3
$\begingroup$

This is a question from Bosnia and Hercegovina mathematical Olympiad $2002$.

There are already many similar questions, which I have done in the past.

For $5$ consecutive integers

A more general case for $3$ consecutive integers.

For $6$ consecutive integers

And similar proves are available for $3$ (Which was once asked in Korean Mathematical Olympiad) and $4$ (Indian national mathematical Olympiad) consecutive integers.

I also know the general result which is here but I am no way going to use that in a competition.

Please try to give a high school argument as supposed to be given in a national Olympiad.

Thanks

$\endgroup$
  • $\begingroup$ I am still thinking about a basic proof. Maybe this will help: Multiplying 10 consecutive numbers will result in a number with last digit 0. Multiplying any number with itself ($x^2$) will have the last digit 0, only if x has the last digit 0. $\endgroup$ – P. Siehr Jul 17 '17 at 15:29
  • $\begingroup$ @P. Siehr Actually at least the last two digits will be $0$, but I am not sure it is a good approach to look only at the divisibility by $2$ and $5$ $\endgroup$ – charmd Jul 17 '17 at 15:31
3
$\begingroup$

Ah got it. Here is a Wordpress article which gives a proof which is not so simple but works fine at Olympiad level.

$\endgroup$
  • $\begingroup$ I was very close to solution $A9$ but before I could complete it you came up with the article. But the other solutions are also very beautiful. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:36
  • $\begingroup$ @астонвіллаолофмэллбэрг, did you use the same argument they are using? If not, then your answer will be gladly welcomed. $\endgroup$ – Vidyanshu Mishra Jul 17 '17 at 15:39
  • $\begingroup$ @VidyanshuMishra No, it's actually very similar, I don't want to post it. But I had something else in mind too, which I thought I could use differently. For example, consider the product of seven consecutive integers, which we can write as $(x-3)(x-2) (x-1)x(x+1)(x+2)(x+3)$. Now, expanding this out, and actually simplifying as a polynomial, we see this is equal to $x(x^2(x-7)^2 + 36)$. Now, if the product is a perfect square, then so is $x(x^2(x-7)^2 + 36)$. If $x$ is co - prime to $36$, then it's coprime to $x^2(x-7)^2 + 36$ as well. But then, each one will have to be a square. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:43
  • $\begingroup$ So either $x$ is not co prime to $36$, or there are two squares differing by $36$, wherein we know all the possibilities. But the question is, what if $x$ isn't co prime. There I was stuck (in the seven case) before I changed approach to counting primes in the interval $(x,x+10)$. Feel free to add thoughts, but I think this is a stuck point, although the polynomial observation is useful. $\endgroup$ – астон вілла олоф мэллбэрг Jul 17 '17 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.