The idea of factorial is very often tested in competitive exams. Have given below two interesting questions on this concept

Questions

1. How many values can natural number n take, if n! is a multiple of 7

^{6}but not 7^{9}?2. How many values can natural number n take, if n! is a multiple of 2

^{20}but not 3^{20}?Correct Answers

1. 14 values

2. 21 values

Explanatory Answers:

1. How many values can natural number n take, if n! is a multiple of 7

^{6}but not 7^{9}?The smallest factorial that will be a multiple of 7 is 7!

14! will be a multiple of 7

^{2}Extending this logic, 42! will be a multiple of 7

^{6}However, 49! will be a multiple of 7

^{8}as 49 (7 * 7) will contribute two 7’s to the factorial. (This is a standard question whenever factorials are discussed). Extending beyond this, 56! will be a multiple of 7^{9}In general for any natural number n,

n! will be a multiple of [n/7] + [n/49] + [n/343] + …..

where [x] is the greatest integer less than or equal to x. A more detailed discussion of this is available on this link

So, we see than 42! is a multiple of 7

^{6}. We also see that 56! is the smallest factorial that is a multiple of 7^{9}. So, n can take values { 42, 43, 44, 45, ……55}There are 14 values that n can take.

2. How many values can natural number n take, if n! is a multiple of 2

^{20 }but not 3^{20}?The highest power of 2 that will divide n! = [n/2] + [n/4] + [n/8] + [n/16]….. and so on. So, let us try to find the smallest n such that n! is a multiple of 2

^{20},If n = 10, the highest power of 2 that will divide n! = [10/2] + [10/4] + [10/8] = 5 + 2 + 1 = 8

If n = 20, the highest power of 2 that will divide n! = [20/2] + [20/4] + [20/8] + [20/16] = 10 + 5 + 2 + 1 = 18

If n = 24, the highest power of 2 that will divide n! = [24/2] + [24/4] + [24/8] + [24/16] = 12 + 6 + 3 + 1 = 22

{Here we can also see that each successive number is just the quotient of dividing the previous number by 2. As in, [12/2] = 6, [6/2] = 3, [3/2] = 1. This is a further shortcut one can use.}

So the lowest number of n such that n! is a multiple of 2^20 is 24

Now, moving on to finding n! that is a multiple of 3. The highest power of 3 that will divide n! = [n/3] + [n/9] + [n/27] + [n/81] and so on,

When n = 20, the highest power of 3 that can divide 20! = [20/3] + {6/3] = 6 + 2 = 8

When n = 35, the highest power of 3 that can divide 35! = [35/3] + {11/3] + [3/3] = 11 + 3 +1 = 15

When n = 45, the highest power of 3 that can divide 45! = [45/3] + [15/3] + [5/3] = 15 + 5 +1 = 21

The lowest number n such that n! is a multiple of 3^20 is 45.

When n takes values from 24 to 44, n! will be a multiple of 2^20 and not 3^20. n can take 21 values totally.