Difference between revisions of "1967 IMO Problems/Problem 3"
Durianaops (talk | contribs) (Created page with "Let <math>k, m, n</math> be natural numbers such that <math>m+k+1</math> is a prime greater than <math>n+1.</math> Let <math>c_s=s(s+1).</math> Prove that the product <cmath>(...") |
Mathboy100 (talk | contribs) (→Solution) |
||
(9 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
Let <math>k, m, n</math> be natural numbers such that <math>m+k+1</math> is a prime greater than <math>n+1.</math> Let <math>c_s=s(s+1).</math> Prove that the product <cmath>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</cmath> is divisible by the product <math>c_1c_2\cdots c_n</math>. | Let <math>k, m, n</math> be natural numbers such that <math>m+k+1</math> is a prime greater than <math>n+1.</math> Let <math>c_s=s(s+1).</math> Prove that the product <cmath>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</cmath> is divisible by the product <math>c_1c_2\cdots c_n</math>. | ||
− | ---- | + | |
+ | ==Solution== | ||
+ | For any <math>m, n</math>, <math>c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)</math>. | ||
+ | |||
+ | We can therefore write the product in the problem as follows: | ||
+ | |||
+ | <cmath>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</cmath> | ||
+ | <cmath> = \left(\prod_{a = 1}^n m+a+k+1\right)\left(\prod{a = 1}^n m+a-k\right).</cmath> | ||
+ | |||
+ | But, the product of any <math>n</math> consecutive integers is divisible by <math>n!</math>. We can prove this as follows: | ||
+ | |||
+ | <cmath>(t-n+1)(t-n+2) \cdots (t) = \frac{(t+n)!}{t!} = n! \frac{(t+n)!}{t!n!} = n! {t+n\choose t}.</cmath> | ||
+ | |||
+ | Therefore, <math>\prod_{a = 1}^n m+a-k</math> is divisible by <math>n!</math>, and <math>(m+k+1)\left(\prod_{a = 1}^n m+a+k+1\right)</math> is divisible by <math>(n+1)!</math>. However, we are told that <math>m+k+1</math> is prime and therefore it is not divisible by any of the numbers <math>1</math> through <math>n+1</math>. Therefore, <math>\prod_{a = 1}^n m+a+k+1</math> is divisible by <math>(n+1)!</math>. | ||
+ | |||
+ | Finally, it is clear that <math>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</math> is divisible by <math>n!(n+1)! = (1 \cdot 2)(2 \cdot 3)(3 \cdot 4) \cdots (n \cdot (n+1)) = </math><math>c_1c_2\cdots c_n</math>. <math>\square</math> | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | ==Solution 2== | ||
+ | We have that <math>c_1c_2c_3...c_n=n!(n+1)!</math> | ||
+ | |||
+ | and we have that <math>c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)</math> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <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 | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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== | ||
+ | |||
+ | {{IMO box|year=1967|num-b=2|num-a=4}} |
Revision as of 11:04, 13 December 2022
Contents
Problem
Let be natural numbers such that is a prime greater than Let Prove that the product is divisible by the product .
Solution
For any , .
We can therefore write the product in the problem as follows:
But, the product of any consecutive integers is divisible by . We can prove this as follows:
Therefore, is divisible by , and is divisible by . However, we are told that is prime and therefore it is not divisible by any of the numbers through . Therefore, is divisible by .
Finally, it is clear that is divisible by .
~mathboy100
Solution 2
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 |