itisSheldon wrote:

How many values can natural number n take, if n! is a multiple of 2^20 but not 3^20?

(A) 11

(B) 21

(C) 16

(D) 20

(E) 23

The problem in this question is the starting point.

What value should we try for n such that it has around 20 2s? Alternatively, we can start the count from 2 , 4, 6, 8 ... counting the number of 2s as we go along but that is a rather painful process. So let's try to guess a value for n.

We get at least one 2 from each multiple of 2. Then there are those multiples that give more than one 2 such as 4 and 12 etc. So let's try to figure out the number of 2s when n is about the 13th multiple of 2. Say when n = 26

26! will have 13 + 6 + 3 + 1 = 23 2s

So we need to go back a few steps and check when we get the 20th 2. We get it at n = 24.

Next, we repeat the process for 20 3s. Since 3s are rarer than 2s, let's try the 15th multiple of 3 i.e. 45

45! will have 15 + 5 + 1 = 21 3s. 45 adds two 3s to the mix so till 44, we have only 19 3s. That is perfect.

So n can take values from 24 to 44 which is 44 - 24 + 1 = 21 values

Answer (B)

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