Difference between revisions of "1967 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | The solution can be found here | + | 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 be natural numbers such that is a prime greater than Let Prove that the product is divisible by the product .
Solution
We have that
and we have that
So we have that We have to show that:
is an integer
But is an integer and is an integer because but does not divide neither nor because is prime and it is greater than (given in the hypotesis) and .
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 |