If 6^k is a factor of (40!), what is the greatest possible value of k?

Question

A. 16
B. 17
C. 18
D. 19
E. 20

Solution:
 6 is not a prime number but we make a 6 by multiplying 2 with 3. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s. If you are a little confused, don’t worry. Look at the solution given below.
 
Let’s find the number of 2s in 40!
 
Step 1: 40/2 = 20
 
Step 2: 20/2 = 10
 
Step 3: 10/2 = 5
 
Step 4: 5/2 = 2
 
Step 5: 2/2 = 1
 
Step 6: 20 + 10 + 5 + 2 + 1 = 38
 
Total number of 2s is 38.
 
Let’s find the number of 3s now.
 
Step 1: 40/3 = 13
 
Step 2: 13/3 = 4
 
Step 3: 4/3 =1
 
Step 4: 13 + 4 + 1 = 18
 
Total number of 3s is 18.

with kind regards why do i have to make the process of the 2s if only the highest prime factor(here 3) is important.

why not finding only the 3s and of course they will be less than the 2s.

fluke · Accepted Answer

Guess, you have a point here. I'd solve it just the same way you did. Just find it for the largest prime factor. There may be some catch though.

PareshGmat · Answer

In 40!, its obvious that number of 3's would be less than the number of 2's..........

We require to find value of k in 6^k = 2^k * 3^k

Counting 3's from 40 to 1

39 ......... 36 .......... 33 .......... 30 ........ 27  ...........  24 ......... 21 ......... 18 ....... 15  .......... 12 .......... 9 ........6 .... 3

1 + 2 + 1 + 1 + 3  +  1 + 1 + 2 + 1  + 1 + 2 + 1 = 8 + 5+5 = 18

LifeChanger · Answer

that's the approach i generally follow..take the highest of the prime factors and get the answer...i don't think there is a catch

anupamadw · Answer

Finally, please tell what is answer ? Thanks

Bunuel · Answer

The answer is 18. For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory . Theory on Trailing Zeros: everything-about-factorials-on-the-gmat-85592.html Hope this helps.

taleesh · Answer

Why are we not taking any other number than 40,10,20,5,2 for (2) and 40,13,4 for (3).
If we have to get 2's and 3's.
Why cant we take 39/3, 3/3,18/3 or 25/3, 19/3
can somebody help to make me understand it .

GMATinsight · Answer

Please find the detailed and quickest solution as attached

cavana · Answer

1) We need to find the number of prime factors 2*3 in the 40!
2) In a consecutive sequence there will always be more 2's than 3's, so the task is to find the number of 3's
3) 40/3=13; 40/9=4; 40/27=1; 13+4+1=18

The correct answer is 18

coolkl · Answer

I am getting answer 21 by thinking in form as - 40! means 39*38*37*....*2*1. So I find numbers in this range which are divisible by 6 which are 6,12,18,24,30,36. Now counting number of time 6 appears in this number we get 1+2+3+4+5+6 = 21. Why is this wrong. What am I missing any ideas ?

coolkl · Answer

Can anyone explain the solution in more detail. I still dont get the reasons behind doing 40/3 = 13. Then for the next division 13/3 = 4 and then 4/3 =1 and finally doing sum of 13+4+1 = 18. I am surely missing a concept. Help appreciated. Thanks.

FB2017 · Answer

6 is not a prime number but we make a 6 by multiplying 2 with 3 which are prime individually. To get the number of 6s in 40!, we just need to find the number of 3s because the number of 3s will be fewer than the number of 2s.
The power of 3 in 40! is 18.

So option with 18.

Scuven · Answer

I solved it in 1 minute and 30 seconds using this approach:

1) Drafted a couple of number series of consecutive integers to see the pattern of numbers divisible by 6
1 2 3 4 5 6 / 7 8 9 10 11 12 -> therefore you can see that every six consecutive integers one is divisible by 6
2) Thus in 40 numbers only 6 numbers are divisible by 6
3) Look after in the answers and find a multiple of 6
4) 18 is the only multiple of 6

anwesharath · Answer

In a similar question: if If 12^k is a factor of (40!), what is the greatest possible value of k?

Do we need to consider the powers 4 in 40 or 2?

Jyoti3380 · Answer

It only needs to be considered in prime factor, hence 2 & 3.

If 6^k is a factor of (40!), what is the greatest possible value of k?

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If 6^k is a factor of (40!), what is the greatest possible value of k?

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