1967 IMO Problems/Problem 3
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 editorial mistakes 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 and
. 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
, except for the fact that whenever we wrote
,
we will now write
. We will not repeat the proof here.
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 |