101$, and $101$ is a prime, the denominator of this fraction ($bdf$) is not a multiple of $101$, but the enumarator is, so we have as the reduced fraction form
$${m \over n}+1 = {101x \over y}$$
with some $x,y \in {\mathbb Z}^+$.
Of course that means
$${m+n \over n} = {101x \over y}$$
and since the right hand side is in reduced fraction form, it follows $m+n \ge 101x \ge 101$, in contradiction to the assumed $m+n<101$.
Also, this proof works the same for any other prime instead of $101$.