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Difference between revisions of "1972 IMO Problems/Problem 3"

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== Solution 2 ==  
 
== Solution 2 ==  
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Let p be a prime, and n be an integer. Let <math>V_p(n)</math> be the largest positive integer <math>k</math> such that <math>p^k|n</math>
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WTS: For all primes <math>p</math>, <math>V_p((2m)!)+V_p((2n)!) \ge V_p(m!)+V_p(n!)+V_p((m+n)!)</math>
 
WTS: For all primes <math>p</math>, <math>V_p((2m)!)+V_p((2n)!) \ge V_p(m!)+V_p(n!)+V_p((m+n)!)</math>
  
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Apply Lemma 2.1 to the problem: and we are pretty much done.
 
Apply Lemma 2.1 to the problem: and we are pretty much done.
  
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P.S. This is my work, but perhaps someone have come up with this method before I did.
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Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.

Revision as of 20:32, 20 November 2018

Let $m$ and $n$ be arbitrary non-negative integers. Prove that \[\frac{(2m)!(2n)!}{m!n!(m+n)!}\] is an integer. ($0! = 1$.)

Solution 1

Let $f(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}$. We intend to show that $f(m,n)$ is integral for all $m,n \geq 0$. To start, we would like to find a recurrence relation for $f(m,n)$.

First, let's look at $f(m,n-1)$:

$f(m,n-1)=\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}=\frac{(2m)!(2n)!(n-1)(m+n)}{m!n!(m+n)!(2n-1)(2n-2)}=f(m,n) \frac{(n-1)(m+n)}{(2n-1)(2n-2)}=f(m,n) \frac{m+n}{2(2n-1)}$

Second, let's look at $f(m+1,n-1)$:

$f(m+1,n-1)=\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!}=\frac{(2m)!(2n)!(2m+1)(2m+2)(n-1)}{m!n!(m+n)!(m+1)(2n-1)(2n-2)}=f(m,n) \frac{(2m+1)(2m+2)(n-1)}{(m+1)(2n-1)(2n-2)}=f(m,n) \frac{(2m+1)}{2n-1}$

Combining,

$4f(m,n-1)-f(m+1,n-1)=f(m,n) (\frac{4(m+n-1)}{2(2n-1)}-\frac{2m+1}{2n-1})=f(m,n) \frac{2m+2n-2-2m+1}{2n-1}=f(m,n)$.

Therefore, we have found the recurrence relation $f(m,n)=4f(m,n-1)-f(m+1,n-1)$.

We can see that $f(m,0)=\frac{(2m)!}{m!m!}$ is integral because the RHS is just $_{2m} C _m$, which we know to be integral for all $m \geq 0$.

So, $f(m,1)=4f(m,0)-f(m+1,0)$ must be integral, and then $f(m,2)=4f(m,1)-f(m+1,1)$ must be integral, etc.

By induction, $f(m,n)$ is integral for all $m,n \geq 0$.

Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html

Solution 2

Let p be a prime, and n be an integer. Let $V_p(n)$ be the largest positive integer $k$ such that $p^k|n$

WTS: For all primes $p$, $V_p((2m)!)+V_p((2n)!) \ge V_p(m!)+V_p(n!)+V_p((m+n)!)$

We know \[V_p(x!)=\sum_{a=1}^{\infty} \lfloor{\frac{x}{p^a}}\rfloor\]

Lemma 2.1: Let $a,b$ be real numbers. Then $\lfloor{2a}\rfloor+\lfloor{2b}\rfloor\ge\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor$

Proof of Lemma 2.1: Let $a_1=\lfloor{a}\rfloor$ and $b_1=\lfloor{b}\rfloor$

$\lfloor{2a}\rfloor+\lfloor{2b}\rfloor=2(a_1+b_1)+\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor$

On the other hand, $\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor=a_1+b_1+(a_1+b_1)+\lfloor\{a+b\}\rfloor$

It is trivial that $\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor\ge\lfloor\{a+b\}\rfloor$

Apply Lemma 2.1 to the problem: and we are pretty much done.

Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.

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