Difference between revisions of "1967 IMO Problems/Problem 3"
Mathboy100 (talk | contribs) (→Solution) |
Mathboy100 (talk | contribs) (→Solution) |
||
Line 8: | Line 8: | ||
<cmath>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</cmath> | <cmath>(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)</cmath> | ||
− | <cmath> = (\prod_{a = 1}^n m+a+k+1)(\prod{a = 1}^n m+a-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: | But, the product of any <math>n</math> consecutive integers is divisible by <math>n!</math>. We can prove this as follows: | ||
Line 14: | Line 14: | ||
<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> | <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> | + | 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)) = c_1c_2\cdots c_n</math>. <math>\square</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)) = c_1c_2\cdots c_n</math>. <math>\square</math> |
Revision as of 22:24, 12 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 |