<s>[INST] Quick solution of an equation in factorials Now I realize it is not that hard but it is a test question and should have a fast way to solve as indicated by the teacher. However, I fail to see it.

If both $m$ and $n$ are two-digit natural numbers and $m! = 156 \cdot n!$ find the value of $m-n$.

 [/INST] $$
\binom{m}{n}=\frac{m!}{n!(m-n)!}=\frac{156}{(m-n)!}
$$
Thus $(m-n)!$ is a divisor of $156=2^2\cdot 3\cdot 13$. This forces $(m-n)!$ to be a divisor of $2^2\cdot3=12$ (the prime $13$ cannot appear for obvious reasons), leaving $m-n=1$, $m-n=2$ or $m-n=3$.


*

*If $m=n+1$, we have
$$
\binom{n+1}{n}=156=\binom{n+1}{1}=n+1
$$

*If $m=n+2$, we have
$$
\binom{n+2}{n}=78=\binom{n+2}{2}=\frac{(n+2)(n+1)}{2}
$$

*If $m=n+3$, we have
$$
\binom{n+3}{n}=26=\binom{n+3}{3}=\frac{(n+3)(n+2)(n+1)}{6}
$$


The first case is dismissed, because it corresponds to $n=155$.
In the third case, as $n>9$, we have $(n+3)(n+2)(n+1)>12\cdot11\cdot 10>156$.
Thus only the second case can hold and $m-n=2$; precisely, $n^2+3n-154=0$ so $n=11$.
By the way, the third equation has no integer solutions so the only cases are $n=11$, $m=13$ and $n=155$, $m=156$.
</s>