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Difference between revisions of "1967 IMO Problems/Problem 3"

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==Solution==
 
==Solution==
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The solution can be found here. [https://artofproblemsolving.com/community/c6h21117p137234]
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We have that <math>c_1c_2c_3...c_n=n!(n+1)</math>
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and we have that <math>c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)</math>
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So we have that <math>(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)=\frac{(m+n-k)!}{(m-n)!}\frac{(m+n+k+1)!}{(m+k+1)!}</math> We have to show that:
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<math>\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\frac{(m+n-k)!}{(m-n)!n!}\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}</math> is an integer
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But <math>\frac{(m+n-k)!}{(m-n)!n!}=\binom {m+n-k}n</math> is an integer and <math>{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}=\binom {m+n+k+1}{n+1}\frac 1{m+k+1}</math> is an integer because <math>m+k+1|m+n+k+1!</math> but does not divide neither <math>n+1!</math> nor <math>m+n!</math> because <math>m+k+1</math> is prime and it is greater than <math>n+1</math> (given in the hypotesis) and <math>m+n</math>.
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The above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [https://aops.com/community/p392191]  
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1967|num-b=2|num-a=4}}
 
{{IMO box|year=1967|num-b=2|num-a=4}}

Revision as of 12:18, 29 January 2021

Problem

Let $k, m, n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1.$ Let $c_s=s(s+1).$ Prove that the product \[(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)\] is divisible by the product $c_1c_2\cdots c_n$.

Solution

We have that $c_1c_2c_3...c_n=n!(n+1)$

and we have that $c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)$

So we have that $(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)=\frac{(m+n-k)!}{(m-n)!}\frac{(m+n+k+1)!}{(m+k+1)!}$ We have to show that:

$\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\frac{(m+n-k)!}{(m-n)!n!}\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}$ is an integer

But $\frac{(m+n-k)!}{(m-n)!n!}=\binom {m+n-k}n$ is an integer and ${(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}=\binom {m+n+k+1}{n+1}\frac 1{m+k+1}$ is an integer because $m+k+1|m+n+k+1!$ but does not divide neither $n+1!$ nor $m+n!$ because $m+k+1$ is prime and it is greater than $n+1$ (given in the hypotesis) and $m+n$.

The above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [1]

See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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