A product of 199 factorials

Let $S = 1!\times 2!\times 3! \times\cdots\times200!.$ We have

$\begin{aligned} S &= 1!\times2!\times3!\times4!\times\cdots\times199!\times200! \\ &= 1!\times(1!\cdot 2)\times3!\times(3!\cdot 4) \times \cdots \times 199! \times (199! \cdot 200) \\ &= (1!\times3!\times\cdots\times199!)^2 \times (2 \times 4 \times \cdots \times 200) \\ &= (1!\times3!\times\cdots\times199!)^2 \times 2^{100} \times 100! \qquad \textbf{(*)} \end{aligned}$

If $n = 100,$ then $\dfrac{S}{n!} = (1!\times3!\times\cdots\times199!)^2 \times 2^{100},$ and if $n = 99,$ then $\dfrac{S}{n!} = (1!\times3!\times\cdots\times199!)^2 \times 2^{100} \times 100.$ Therefore, $99, 100$ are values of $n.$ We prove that there are no other values.

Assume that there are other values of $n.$ Let $\dfrac{S}{n!} = m^2$ for some integer $m,$ and let $S = k^2 \cdot 100!$ where $k^2$ is the perfect square from $(*).$ Then we have $k^2\cdot 100! = m^2 \cdot n!$ If $1 \le n \le 96,$ the right-hand side of this equation will contain an even number of factors of the prime $97,$ while the left-hand side will contain an odd number of factors of $97$ (since $100!$ is divisible by $97$ only once). This is a contradiction, therefore no value of $n < 97$ will work. By inspection, $n = 97, 98$ do not work, and we found already that $n = 99, 100$ are solutions, so we now have to consider $n \ge 101.$

Take the equation $k^2 \cdot 100! = m^2 \cdot n!$ once again. If $101 \le n \le 201,$ then $n!$ is only divisible once by the prime $101,$ so the right-hand side of the equation contains an odd number of factors of $101.$ We know that $101 \not \mid 100!,$ so the left-hand side has an even number of factors of $101,$ again a contradiction. Therefore, any other values of $n$ must satisfy $n \ge 202.$

Take $n \ge 202.$ By Bertrand's Postulate, there exists a prime $p$ between $\lfloor n/2 \rfloor$ and $n.$ Then $n!$ is only divisible by $p$ once, and $p \not \mid 100!.$ So, $p$ appears an odd number of times in $m^2 \cdot n!$ but appears an even number of times in $k^2\cdot 100!,$ another contradiction.

Therefore, no other values of $n$ exist such that $\dfrac{S}{n!}$ is a perfect square. The answer is $100 + 99 = \boxed{199}.$

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Note: An alternate way to check $n \ge 202$ would be to use the prime $211$ and then check $n = 202, 203, \ldots, 210$ ; I used a different way to avoid excessive computation.

We work first with the numerator. Rewrite the expression by factoring the highest even term from each even factorial term. $(1!)^2 * 2 *(3!)^2 * 4 * ... * (199!)^2 * 200$

Then regroup the terms,

$= (1!*3!*...*199!)^2* (2*4*...*200)$

Then factor out 2^100,

$= (1!*3!*...*199!)^2) * 2^{100} *100!$

Removing the largest square part, the question reduces to:

When is $100!/n!$ a perfect square?

Clearly for n = 100 and n = 99, we have perfect squares. We shall verify there are no others.

For n = 98, we have (99)(100) is not a perfect square.

For n = 97, we have (98)(99)(100) is not a perfect square.

For n < 97, we have 97 is a prime factor of 100!/n! with multiplicity one. Hence 100!/n! is not a perfect square.

The answer is then 99 + 100 = 199.

Let $A = \frac{1! \times 2! \times 3! \times \cdots \times 200!}{n!}$ .

One of the requirements for $A$ to be a perfect square, is that $A$ is integer. For $n \geq 211$ , the denominator is divisible by the prime number $211$ , whereas the numerator isn't (it can be trivially decomposed into factors that are all smaller). Therefore, $A$ is then not integer, so we must have $n < 211$ . Note that this is a necessary condition, but not necessarily a sufficient one (it is, but I won't bother proving it). For the moment, let us not worry about this, and see for what values $n < 211$ , $A$ is the square of a rational number.

Note that, in order to test whether a number is a rational square, we can freely divide and multiply by any square: the resulting number is a rational square if and only if the original number is. We will make full use of this.

First off, note that for instance $59! \times 60!$ can be written as $60 \times (59!)^2$ , so we can do away with the square and be left with $60$ . Generalizing this, if we divide $A$ by $(1!)^2$ , $(3!)^2$ , $(5!)^2$ , and all the way up to $(199!)^2$ , $A$ reduces to

$A_1 = \frac{2 \times 4 \times 6 \times \cdots \times 200}{n!}$

We can divide this by $(2^{50})^2$ , and be left with

$A_2 = \frac{100!}{n!}$

Finally, let us multiply by $(n!)^2$ , to get $A_3 = 100! \times n!$ .

An integer (like $A_3$ ) is a square if and only if every prime number divides it an even number of times (this is easy to prove using unique prime factorization).

Let us consider the prime factor $97$ (not coincidentally the largest prime number below 100). It divides $100!$ once, and $n!$ is divisible $\lfloor n/97 \rfloor$ times. Clearly we must then have that $\lfloor n/97 \rfloor$ is odd, which together with the restriction $n < 211$ , means that

$97 \leq n < 2\times97$

Now consider the prime factor $101$ . This does not divide $100!$ , so $\lfloor n/101 \rfloor$ must be even. This (together with $n < 211$ ) means that either $n < 101$ or $n \geq 202$ .

Combining the two regimes, we are left with the possibilities $n \in \{97,98,99,100\}$ . For $n=97$ and $n=98$ , we find $A_3 = 98\times99\times100$ and $A_3 = 99\times100$ respectively, which are both obviously not squares. For $n=99$ and $n=100$ , we find $A_3 = 100$ and $A_3 = 1$ , which are squares. Hence, in these cases, and only in these cases, is $A$ the square of a rational number, and potentially integer.

Finally, we should remember to check whether $A$ is integer in these cases. This is however trivially true for all $n \leq 200$ .

The only solutions are thus $n=99$ and $n=100$ , and the final answer is $99+100 = \boxed{199}$

Let $P = 1! \times 2! \times 3! \times ... \times 200!$ . Then, using the identity $n! = (n-1)! \times n$ , we obtain

$\begin{aligned} P &= 1! \times 2! \times 3! \times ... \times 200! \\ &= 2 \times 3! \times (3! \times 4) \times 5! \times (5! \times 6) \times ... \times 199! \times (199! \times 200) \\ &= (3! \times 5! \times ... \times 199!)^2 \times (2 \times 4 \times ... \times 200) \\ &= (3! \times 5! \times ... \times 199!)^2 \times 2^{100} \times 100! \textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ }\textrm{ } (*) \end{aligned}$

We want to divide $P$ by $n!$ to obtain a square. Observe that the first two terms in $(*)$ are both squares, so we can divide $(*)$ by $100!$ to obtain a square. Since $100! = 99! \times 10^2$ , we can also divide $(*)$ by $99!$ to obtain another square.

Now, we shall show that there are no other possible $n$ . Observe that $211$ is a prime, so if $n \geq 211$ , then $211$ appears as a prime factor in the denominator but not in the numerator, and we would have $\frac{P}{n!}$ being a non-integer. So $n < 211$ . Furthermore, since $199$ is prime, so we can easily see from $(*)$ that $199$ appears exactly twice as factors of $P$ . If $199 \leq n < 211$ , then $n!$ would have exactly one factor of $199$ ; thus if $\frac{P}{n!}$ is an integer, then it will have exactly one factor of $199$ and cannot be a square. So $n < 199$ .

Now, because $101$ is prime, the only times it appears as factors of $(*)$ are in $101!, 103!, ..., 199!$ , which are all squared. Therefore, $101$ appears an even number of times. If we choose $101 \leq n \leq 198$ , then $n!$ would have exactly one factor of $101$ and thus $\frac{P}{n!}$ will have an odd number of factors of $101$ and cannot be a square. So $n < 101$ . But $97$ is also prime, and besides appearing as a factor in the squared term of $(*)$ , it also appears exactly once in the factor $100!$ . So $97$ appears as a factor of $P$ an odd number of times, so we would need to cancel out at least one factor of $97$ in the denominator. So $97 | n! \Rightarrow n \geq 97$ . Thus, we have $97 \leq n \leq 100$ .

We may easily check that $n = 97$ or $n = 98$ do not make $\frac{P}{n!}$ a square, by observing that $\frac{P}{97!} = (3! \times 5! \times ... \times 199!)^2 \times 2^{100} \times 98 \times 99$ and that $\frac{P}{98!} = (3! \times 5! \times ... \times 199!)^2 \times 2^{100} \times 99$ . Clearly, $99$ and $98 \times 99$ are not squares, so $\frac{P}{n!}$ is not a square for $n = 97$ and $n = 98$ .

Therefore, the only possible $n$ are $n = 99$ and $n=100$ , so the sum of all $n$ is $199$ .

By taking the number of times each number occurs across the factorials, the numerator can be re-written: $2^{199} * 3^{198} * 4*{197} * ... * 199^{2} * 200^{1}$ .

Now, since even bases have odd powers, the numerator is a perfect square times $2*4*6*8*...*200 = 2^{100} * 100!$

Thus, the fraction is a perfect square only when n! has the same primes with odd powers as 100!. Since 100! is divisible by the primes 53, 59, 61, ... 89, 97, exactly once, n! must have an odd number of all of them. This can only occur when $(2k-1)*97 \leq n < (2k*53)$ , for some integer k. Furthermore, if $n \geq 211$ , n! will contain primes the denominator does not, and therefore the original fraction won't even be an integer. Combining these, we see $97 \leq n < 106$

For $n < 100$ , $\frac{100!}{n!}$ is a perfect square iff n! has the same prime with odd powers as 100!.

For n = 97, $\frac{100!}{97!} = 100*99*98$ , which is not a perfect square.

For n = 98 $\frac{100!}{98!} = 100*99$ , which is also not a perfect square.

For n = 99 $\frac{100!}{97!} = 100$ , a perfect square, so n = 99 is a solution

For $100 < n < 106$ , 101 divides n! exactly once, so $\frac{n!}{100!}$ cannot be a perfect square.

The only solutions are 100 and 99, giving a sum of 199

By the identity: $n!(n+1)! = (n!)^2(n+1)$ for all $n\in\mathbb{N},$ the product of the factorials $1!\cdot 2! \cdots 200!$ can be expressed as $(1!)^2 2 \cdot (3!)^2 4 \cdot (5!)^2 6 \cdots (197!)^2 198\cdot (199!)^2 200.$ Since $2\cdot 4 \cdots 200 = ( 2\cdot 1)(2\cdot 2)\cdots(2\cdot 100) = 2^{100}100!,$ the product above becomes $100! \cdot 2^{100}A^2=100!(2^{50} A)^2,$ where $A = 1!\cdot 3!\cdot 5! \cdots 199!.$ Let $n$ be a positive integer such that $\frac{100! \cdot (2^{50}A)^2}{n!}\quad\quad (*)$ is a perfect square. Notice that $n\le 210.$ For otherwise the numerator of $(*)$ would contain the prime factor $211$ , which is impossible. Since there are an even number of prime factors that are larger than $100$ in the numerator of $(*)$ (as factors of $A^2$ ) while each prime factor between $100$ and $210$ appears only once in $n!$ , it follows that $n \le 100.$ Since there is an odd number of prime factor $97$ in the numerator of $(*)$ , it follows that $n \ge 97.$ Hence there are only $4$ cases to consider for the value of $n$ : $97, 98, 99,$ and $100.$ We note a basic fact from number theory that for all positive integers $a$ and $b$ if $ab^2$ is a perfect square, then $a$ is a perfect square.

$\textbf{case I}$

If $n=97$ then $(*)$ becomes $98\cdot 99 \cdot 100 (2^{50}A)^2 = (98)(99)(10\cdot 2^{50}A)^2.$ Since $(98)(99),$ ending in a $2,$ is not a perfect square, this case is impossible.

$\textbf{case II}$

If $n=98$ then $(*)$ becomes $99\cdot 100 (2^{50}A)^2= 99(10\cdot 2^{50}A)^2.$ Since $99$ is not a perfect square, this case is impossible.

$\textbf{case III}$

If $n=99$ then $(*)$ becomes $100 (2^{50}A)^2=(10\cdot 2^{50} A)^2,$ which is a perfect square, as required.

$\textbf{case IV}$

If $n=100$ then $(*)$ becomes $(2^{50}A)^2,$ which is a perfect square, as required.

Hence there are only two values of $n$ namely $\boxed{n=99}$ and $\boxed{n=100}$ for which $(*)$ is a perfect square. The sum of these $n$ 's equal $199.$

first consider the numerator n! * (n+1)!=(n!)^2 (n+1) using this we pair (1! * 2!) (3! * 4!) (5! * 6!) *(7! * 8!) ........... (197! *198!) *(199! * 200!) we pair up all of the squares as a single one say k^2 then numerator reduces to (k^2) 2 * 4 * 6 * 8 *......198 * 200 taking 2 common and clubbing it with the square part we get numerator as K^2 *100! it can be easily seen that can be either only 100 or 99. since if n>100 then denominator after cancellation with 100! from numerator will not be a perfect square. and also if n<99 then it leaves 99 in the numerator which cannot be reduced .

To begin with, let us modify the expression $1!\times2!\times3!\times\dots\times200!$ a bit.

$1!\times2!\times3!\times\dots\times200!=$

$=1\times2\times3!\times(3!\times4)\times5!\times(5!\times6)\times\dots\times199!\times(199!\times200)=$

$=(3!\times5!\times7!\times\dots\times199!)^{2}\times2\times4\times6\times\dots\times200=$

$=(3!\times5!\times7!\times\dots\times199!)^{2}\times2\times1\times2\times2\times2\times3\times\dots\times2\times100=$

$=(3!\times5!\times7!\times\dots\times199!)^{2}\times2^{100}\times(1\times2\times3\times4\times\dots\times100)=$

$=(3!\times5!\times7!\times\dots\times199!)^{2}\times2^{100}\times100!=$

$=(3!\times5!\times7!\times\dots\times199!\times2^{50})^{2}\times100!$

Therefore, the original expression will now be

$\frac{(3!\times5!\times7!\times\dots\times199!\times2^{50})^{2}\times100!}{n!}$

We have now achieved a complete square with the exception of $100!$ .

Speaking of which, let us see if any factorial can be a perfect square. If the largest prime factor $p$ of the factorial $m!$ (where $m$ is a positive integer) does not occur in the factorial again, no matter how you expand it essentially means that the factorial can not be a perfect square. If you do not encounter integers of the type $2p$ , $3p$ or any $tp$ (where $t$ is a positive integer), in the expansion of the factorial, then the best case scenario we could possibly get would be to have a perfect square times a prime number, which is in no case a perfect square.

Consequently, we learn that $100!$ is in no way a perfect square because $97$ is the largest prime in the expansion and it is not a divisor of $98,99$ or $100$ . This means that if we get rid of it the expression is going to be a perfect square. Thus, one possible solution for $n$ is $n=100$ .

However, note that $100=10^{2}$ and $100!=99!\times100$ . This allows us to tweak the original expression even further, giving us

$\frac{(3!\times5!\times7!\times\dots\times199!\times2^{50}\times10^{2})^{2}\times99!}{n!}$

which essentially provides us with the second solution for $n$ , which is $n=99$ .

The only thing left at this point is to see if there are any other solutions for $n$ . Since $99$ and $98$ are not a perfect squares, we may not use the method we did for getting $n=99$ . Furthermore, since $97$ is a prime integer it is evident that $\frac{99!}{n!}$ can not be a square for any $1\leq n\leq99$ , again due to the fact that it is not a divisor of $98$ and $99$ . Thus, any sum left of $99!$ for the fraction can not be a square because $97$ does not repeat for any expansion.

We can now see that the only solutions are $n_{1}=100$ and $n_{2}=99$ . Therefore, the required sum is $n_{1}+n_{2}=199$ .

Note that $(2k)! (2k-1)! = {(2k-1)!}^2 \cdot 2k$

So we can rewrite the original expression $X$ as: $\begin{aligned} X &= \frac{1}{n!} (\prod_{k=1}^{k=100}(2k-1)!)^2 \cdot \prod_{k=1}^{k=100}(2k) \\ &= \frac{1}{n!} (\prod_{k=1}^{k=100}(2k-1)!)^2 \cdot 2^{100} \cdot 100! \end{aligned}$

The terms in the numerator are a perfect square except for $100!$ . So in order for $X$ to be a perfect square, it is necessary and sufficient that $Y = \frac{100!}{n!}$ be a perfect square.

This can't happen if $n \gt 100$ : $n!$ has $101$ as a prime factor, so it doesn't divide evenly into $100$ .

Now, if $n \lt 97$ then $Y$ has exactly one factor of $97$ in it (a prime number). So it can't be a square.

And if $97 \le n \lt 99$ then $Y$ has exactly one factor of $11$ in it, so it can't be a square either.

All that remains is to check $n=100$ , where we get $Y = 1$ , and $n=99$ we get $Y = 100$ . These are both squares, so these are the only two solutions. They sum to $\boxed{199}$ .

the answer is given in the title of the question -_-

We can rewrite the given number as: $\frac{(1!)^2\cdot (3!)^2\cdot (5!)^2 \cdots (199!)^2 \cdot 2\cdot 4\cdot 6\cdots 200}{n!} =$ $\frac{(1! \cdot 3! \cdot 5! \cdots 199!)^2 \cdot 2^{100} \cdot 100!}{n!} =$ $(1! \cdot 3! \cdot 5! \cdots 199! \cdot 2^{50})^2 \cdot \frac{100!}{n!}$

Now we have to find some $n$ for which $(1! \cdot 3! \cdot 5! \cdots 199! \cdot 2^{50})^2 \cdot \frac{100!}{n!}$ is a square.

We can see that the only available numbers for $n$ are $100$ and $99$ .

To understand why we have to notice that $101$ and $97$ are prime numbers, so $n$ can not be smaller than $97$ in order to have an even exponent of $97$ in the prime factorization of the given number. If $n$ is greater of equal to $101$ it has to be bigger than $202$ in order to have an even exponent of $101$ in the prime factorization of the given number. But by induction $n$ must be infinite. In fact we know that between $x$ and $2x$ with $x>1$ there is always a prime number, therefore there will be a prime between $101$ and $202$ . In order to have an even exponent of that prime $n$ must be bigger that $2$ times the prime... and so on...

The only possible values of $n$ are: $100$ , $99$ , $98$ and $97$ . And the only that actually works are $100$ and $99$

So the answer is $\fbox{199}$

• 1!×2!×3!×⋯×200!/n! = 2x4x8x10x................200(1!3!5!................199!)^2/n! = (2^100)100!(1!3!5!................199!)^2/n! Now n=100 or 99 otherwise the number isn't a perfect square. Therefore the sum is 199