find the sum of all positive value of n such that (1!x2!x3!.....200!)/n! this is perfect square?

Question

the answer is between 0 to 999

Answer 1

My approach is to see what restrictions can be placed on n:



Any prime that is close to 200 will probably impose a significant restriction on n. 199 is prime (clearly not divisible by 2,3,5,11 and can be checked to be indivisible by 7 and 13). 199 appears in the numerator product twice (199! and 200!). Therefore, n cannot be 199 because 199! would cancel out only one of the factors of 199 making the expression not a square. But we can do better. Any prime p such that 101 <= p < 200 is odd and appears as a factor an even number of times in the numerator. In particular 101 appears as a factor an even number of times in the numerator (100 times to be exact). Therefore, n cannot be any number between 101 and 200 because n! would cancel out exactly one factor of 101 resulting in the expression not being a perfect square. Thus n > 200 or n <= 100.



Now if n > 200, we can place a restriction on n because if n! is too large it will include a prime that is that will not cancel with the numerator resulting in the fraction not being an integer. The smallest prime greater than 200 is 211 (clearly not divisible by 2,3,5,7,11 and we can check indivisibility by 13). Thus we have that n <= 100 or 201 <= n <= 210.



However, if 201 <= n <= 210, n! will still cancel out exactly one factor of 199 which will result in the expression being not a square. Therefore, we are down to the only possibilities for n being n <= 100.



Now for n <= 100, clearly the expression is an integer. Also, from our reasoning above, it should be clear that every prime p >= 101 appears an even number of times in our expression. We need to find out more about the primes p < 100 and determine restrictions on n to see what values will ensure that all such primes appear an even number of times.



Now if p is a prime such that 2p < 200, but such that 3p > 200, then we have 66 < p < 100. Such a prime appears p + 2(200 - 2p + 1) times in the numerator (the p comes from the factors p! through (2p - 1)!, the 2(200 - 2p + 1) comes from the factors (2p)! through 200!. Now p + 2(200 - 2p + 1) = 402 - 3p which is odd. Therefore n! must cancel at least 1 such prime factor from the numerator. Since the largest such prime is p = 97, we must have 97 <= n <= 100.



98! = 98 x 97! = 2 x 7^2 x 97!

99! = 99 x 98! = 2 x 3^2 x 7^2 x 11 x 97!

100! = 100 x 99! = 2^3 x 3^2 x 7^2 x 11 x 97!



Since n! for 97 <= n <= 100, differ by an even number of factors of 7 and 3 which will not affect the squareness of our expression, clearly our best bet would be to investigate the cases p = 2 and p = 11 in more detail.



For p = 11, the numerator contains

0 x 10 + 11 x (1 + 2 + 3 + ... + 10) + 11 x (12 + 13 + 14 + ... + 18) + 19 x 3 = 11x11x5 + 11x(odd) + odd = an odd number of factors of 11. Therefore n! must contain an odd number of factors of 11. Since 99! contains 9 factors of 11, we must have n = 99 or n = 100.



Since 100! and 99! differ by a factor of 100, a perfect square, either they both work or neither works.



Assuming they both work, the answer to your problem is 99 + 100 = 199.



However, to be thorough, we need to prove that one of these values actually results in producing a perfect square, e.g., n = 100, by confirming that the expression in question has an even power of p for every prime p.



Since the parity of the power of a prime p in (2n - 1)!(2n)! = (2n - 1)!^2 (2n) is just the same as the parity of the power of p in (2n), we have that the parity of the power of a prime in the numerator is simply the parity of the power of p in the product 2x4x6x..x200 = 2^100 x 100!. Note that if p > 2 then this is the same as the parity of the power of p in 100!. If p = 2, then since 2^100 contains an even number of factors of 2, the parity of the power p in the numerator is still the same as the parity of the power of p in 100!. This proves that the parity of the power of p in (1!x2!x3!.....200!)/100! is always even and thus that (1!x2!x3!.....200!)/100! is a perfect square. Since (1!x2!x3!.....200!)/99! = (1!x2!x3!.....200!)/100! x 100, n = 99 is also a solution.

Answer 2

The values of n are 99 and 100; the sum is 199.



Proof:

Definition: The "square-free" value of a number is that number divided by its largest square factor. For example: the square-free value of 180 is 5, since 180/36=5.



Lemma 1: If A is a multiple of 𝑘², then A/𝑘² has the same square-free value as A; in other words, we can "cancel out" squares.

Proof of Lemma 1: Let's say that the square-free value of A is 𝑥. Hence, there is a 𝑗 such that:

A = 𝑥𝑗²

Now:

A/𝑘² = 𝑥𝑗²/𝑘² = 𝑥(𝑗/𝑘)²

Hence the square-free value of A/𝑘² is 𝑥, the same as that of A.



Lemma 2: If A/B is a square, then A and B have the same square-free value.

Proof of Lemma 2: Let's say that the square-free value of A is 𝑎, and the square-free value of B is 𝑏. Hence, there are a 𝑗 and 𝑘 such that:

A = 𝑎𝑗²

B = 𝑏𝑘²

Now:

A/B = (𝑎/𝑏)(𝑗/𝑘)²

This means that for A/B to be a square, so does 𝑎/𝑏. But 𝑎 and 𝑏 are both "square-free" - they have no square factors except for 1! Hence, 𝑎/𝑏=1 and 𝑎=𝑏.



Lemma 3: The square-free value of (1!x2!x3!.....200!) is equal to 100!.

Proof of Lemma 3: When you multiply all of those together, it's not hard to see that it is equal to:

N = 1^200 * 2^199 * 3^198 * 4^197 ... * 200^1

1^200 is a square. 2^199 is equal to 2 times a square. 3^198 is a square, 4^197 equals 4 times a square. Using these facts, we can cancel out a lot of squares and show that the square-free value of N is equal to the square-free value of:

2 * 4 * 6 * 8 * ... * 200

Factoring a two out of all of those, we get:

2^100 * 1*2*3*4...*100 = 2^100 * 100!

Canceling out 2^100 (a square), we see that the square-free value of (1!x2!x3!.....200!) is equal to 100!.

Note: 100! = 100*99!, and so 100! and 99! have the same square-free value.

Since N and 100! have the same square-free value, N/100! is a square. Since N has the same square-free value as 99!, too, N/99! is a square.



Lemma 4: 100! does not have the same square-free value as any other factorial from 1! to 210!, except for 99!.

Proof of Lemma 4: Let's say there is a factorial, M!, that has the same square-free value of 100!. Hence 100!/M! (if M<100), or M!/100! (if M>100) would be a square.

M cannot be 98, 97, or 96, since 100!/98!, 100!/97!, or 100!/96! (100*99, 100*99*98, 100*99*98*97 respectively) are not squares. For any value of M less than 96, 100!/M! would have the prime factor 97 there once and only once in its prime factorization, since no number less than 97 has 97 as a prime factor. No square has a prime factor only once in it's prime factorization. So M cannot be less than 96.

M cannot be 101, since 101!/100! (101) is not a square. M cannot be anything more than that but less than 200 either, since no number higher than 101 and less than 200 has 101 as a prime factor, meaning that M!/100! has 101 as a prime factor only once in its prime factorization and hence not a square.

M cannot be anywhere from 201 to 210, as M!/100! would have the prime factor 199 once.

That means that M can only be 100 or 99.



Lemma 5: 211! is not a factor of (1!x2!x3!.....200!).

Proof of Lemma 5: The prime factor 211 is never seen in "(1!x2!x3!.....200!)".





Now, let's answer the question.

Because of Lemma 5, the "n" in your question (I'm going to call it 𝑛) is from 1 to 210.

Because of Lemma 2, 𝑛! has the same square-free value as (1!x2!x3!.....200!).

Because of Lemma 3, 𝑛 can be 100. It is also shown in Lemma 3 that 𝑛 can be 99.

Because of Lemma 4, 𝑛 cannot be anything other than 99 and 100.



Hence, 𝑛 is 99 or 100; the sum is 199.

Answer 3

n = 100