Factorial Product Is Square

If $n=k+x$ we can write $n!$ as $(k+x)(k+x-1)...(k+1)k!$ then $n!k!=(k+x)(k+x-1)...(k+1)(k!)^2$ .

So $(k+x)(k+x-1)...(k+1)$ is a perfect square if and olny if $n!k!$ is a perfect square too.

In particular if $n$ is a perfect square $n!(n-1)!$ is a perfect square too.

We want $n<=200$ . We check that $196$ is the largest perfect square before $201$ so it works.

Now we just need to check if there is any number greater than $196$ that works.

$197$ and $199$ are primes so they don't work because $n!k!$ just have the prime factor ( $197$ or $199$ ) once since $k<n$

$200$ and $198$ don't work because they have the prime factors $199$ and $197$ so we need $200!199!$ or $198!197!$ to have the prime factor twice but they don't work since $200$ and $198$ are perfect squares.

So $196$ it's the answer.

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According to the problem ,

We want $n!k!$ as a perfect square , where $k < n$

Since we have the freedom of choosing the " $k$ " ,

We can set $k = (n-1)$ .

Well , you'll say - "Why do we do this ?"

Okay . Observe that

$n! = n \cdot (n-1)!$

Now, we'll have :

$n! \cdot k! = \left[ n \cdot (n-1)! \right] \times (n-1)!$

And so ,

$n! \cdot k! = n \times \left[ (n-1)! \right]^2$

Therefore , by setting $k = (n-1)$ , we get the following conditions -

For $n!k!$ being a perfect square ,

$n$ must be a perfect square too !

Therefore , we need to find the largest value for $n \leq 200$ where $n$ is a perfect square .

Hence we get

$14^2 < 200 < 15^2$

Therefore ,

$n = 14^2 = \boxed{196}$

Cheers !

When we have ${n!} \times {k!}$ , then ${n!} \times {k!} = n \times {(n-1)} \times {(n-2)} \times \ldots \times {k!^2}$ . So, if we want perfect squares, $\frac {n!}{k!}$ need to be a perfect square too. The greatest integer which is smaller than 200 is $196 = 14^2$ . Then, the answer is 196.

I'm sorry if my explanations wasn't good or my english was incomprehesible.

$196 = 14^2 \Rightarrow 196!195! = (14 \cdot 195!)^2$ .

From the question, since $k<n$ , so to make our life easier, let $k=n-1$ . Since $n!(n-1)! = n((n-1)!)^{2}$ , so $n$ must be a perfect square. This implies to $n=\boxed {196}$ , since $196$ is the biggest perfect square number which is smaller to $200$ .

'n' is equal to or less than 200 ..........................Given.

n! is equal to (n)(n minus 1)(n minus 2)... and so on.

'k' is less than n .......................................................Given.

n!k! is a perfect square ..........................................Given.

So if 'n' is a perfect square and 'k' is one less than 'n' then the condition could be satisfied. But 'n' is less than or equal to 200. Hence largest square uptill 200 is equal to 'n'.

And hence 196 is the answer!!!

For the product of factorials to be a square, they each have the same numbers. There fore n! x (n-1)! will be a perfect square, ie., (n-1)! However, for the entire product to be a perfect square, 'n' should also be a perfect square. The largest square under 200 in 14^2 i.e 196. So technically, the complete solution would be n=196 and k=195

1.Firstly just look at the last integers (or n) because the integers before it are perfect square if you optimize the k 2. Then look at the biggest possible square from 0 to 200. Then voila I've found 14.

The product of n!k! is a perfect square. That means that n must be a perfect square so that n!k! is a perfect square. The largest integer n is 196 .

For n!k! to be a perfect square, n has to be one more than k and n has to be a perfect square. That way, all the numbers that are contained in k! are squared and n is left and is also a perfect square, thus the number is a perfect square since all its factors are repeated at twice. The largest square under 200 is 196, which is the answer.

largest square less than 200 is 196=14 * 14 .....................thus 199 * 198fact * 198fact is a square

n!=1 2 3*..n

k!=1 2 3*..k

where k<n.

If k=n,n!k! is a perfect square.

But since k<n,n!/k! is a perfect square.The largest perfect square less than or equal to 200 is 196.

Therefore n!/k!=196 where n=196,k=195.

n!k! is always a square number when n is a square and k=n-1 Therefore n!=n.(n-1)! or n!=n.k! Thus n!.k!= n.(k!)^2 Thus its a square when n is a square. The greatest square number below 200 is 196.

'n' is equal to or less than 200 ..........................Given.

n! is equal to (n)(n minus 1)(n minus 2)... and so on.

'k' is less than n .......................................................Given.

n!k! is a perfect square ..........................................Given.

So if 'n' is a perfect square and 'k' is one less than 'n' then the condition could be satisfied. But 'n' is less than or equal to 200. Hence largest square uptill 200 is equal to 'n'.

And hence 196 is the answer!!!

n!=n [n-1]! WE WANT THE GREATEST n, SO LET US ASSUME THAT k=[n-1] THEREFORE,n!k!=n[n-1]! [n-1]! =n[n-1]!^2 IF,WE WANT n!k! AS PERFECT SQUARE,THEN n SHOULD BE A PERFECT SQUARE BECAUSE [n-1]! IS ALREADY A SQUARE. THEREFORE LARGEST n<200 WILL BE 196

n!=1x2x3x4...(n-1)xn. k!=1x2x3x4....(k-1) xk. Since k<n,therefore largest possible value of k will be n-1. k!n!=1^2 x 2^2 x 3^2 x 4^2....(n-1)^2 x n To make it a perfect square 'n' should be a square and largest value of n<200 which is a square is 196.

let k = n - 1 Now n!k! = n x k! x k! k! x k! is always a perfect square. So, we require n such that "n" is a perfect square. Largest such value is 196.

n must be perfect square if n!k! is perfect square. If k takes n-1, n!k! = n x (n-1)^2, so n must be perfect square, and the largest n is 196