If K is the product of all integers from 1 to 100, both inclusive,

Question

If K is the product of all integers from 1 to 100, both inclusive. What is the highest power of 4! that can divide K?
 
A. 10
B. 25 
C. 32
D. 48
E. 97

garcmillan · Accepted Answer

This is a very difficult question in case you don't the shortcut about divisibility for factorial numbers. So firstly let me show you the concept with example numbers:

1) METHDOLOGY TO ANSWER THESE TYPES OF QUESTIONS

To find out k of the factor 2^k in the factorial number 20!, we need to find the TOTAL number of 2s in 20! (which is the same as finding the maximum value of k that makes 2^k divisible by 20!)

I will break down the concept first so that you can understand what we have behind the same "mechanics" we will be applying to solve this kind of questions

20! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20

Of the numbers above, these are the numbers that have 2 as a factor: 2, 4 (two of them), 6, 8 (three of them), 10, 12 (two of them), 14, 16 (four of them), 18, 20 (two of them)

This makes 1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 = 18

And this is the method to solve it

1º: 20/2 = 10
2º: 10/2 = 5
3º: 5/2 = 2
4º: 2/2 = 1

Stop when quotient would give 0 (next one would be 1/2, whose quotient is zero)

So k = 10 + 5 + 2 + 1 = 18

The greatest possible value of k is 8.

The logic behind this is that each alternate number in 20! will have a 2. Out of 20 numbers, 10 numbers will have a 2 (therefore 1º: 20/2 = 10) (these numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20)

Now, out of these 10 numbers, every alternate number will have another 5 numbers that will have a 2 (therefore 2º: 10/2 = 5)

Basically, same thing with 3º and 4º

Now, to account for all the 2s we add the quotient we've been obtaining 10 + 5 + 2 + 1 = 18 (These are the number of 2s in 20!)

By doing this, you can find the maximum power of any number in any factorial, but we need to know one extra thing for non-prime numbers

If I ask you now if 6^k is a factor of (78!), what is the greatest possible value of k? What would you tell me?

factoring 6 we get 2*3, so which factor should we use to do our method? Easy answer. The one that is more restrictive, in other words, the highest factor. In our case 3. So we would proceed as follows:

Solution:

1º 78/3 = 26

2º 26/3 = 8

3º 8/3 = 2

Greatest possible value of k is 26 + 8 + 2 = 36 (This would be for both 3 and 6)

So now let's tackle the given question

2) ANSWERING GIVEN QUESTION

First of all, we need to realize that integers from 1 to 100 multiplies is the same as 100!.

the number for which we need to find k is 4!=24=3*2^3

As with the example I provided with 6, the most restricitive number in this case is 2^3 (cause 8 is higher than 3)

Let's do the successive divisions:

100/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1

50 + 25 + 12 + 6 + 3 + 1 = 97

as we had 2^3, k is not 97 but 97/3 because (2^k)^3

So k is 97/3 = 32 (only take quotient)

OPTION C

Please give me kudos if you definitely learnt the way to approach this kind of questions

Kinshook · Answer

Given:  K is the product of all integers from 1 to 100, both inclusive.

Asked:  What is the highest power of 4! that can divide K?

What is the highest power of 4! = 24 that can divide K = 100!
 24 = 2^3 * 3

Highest power of 2 in 100! = 50 + 25 +12 + 6 + 3 +1 = 97
Highest power of  2^3  in 100! =   = 32
Highest power of 3 in 100! = 33 + 11 + 3 + 1 = 48

Highest power of  4!=24=2^3*3  in 100! = min (32,48) = 32

IMO C

AcetheGMAT2019 · Answer

Concept 1
Example - What is the highest power of 2 in 100!
100/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
SUM = 97
Thus highest power of 2 in 100! is 97

Concept 2
Example - What is the highest power of 3 in 100!
100/3 = 33
33/3 = 11
11/3 = 3
3/3 = 1
SUM = 48
Thus highest power of 3 in 100! is 48

Similarly, the highest power of 4! in 100! would be the highest power of 24 in 100!
We know that 24 = 3*8 
 = 3*2^3

Now the highest power of 2^3 in 100! is (2^3)^32.33 or 8^32 i.e 2^97 as seen in Concept 1
Now the highest power of 3 in 100! is 3^48 as seen in Concept 2
Thus (2^3)^32 is the limiting factor. This essentially means that although we have more 48 3's in 100!, we have only 32 8's in 100!
To make 24 we need same number of 3's and 8's. Thus the highest power of 4! or 24 in 100! is 32.

EgmatQuantExpert · Answer

Given

• K is the product of all integers from 1 to 100, both inclusive.

To Find

• What is the highest power of 4! that can divide K.

Approach and Working Out

• We know 4! = 24 =  2^3  × 3

• To find the power of 4! we need to find the number of 2s and 3s in 100!

• The number of 2s can be found as follows,
 o That will be 100/2 + 100/4 + 100/8 + 100/16 + 100/32 + 100/64 
 o = 50 + 25 + 12 + 6 + 3 + 1
o = 97
  Here we need to take only the quotient.

• The number of 3s can be found as follows,
 o That will be 100/3 + 100/9 + 100/27 + 100/81
o = 33 + 11 + 3 + 1
o = 48 
   Here also we need to take only the quotient.

• The number of 3 is 48 and the number of  2^3  is 97/3 = 32
 o 32 will be the answer.

Correct Answer: Option C

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If K is the product of all integers from 1 to 100, both inclusive,

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