Checking the divisibility of a factorial

When we first see divisibility questions like this (especially with factorials), one idea to approach the question could be using primes. This is because for a factorial $n!$ , if prime $p>n$ , then $p \nmid n!$ as there are no prime factors $p$ in $n!$ . (This is one case of our question too!)

So, firstly, as many other number theory questions, let's think of cases to split our solution into: (the bold sentence/phrase indicates the Case)

Case 1 A similar idea as mentioned above: Let $n$ be a prime. We know $n > n-1$ , hence it follows that $n^2 \nmid (n-1)!$ . Keeping in mind the condition of $2 \leq n \leq 100$ . There are $25$ primes less than $100$ .

Case 2 * Let $n = 2 \times p$ where $p$ is a prime *. Then, observe that there will only be $1$ factor of $p$ in $(n-1)!$ since $n-1 < n = 2p$ . Thus, $n^2 = 2^2 \times p^2 \nmid (n-1)!$ . There are $15$ primes such that $2 \leq n = 2p \leq 100$ .

Case 3 There are $2$ exceptions that satisfy the condition of the question but do not belong to either Case 1 or 2. They are $n = 8, 9$ , thus $2$ solutions in Case 3. When $n=8 = 2^3$ , $(n-1)! = 7!$ has $\lfloor\frac{7}{2}\rfloor + \lfloor\frac{7}{4}\rfloor = 3$ factors of $2$ , hence, $n^2 = 8^2 = 2^6 \nmid 7!$ . Similarly, for the solution $n=9 = 3^2$ , $(n-1)! = 8!$ has only $\lfloor\frac{8}{3}\rfloor = 2$ factors of $3$ . Hence, $n^2 = 9^2 = 3^4 \nmid 8!$

(I tested out the first $10$ values of $n$ and observed the exceptions. So, how do I know that there are only $2$ exceptions? Well, testing the next powers of $2$ and $3$ which are $n=2^4 = 16$ and $n=3^3 =27$ respectively, will show that they do not satisfy the conditions.)

Hence, the number of integers $n$ satisfying all conditions in the question

$= 25 +15 +2 = \boxed{42}$ .

Given an integer $n$ , let $\alpha:=v_p (n)$ be the number of factors of a prime $p$ in $n$ . Then $v_p (n^2)=2\alpha$ . We want to calculate $v_p ((n-1)!)$ and in order to do this we use the Legendre-De Polignac identity and we get $\displaystyle v_p ((n-1)!)=\sum_{i\ge1}\left\lfloor\frac {n-1}{p^i}\right\rfloor$ but we have the trivial inequality $n\ge p^\alpha$ , so $\displaystyle v_p ((n-1)!)\ge\sum_{i\ge1}\left\lfloor\frac {p^\alpha-1}{p^i}\right\rfloor=\sum_{i=1}^{\alpha-1} p^{\alpha-i}-1$ . We can transform the last sum into $\displaystyle\frac {p^\alpha-1}{p-1}-\alpha$ by the formula for the sum of a geomteric progression. The numbers that satisfy the statement in the text are those for which $v_p (n^2)> v_p ((n-1)!)$ , which is $2\alpha>\frac {p^\alpha-1}{p-1}-\alpha$ , ie $3\alpha>\frac {p^\alpha-1}{p-1}$ If $\alpha=1$ this inequality holds for every prime. Then if n is a prime, it satisfies the problem; if $n=kp$ with k not divisible by p, then $n^2$ has 2 factors p, while $(n-1)!$ has at least k-1 factors p; so the condition of the problem is satisfied if and only if k is 2 (or 1).

If $\alpha=2$ then p can be only 2 or 3 and it is easily checked that 4 and 9 are ok, while 4k and 9k aren't ok if $k\ge2$

If $\alpha=3$ then p can be only 2, and we see that 8 is ok, while 8k isn't.

If $\alpha\ge4$ then there are no solutions. So, the only values for n are p, 2p, 4, 8, 9; the primes under 100 are 25, the primes such that $2p\le100$ are 15. So there are $25+15+2=\boxed {42}$ values for n that satisfy the problem (4 is already counted as 2p).

Hi, I am beginner in python

I try to solve this problem use python

thisi is my code

   import math

   for n in range (2,101):

if math.factorial(n-1)%n**2==0:

    print n

Output:

12 15 16 18 20 21 24 25 27 28 30 32 33 35 36 39 40 42 44 45 48 49 50 51 52 54 55 56 57 60 63 64 65 66 68 69 70 72 75 76 77 78 80 81 84 85 87 88 90 91 92 93 95 96 98 99 100

But, I can't count direct the answer, and I count one by one.

Can you help me, guy?s??

Can somebody help me find solutions to this question?

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Well!. This is a oldProblem. Was also asked in INMO. See here, INMO

Clearly if $n$ is prime then the condition is met, however if $n=2p$ for some prime $p$ then the condition is also met because there is only one multiple of $p$ in $\left(2p-1\right)!$ thus $4p^2$ will never divide $\left(2p-1\right)!$

Now suppose either $n = pk$ for some prime $p$ and some non-negative integer $k > 2$ . Clearly since there are at least two multiples of $p$ in $\left(2p-1\right)!$ then $k^2p^2$ must be able to divide $\left(2p-1\right)!$ . Same argument can be made $k$ -centric.

Thusly no composite number is eligible and so either $n$ is prime or the double of a prime for $2 \leq n \leq 100$ which means there are $42$ values of $n$ for which $n^2$ does not divide $\left(n-1\right)!$