What is the value of $n$ for which $n!=2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23 $

Question

What is the value of $n$ for which $n!=2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23 $

The way I am approaching this problem is just to find the highest number possible and to get all the numbers in between $2$ and $23$ because I thought it might be $23!$

but I had the remaining numbers:

$$2^{10},3^5,5^3,7, 11, 13 $$

So i know $n$ is higher than $23!$ My next method would be to exhaust all numbers. However, my question is would anyone do this question a different besides exhausting all the numbers? I'm just curious to see if there is any method in solving this question.

This might be pointless but just throwing it out there, im not looking for the "using the calculator" method.

Answer 1

You know $n$ is between $23$ and $28$, inclusive, because there is a $23$ in the prime factorization of $n!$ but no $29$. There are $4$ factors of $7$, so these must be from $7$, $14$, $21$, and $28$.

Therefore $n=28$.

Answer 2

$n!=2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23$

Count the prime factors:

For $2^n!$ there are $2^{n-1}$ multiples of 2, $2^{n-2}$ multiples of 4, $2^{n-3}$ multiples of 8, and so on the $1 + 2 + 4 + .... + 2^{n-1} = 2^n - 1$ powers of 2. I.E. $2^{2^n - 1}|2^n!$ but $2^{2^n}$ does not.

Now 25 = 15 + 7 + 3. So $2^{15}|16!$, $2^7 | 17*18... 24$, $2^3 | 25*...*28$. So $2^{25} | 28!$ and $28 \le n < 30$.

Similarly and for the exact same reasons $3^n!$ has $3^{n -1} + 3^{n-2} + ... + 3 + 1$ powers of 3.

So $3^{13}| n!$ and 13 = 1 + 3 + 9. So $3^3 = 27 \le 30$.

Same for 5. $6 = 5 + 1$ so $5^2 \le n \le 30$.

There are four powers of 7. 4 < 7 + 1 so $n < 7^2$. $4*7 = 28 \le n < 5*7=35$.

There are 2 powers of 11 so $22 \le n < 33$.

13: $26 \le n < 39$.

17: $17 \le n < 34$

19: $19 \le n < 38$

23: $23 \le n < 46$

29: There are no powers of 29. So $n < 29$.

Put these all together and n = 28.

Okay, that was longer and tedious an far more than we neededt to do. But I wanted to give ideas of how to solve these in general.

It'd be easier to note: n < 29 as there are no 29 powers and 4 seven powers mean $n \ge 28$. Then we'd predict 28! would have the correct powers for all the rest.

But this gives us a fast method to factorize factorials. For example 50! =

$2^{31+15+1}*3^{(9+3+1)+2(3+1) + 1}*5^{2(5 + 1)}*7^{1+7}*11^4*13^3*17^2*23^2*29*31*37*41*43*47$