What is the largest integer \(n \) such that \(7^n\) divides \(1000!\)? Btw the answer isn't 163.

Guest Jul 21, 2019

#1**+1 **

We need to find how many factors of seven all the integers from 1 to 1000 have.

Each integer from 1 to 1000 can be a factor of 7, 49, or 343, giving it 1 factor of 7, 2 factors of 7, or 3 factors of 7.

There are \(\lfloor\frac{1000}{7}\rfloor=142\) numbers that have 1 factor of 7.

There are \(\lfloor\frac{1000}{49}\rfloor=20\) numbers that have 2 factors of 7.

There are \(\lfloor\frac{1000}{343}\rfloor=2\) numbers that have 3 factors of 7.

So, the answer is \(2\cdot3+20\cdot2+142\cdot1\).

But, wait!

We are counting numbers that have 2 factors in the numbers that have 1 factor of 7!

And, we are counting numbers that have 3 factors in the numbers that have 2 factors and the numbers that have 1 factor!

So, the answer is \(2\cdot3+20\cdot2+142\cdot1-2-2-20=2+20+142=164\)!

Davis Jul 22, 2019