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 $m-n$.
$m\cdot(m-1)\cdots(n+1) = 156$ and $m$ and $n$ are two digit numbers. So we are looking for the product of some number of consecutive two digit numbers equaling 156. One two digit number is going to be less than 100, and three is going to be more than 1000, so it's got to be two consecutive numbers. If the problem has a solution, it is therefore 2.
It turns out that $\sqrt{156}\approx 12.5$, so a reasonable guess is 12 and 13. The product of these two numbers is indeed 156. So we can deduce that $m=13$ and $n=11$, and the difference is indeed 2.
The prime factorization of $156$ is $2^2 \times 3 \times 13$. Since it is given that $m$ and $n$ are two digit natural numbers with $m! = 156 n!$, this implies that $m>n$ and $m\times (m-1)\times ... \times (n+1) = 156$. Therefore, we must factor $156$ as the product of consecutive natural numbers among which atleast one should be of two digits. Since $13$ is a prime factor, it must be one of these natural numbers and the only adjacent natural number that can be formed from the remaining factors of $156$ is $12$. This shows that $m=13$ and $m-1=n+1=12 \Rightarrow n=11$. So, $m-n=2$.
$$ \binom{m}{n}=\frac{m!}{n!(m-n)!}=\frac{156}{(m-n)!} $$ 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 $m-n=3$.
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 $n=155$, $m=156$.
