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 $m-n$.

  • 1
    $\begingroup$ Hint $156=2^2 \cdot 3 \cdot 13$. $\endgroup$ – dxiv Nov 16 '16 at 0:10
  • $\begingroup$ Divide both sides by n! and you have (n+1)...... (m)=156 $\endgroup$ – fleablood Nov 16 '16 at 0:11
  • $\begingroup$ If you drop the requirement that $m$ and $n$ are two-digit numbers and just assume they are natural numbers greater than one, you can still solve the problem just as easily. $\endgroup$ – Jeppe Stig Nielsen Nov 16 '16 at 1:30
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    $\begingroup$ @JeppeStigNielsen without that requirement there is another possibility, namely $(n,m)=(155,156)$. $\endgroup$ – stewbasic Nov 16 '16 at 5:00
  • $\begingroup$ @stewbasic Good point, thanks. $\endgroup$ – Jeppe Stig Nielsen Nov 16 '16 at 6:55

$m\cdot(m-1)\cdots(n+1) = 156$ 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.

  • 1
    $\begingroup$ I was writting almost word by word the same anser!! O.o (+1)! $\endgroup$ – user378947 Nov 16 '16 at 0:14
  • $\begingroup$ Thank you. Can I ask your opinion on something? Do you think such a question is a fair test question or is it a time-killer? $\endgroup$ – SarpSTA Nov 16 '16 at 0:21
  • $\begingroup$ Totally fair question, IMHO. Factorials require a slightly special way of thinking, but with practice it becomes quick. $\endgroup$ – Dr Xorile Nov 16 '16 at 0:24

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, $m-n=2$.

  • $\begingroup$ I marked the other question since it was earlier but yours is equally good and explaining. Thank you! $\endgroup$ – SarpSTA Nov 16 '16 at 1:09

$$ \binom{m}{n}=\frac{m!}{n!(m-n)!}=\frac{156}{(m-n)!} $$ 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 $n=155$, $m=156$.


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