Difference between revisions of "1967 IMO Problems/Problem 3"
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+ | ==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] | ||
+ | |||
+ | |||
+ | ==Remark and correction (added by pf02, September 2024)== | ||
+ | |||
+ | 1. The two solutions are essentially the same. They are | ||
+ | just different expressions of the same idea. Unfortunately, | ||
+ | both are written very sloppily, but the interested reader | ||
+ | can easily correct the mistakes, improve on the arguments, | ||
+ | and make sense of the proofs. | ||
+ | |||
+ | 2. Both solutions are incomplete. They both assume (without | ||
+ | saying it) that <math>m + 1 > k</math>. Maybe the problem meant to say | ||
+ | this, but it didn't, so we have to see what happens if | ||
+ | <math>m + 1 \le k</math>. There are two cases: | ||
+ | |||
+ | First, assume <math>m + 1 \le k \le m + n</math>. In this case, | ||
+ | <math>(c_{m + 1} - c_k)(c_{m + 2} - c_k) \cdots (c_{m + n} - c_k) = 0</math>, | ||
+ | and we can say that <math>0</math> is divisible by anything. | ||
+ | |||
+ | Next, assume <math>m + n < k</math>. In this case, each factor of the | ||
+ | first product is negative, and we will consider the product | ||
+ | |||
+ | <math>P = (c_k - c_{m + 1})(c_k - c_{m + 2}) \cdots (c_k - c_{m + n})</math> | ||
+ | |||
+ | for the sake of working with positive factors. <math>P</math> differs from | ||
+ | the product in the problem by at most a sign, so it is enough to | ||
+ | show that <math>P</math> is divisible by <math>c_1c_2 \cdots c_n</math>. | ||
+ | |||
+ | The proof of this fact goes along the same lines as the proof when | ||
+ | <math>m + 1 > k</math>. We will give the proof for the sake of having a | ||
+ | complete, properly written proof. | ||
+ | |||
+ | Using <math>p(p + 1) - q(q + 1) = (p + q + 1)(p - q)</math> we get that | ||
+ | <math>c_1c_2 \cdots c_n = (n + 1)! \cdot n!</math>. We also get that | ||
+ | <math>P = \prod_{p = 1}^n (k + m + p + 1) \cdot \prod_{p = 1}^n (k - m - p)</math>. | ||
+ | |||
+ | We have that <math>n!\ |\ \prod_{p = 1}^n (k - m - p)</math> (i.e. the product is | ||
+ | divisible by the factorial) because | ||
+ | |||
+ | <math>\frac{\prod_{p = 1}^n (k - m - p)}{n!} = | ||
+ | \frac{(k - m - 1)!}{(k - m - 1 - n)! \cdot n!} = {k - m - 1 \choose n}</math> | ||
+ | |||
+ | which is an integer. (Note that this all makes sense because we | ||
+ | are in the case when <math>m + n < k</math>.) | ||
+ | |||
+ | Let us now show that <math>(n + 1)!\ |\ \prod_{p = 1}^n (k + m + p + 1)</math>. | ||
+ | Let <math>P_1 = \prod_{p = 1}^n (k + m + p + 1)</math>. We have that | ||
+ | |||
+ | <math>\frac{(k + m + 1) \cdot P_1}{(n + 1)!} = | ||
+ | \frac{(k + m + 1) \cdot \prod_{p = 1}^n (k + m + p + 1)}{(n + 1)!} = | ||
+ | \frac{\prod_{p = 0}^n (k + m + p + 1)}{(n + 1)!} = | ||
+ | \frac{(k + m + n + 1)!}{(k + m)! \cdot (n + 1)!} = | ||
+ | {k + m + n + 1 \choose n + 1}</math> | ||
+ | |||
+ | which is an integer. | ||
+ | So <math>(n + 1)!\ |\ (k + m + 1) \cdot P_1</math>. But none of the factors | ||
+ | of <math>(n + 1)!</math> is a divisor of <math>(k + m + 1)</math> because <math>(k + m + 1)</math> | ||
+ | is prime, and <math>n + 1 < k + m + 1</math>. So | ||
+ | <math>P_1 = \prod_{p = 1}^n (k + m + p + 1)</math> must be divisible by | ||
+ | <math>(n + 1)!</math>. | ||
+ | |||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1967|num-b=2|num-a=4}} |
Latest revision as of 16:39, 7 September 2024
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]
Remark and correction (added by pf02, September 2024)
1. The two solutions are essentially the same. They are just different expressions of the same idea. Unfortunately, both are written very sloppily, but the interested reader can easily correct the mistakes, improve on the arguments, and make sense of the proofs.
2. Both solutions are incomplete. They both assume (without saying it) that . Maybe the problem meant to say this, but it didn't, so we have to see what happens if . There are two cases:
First, assume . In this case, , and we can say that is divisible by anything.
Next, assume . In this case, each factor of the first product is negative, and we will consider the product
for the sake of working with positive factors. differs from the product in the problem by at most a sign, so it is enough to show that is divisible by .
The proof of this fact goes along the same lines as the proof when . We will give the proof for the sake of having a complete, properly written proof.
Using we get that . We also get that .
We have that (i.e. the product is divisible by the factorial) because
which is an integer. (Note that this all makes sense because we are in the case when .)
Let us now show that . Let . We have that
which is an integer. So . But none of the factors of is a divisor of because is prime, and . So must be divisible by .
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 |