f(n) is defined as multiplication of Integers from 1 to n (both inclus

hi chetan2u,

How did u calculate the number of 3's for 100!?

thanks GMATInsight!...i did know this rule, could not recall it!..thanks again!

I guess the tradition of Thanking on GMAT CLUB is by pressing the Kudos button.

Hi,
It seems you have got the reply for your Question...
i'll just explain you the rason behind it so that even if one forgets the formulae, one can understand the reasoning and get to the answer..
Since 100! is multiple of all positive integers till 100...
when we divide this number by 3, we get the numbers that are divisible by 3 as every third number should be div by3=33
similarly for 5,7 etc
now every 9th number will be div by 9 which adds one extra powe to 3.. so 100/9=11
also every 27 th number will add one more power in addition to one by 3 and one by 9, which have already been calculated above=100/27=3
finally 81st number, as beyond that it will go above 100, will give 100/81=1..
so total power of 3=33+11+3+1=48
hope it helped

f(n) = n!

For finding the value of x, we need to remember that the formula for calculating the number of powers of k in n! = Integer value of [n/k + n/ k^2 + n/k^3 until n/k^r >1 ]

Thus powers of 2 in 100! = 100/2 + 100/4 + 100/8 + 100/16+100/32+100/64 = 97

Power of 3 in 100! = 100/3 + 100/9 + 100/27+ 100/81 = 48

Now 72 = 2^3 * 3^2 and for 72^x to evenly divide 100!, it should have atleast 1 pair of 2^3 and 3^2. Also, 3^2 will be a scarcer than 2^3.

Thus we have 48/2 sets of 3^2 in 100!. The answer is 48/2 = 24 , D is the correct answer.

A small correction in highlighted part,

Thus powers of 2 in 100! = 100/2 + 100/4 + 100/8 + 100/16+100/32+100/64 = 97

Thanks. Edited the typo.

Since we require number of 3's in the Product 1 to 100 inclusive. There is a method to calculate number of 3's in a product of consecutive number starting from 1

step 1: 100/3 = 33 ==> There are 33 numbers which are multiple of 3 i.e there 33 have one 3

Step 2:
In that 33
33/3 = 11 ==> This 11 will have two 3's one we have counted above

Step3:
In that 11
11/3 = 3 ==> there are 3 numbers which have three 3's two we have counted in step 1 & 2

Step 4
3/3 = 1 ==>. There are 1 number which have four 3's there we have counted in steps 1,2, &3.

Adding - we will get 48.

You can use this to calculate for any number. Provided the number should be a product of consecutive numbers from 1 and you have to start with largest number not with the product

I completely forgot about this formula - good looking out - as it makes solving the problem easy

Hi Chetan,

How do you deduce "72x means to find 3s in it.." I've reread every post here, can't seem to understand this bit! Very confused! Please help!

Thanks!

The stem tells us that f(n) = n!. So f(100) = 100!, and is divis by 72^x. Prime factorization of 72 = 2^3*3^2.

The power of 3s in 100:
100/3 = 33
100/3^2 = 11
100/3^3 = 3
100/3^4 = 1
Total = 48. Since we had 3^2 in 72, we divide 48 by 2. Answer = 24, D

I dont understand what it means by being 'evenly divisible'. Secondly, Id appreciate if any one can tell me why did we divide the total number of powers of 3 by 2 to get the final answer?

Could you please expand on why 3^2 is taken into consideration and not 2^3?

Evenly Divisible Means Completely divisible without leaving any remainder


SINCE 72 = 2^3 *3^2

So we need the total groups of 2^3

Total Powers of 2 available with us in f(100)=97

So total groups of 2^3 = Integer value of (97/3) = 32

I hope this helps!!!

There are 32 groups of 2^3 and 24 groups of 3^2

But we need pairs of (2^3) and (3^2)

so we get a total of 24 pairs of (2^3) and (3^2) because after that all powers of 3 get exhausted.

I hope this helps!!!

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