In the prime factorization of 109!, what is the exponent of 7? (Reminder: The number n! is the product of the integers from 1 to n. For example, 5! = 5 * 4 * 3 * 2* 1 = 120.)

Guest Dec 31, 2020

#1**+1 **

Hello Guest!

We're kinda looking for the number of 7s in the numbers 1-109.

The amount of numbers divisable by 7 from 1-109 is 15 (7, 14, 21, 28, ..., 105).

However, some numbers are divisable 49 which is made out of 2 7s, 7^2.

The amount of numbers divisable by 49 from 1-109 is 2 (49, 98).

So there are 15 numbers divisable by 7, 2 of which are divisable by 49.

There are 13 numbers divisable by 7 and not 49, and 2 number divisable by 49.

Thus, our answer is 13 + 2*2= 17.

A cool way to do this is to just keep dividing.

Btw, floor is just the highest integer lower than the answer. For example floor(3.5) = 3.

floor(109/7) = 15

floor(15/7) = 2

floor(2/7) = 0

We stop dividing when our final answer is less than 1.

Then, we just add everything, 15 + 2 = 17.

This works because dividing by 7 two times is just like dividing by 49.

I hope this helped. :)))))

=^._.^=

catmg Dec 31, 2020

#1**+1 **

Best Answer

Hello Guest!

We're kinda looking for the number of 7s in the numbers 1-109.

The amount of numbers divisable by 7 from 1-109 is 15 (7, 14, 21, 28, ..., 105).

However, some numbers are divisable 49 which is made out of 2 7s, 7^2.

The amount of numbers divisable by 49 from 1-109 is 2 (49, 98).

So there are 15 numbers divisable by 7, 2 of which are divisable by 49.

There are 13 numbers divisable by 7 and not 49, and 2 number divisable by 49.

Thus, our answer is 13 + 2*2= 17.

A cool way to do this is to just keep dividing.

Btw, floor is just the highest integer lower than the answer. For example floor(3.5) = 3.

floor(109/7) = 15

floor(15/7) = 2

floor(2/7) = 0

We stop dividing when our final answer is less than 1.

Then, we just add everything, 15 + 2 = 17.

This works because dividing by 7 two times is just like dividing by 49.

I hope this helped. :)))))

=^._.^=

catmg Dec 31, 2020