A 37

Peter

#1

A 37

**Posted:** Fri May 25, 2007 2:24 am

nkouevda

#2

The largest power of $2$ that divides $\frac{(n+10)!}{n!}$ is $32$ , so it cannot be a perfect square.

**Posted:** Thu Sep 13, 2007 8:45 am

scorpius119

#3

That doesn't quite work ( $n = 54$ , for example).

Suppose the product is a perfect square. Among the ten terms in the product:

- five are even
- at most two are odd and divisible by 3
- at most one is odd and divisible by 5
- at most one is odd and divisible by 7

So at least one term is not divisible by any of 2,3,5,7. This term is relatively prime with the others since any common divisor of two terms is at most 9. So this term is a perfect square, and at least $11^{2}$ .

Now note that this is the only square in our product, otherwise we would have a square greater than or equal to $11^{2}$ differing from another square by at most 9.

In particular, there must be exactly one term not divisble by any of 2,3,5,7. So we must have equality in all of our counts, and no odd term can be divisible by 15, 21, or 35.

Consider the odd terms divisible by 3. Neither are divisible by 2,5,7, and the gcd between either of these terms and any another one is at most 9, therefore a power of 3.

So we can factor out the largest power of 3 from each of these terms. Each quotient that results is relatively prime to all other terms and are both squares. So it follows that each odd term divisible by 3 is either a square or three times a square.

We know from above that neither can be squares, so both are three times a square. But this also produces two squares that are too close (in particular, they differ by 2). This is a contradiction, so the product cannot be a square.

**Posted:** Thu Sep 13, 2007 4:38 pm

pluricomplex

#4

We can prove it by using problem A9 in here.

From ten consecutive numbers we have one which is relatively prime to others. Remove this number from the list we have nine numbers which we can chose five from them which are consecutive. For five consecutive we will prove that there's one prime to others else. Indeed, there are at most two ones which is divisible by $3$ , and if there're two ones then one is odd and one is even. In case we remove one which is even, so one left there're at most two even numbers from $4$ number left or $3$ for the left case. Remove them we have one odd left or two odd but in which there's one is not divisible by $3$ . So there's one number which is odd, not divisible by $3$ . This number from list $5$ consecutive numbers is relatively prime to others else. Remove it we get four numbers which always find two are consecutive, and then of cause relative each other. So we could find three numbers which is relatively to any $9$ numbers else. Now note that if $a$ is square then $a+1,a+2$ are not and $a-1,a-2$ are not of $a>2$ . So in ten consecutive numbers it could not contains more than two squares. This proves the problem.

**Posted:** Fri Sep 14, 2007 12:44 pm

linboll

#5

**Posted:** Mon Feb 16, 2009 7:47 am

mszew · Riemann Hypothesis Offline Joined: 26 Apr 2007 Posts: 471

#6

Re: A 37

**Posted:** Mon Feb 16, 2009 1:31 pm

jevgeniy

#7

There was such a problem at Latvian Open Olympiad in Mathematics:

It is given that $n$ is a natural even number. Let $R = n\left(n+1\right)\left(n+2\right)\left(n+3\right)$ .

a

$R$

b

$R$

**Posted:** Tue May 26, 2009 1:47 pm