Divisibility of $^{72}C_{36}$

Question

$^{72}C_{36}$ is divisible by

(A) $7$

(B) $5$

(C) $11$

(D) $13$

My approach is as follow \begin{align}^{72}C_{36}&=\frac{72!}{36!\cdot36!}\\ &=\frac{72\cdot71\cdots37}{36!}\\ &=2^{18}\times\frac{71\cdot69\cdot67\cdots37}{18!}\end{align}

After this step I am getting confused

Answer 1

This can be solved using Kummer's theorem. For instance, for $p=5$, we have $72_{10}=242_5$ and $36_{10}=121_5$. There is no carry in the addition $121_5+121_5=242_5$, so $5$ does not divide $\binom{72}{36}$.

Answer 2

The answer is 13. Use if $n!=p^m.k$ then maximum value of m is given by $$m=\lfloor \frac{n}{p}\rfloor+\lfloor \frac{n}{p^2}\rfloor +....$$ Where $n,p,m,k \in N \,\,and\,\,p$ is a prime number. For detailed solution you can check this. https://www.mathsdiscussion.com/discussion-forum/topic/binomial-divisibility-theorem/?part=1#postid-104

Answer 3

Which factors of 36 are divisible by 11?

$11,22,33$

$11^6|(36!)^2$

And $72$ divided by $11 = 6$ plus a remainder

$11^6| 72!$ but no greater power of 11 will divide $72!$

$5$ is a little slightly more complicated, since $25$ (and $50$) are both factors of $36!$ and $72!$

$5^8 |36!$ and $5^{16}|72!$

Both $7$ and $13$ divide ${72\choose 36}$

Answer 4

For finding power of particular of prime 'p' in prime farctorization of n! is $\sum_{r=1}^{\infty} [\frac{n}{p^r}]$. (where '[x]' represent greatest integer less than x, once p$^k$ becomes greater than n for first time then no need to check further because then all the terms such that $r\ge k$ are 0 meaning no multiple of $\ $ p$^r$ till n. )

Number of multiples of p from 1 to n = $[\frac{n}{p}]$.

same way for multiples p$^r$ is = $[\frac{n}{p^r}]$

So a multiple of p$^n$ gives p 'n' times in prime factorization of n! , which is counted once in $[\frac{n}{p}]$ - once in $[\frac{n}{p^2}]$-.......once in $[\frac{n}{p^n}]$.

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For example -:Power of 2 in 72! Is 70 , in 36! is 34 , so overall power of 2 in our number is 2($70 -2\times34$)

Same way for 5 in 72!=16 and 36!=8 , so our number not divisible by 5

For 11, 72! = 6 and 36!=3, so our number not divisble by 11.

For 13 , 72!= 5 and 36!=2 , so our number is divisible by 13.

For 7 , 72! = 11 and for 36!=5 , hence our number is divisble by 7