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$\binom{74}{37}-2$ is divisible by :

a) $1369$

b) $38$

c) $36$

d) $none$ $of$ $ these$

I have no idea how to solve this...I tried writing $\binom{74}{37}$ in some useful form but its not helping...any clues?? Thanks in advance!!

Did
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SirXYZ
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  • Hi; I agree, did not read the problem correctly, I have removed the errant comment. Sorry to all. – bobbym Dec 03 '16 at 07:31
  • It is in fact divisible by $37^2=1369$. I had hoped to find $\binom{2n}{n}-2$ is divisible by $n^2$ for each $n$, but that doesn't appear to be the case. (sorry for earlier comment). $\binom{12}{6}-2=922$ is not divisible by $36$, so the pattern doesn't continue. Perhaps it is still true however for prime numbers specifically. The search remains for an elegant way to prove that $\binom{74}{37}-2$ is divisible by $37^2$. – JMoravitz Dec 03 '16 at 07:38
  • See also http://math.stackexchange.com/questions/726845/proof-that-2p-choose-p-equiv-2-pmod-p and http://math.stackexchange.com/questions/817556/check-let-p-be-a-prime-number-show-that-binom2pp-is-congruent-to-2-mo – Gerry Myerson Dec 03 '16 at 07:57

2 Answers2

2

The first line of the following is true by a combinatorial argument (among other arguments) where you count how many ways to choose $p$ marbles from a collection of $2p$ marbles, where half are red and half are blue. $$\begin{align} \binom{p+p}{p} &=\sum_{k=0}^p\binom{p}{k}\binom{p}{p-k}\\ &=2+\sum_{k=1}^{p-1}\binom{p}{k}\binom{p}{p-k}\\ \end{align}$$

Each binomial coefficient in each term in the sum is divisible by $p$ (if $p$ is prime). So mod $p^2$, $$\binom{2p}{p}\equiv2$$ With $p=37$, this shows $1369=37^2$ divides $\binom{74}{37}-2$.

On a multiple choice test, I would expect that the intention is to check if a person has simply memorized that $\binom{2p}{p}\equiv2$ mod $p^2$.

2'5 9'2
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Note that $\binom{74}{37}=\frac{74!}{37!37!}$

Note also that $n!$ contains $\sum\limits_{k=1}^\infty \left\lfloor\frac{n}{p^k}\right\rfloor$ factors of a prime $p$

Counting the number of factors of $19$ in the numerator, we get $\lfloor\frac{74}{19}\rfloor+\lfloor\frac{74}{19^2}\rfloor+\dots=3+0+0+\dots=3$

The number of factors of $19$ in the denominator however, we get each $37!$ has $\lfloor\frac{37}{19}\rfloor+\lfloor\frac{37}{19^2}\rfloor+\dots=1+0+0+\dots=1$ so the denominator has $2$ factors of $19$.

Thus, one of the factors from the numerator remains uncancelled and so $\binom{74}{37}$ is divisible by $19$. This implies $\binom{74}{37}-2$ is not divisible by $19$ and therefore cannot be divisible by $38$.

Similarly, running the same argument for counting the number of factors of $2$, we get a total of $71$ factors of two on the numerator and $68$ factors of two on the denominator, implying that $\binom{74}{37}$ is divisible by four. Thus $\binom{74}{37}-2$ is not divisible by $4$ and therefore cannot be divisible by $36$.

For showing $\binom{74}{37}-2$ is divisible by $37^2$, we note first that $37$ is prime. We conjecture that $\binom{2p}{p}-2$ is divisible by $p^2$ for each prime $p$.

We notice first that $\binom{2p}{p}=\sum\limits_{j=0}^p\binom{p}{j}^2$ and also that $\binom{p}{j}\equiv 0\pmod{p}$ for $0<j<p$ and $\binom{p}{0}=\binom{p}{p}=1$ otherwise. (To see this, use a similar technique as above, noting that there is exactly one factor of $p$ in the numerator and no factors of $p$ in the denominator of $\frac{p!}{j!(p-j)!}$ for each $0<j<p$)

As the terms in the summation are being squared, we notice further that $\binom{p}{j}^2\equiv 0\pmod{p^2}$ for each $0<j<p$ and $\binom{p}{0}^2=\binom{p}{p}^2=1$

Thus $\binom{2p}{p}\equiv \sum\limits_{j=0}^p\binom{p}{j}^2\equiv 1+(0+0+\dots+0)+1\equiv 2\pmod{p^2}$

Finally, $\binom{2p}{p}-2\equiv 0\pmod{p^2}$ so the claim is true.

This proves in particular that $\binom{74}{37}-2$ is divisible by $37^2$

JMoravitz
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