Let $$a=\frac{72!}{(36!)^2}-1$$
Find whether
- $a$ is odd.
- $a$ is even.
- $a$ is divisible by 71.
- $a$ is divisible by 73.
Multiple answers can be correct.
I was able to find whether $a$ is even or odd in the following way:
There is a method of finding the power of a prime factor in the factorial of a number. If $p$ is a prime factor of any number $q\;!$, and the power of $p$ in $q\;!$ is $n$, then $$n=\left[\frac {q}{p}\right]+\left[\frac {q}{p^2}\right]+\left[\frac {q}{p^3}\right]+\cdots$$ where $\bf{[\;.]}$ denotes Greatest Integer Function. So, in the above expression, putting $q=72$ and $p=2$, we get $n=70$. Putting $q=36$ and $p=2$, we get $n=34$. So, the power of $2$ in the numerator is $70$, while in denominator, it is $68$. Thus, $a+1$ is an even number, making $a$ an odd number.
However, I am not able to check the divisibility of $a$ by $71$ or $73$. What is the method to do that for factorial numbers? I am to solve this by using concepts of binomial theorem.
Since $71$ is prime, $72!$ is divisible by $71$ whereas $36!^2$ is not. Hence $72!/(36!^2)$ is divisible by $71$ and $72!/(36!^2)-1$ is not.
For divisibility by $73$, notice that $$36!^2 = \prod_{k=1}^{36} k^2 \equiv \prod_{k=1}^{36} (-k(73-k)) = \prod_{k=1}^{36} k(73-k) = 72! \mod 73,$$ so the quotient $72!/(36!^2)$ is $1$ modulo $73$ (note that $73$ is prime, thus $72!$ is not divisible by $73$). Since $72!/(36!^2) \equiv 1 \mod 73$, we find that $72!/(36!^2)-1$ is divisible by $73$.