(a) What is the largest integer $n$ such that $7^n$ divides $500!$ ?

(b) What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

(c)The expression \[\frac{1000!}{(500!)^2}\] is an integer. What is the largest integer $n$ such that $7^n$ divides this integer?

Mellie
Jul 1, 2015

#1**+20 **

Here's my attempt at (a).....

Notice that .....

500/7 = 71 numbers that have at least one factor of 7

500/49 = 10 more numbers that have an additional factor of 7

500/343 = 1 more number that adds an additional factor of 7

So.....71 + 10 + 1 = 82

So ..... 7^82 will divide 500!....thus, n = 82

CPhill
Jul 2, 2015

#1**+20 **

Best Answer

Here's my attempt at (a).....

Notice that .....

500/7 = 71 numbers that have at least one factor of 7

500/49 = 10 more numbers that have an additional factor of 7

500/343 = 1 more number that adds an additional factor of 7

So.....71 + 10 + 1 = 82

So ..... 7^82 will divide 500!....thus, n = 82

CPhill
Jul 2, 2015

#2**+5 **

That is really good logic I think Chris

I forgot the 'divide by 49' one when i did it.

Melody
Jul 2, 2015

#3**+10 **

Here's (b)....it's pretty much an extension of (a)

1000/7 = 142 numbers that have at least 1 factor of 7

1000/49 = 20 numbers that have at an additional factor of 7

1000/343 = 2 numbers that have a third factor of 7

So......... 142 + 20 + 1 = 163

And 7^163 will divide 1000! ...so n = 163

CPhill
Jul 2, 2015

#5**0 **

(c) Based on the other two, (a) and (b), we know that there are 164 powers of 7 that divide into 1000!, and there are 82 factors of 7 that divide into 500!. Since we are dividing the same numbers that have different powers, we subtract the denominator's exponent (82) from the numerator's exponent (164). However, the denominator is SQUARED, meaning that we subtract 82 TWICE from 164.

\(164-82-82=0\)

To put it mathematically, all the powers of 7 are canceled, and that leaves 7^0 = 1. N, the exponent of 7, is 0.

Guest Apr 9, 2016

edited by
Guest
Apr 9, 2016