12 Days of Christmas GMAT Competition - Day 3: What is the highest int

Question

12 Days of Christmas GMAT Competition with Lots of Fun

What is the highest integer power of 3 in P!, where P is a positive integer ?

(1) The highest integer power of 6 in P! is 9.
(2) The highest integer power of 15 in P! is 4.

Bunuel · Accepted Answer

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brains · Answer

The answer should be A:

Statement 1: Since 6 has 2 and 3 as prime factors and for any factorial of positive integer >2, number of 3's will be less than that of 2's, so finding number of 3's will ensure that P! has same number of power raised . Hence P! will have also 9 numbers of 3's . So sufficient:

Statement 2: Insufficient: Since 15 has 3 and 5 as prime factors, and number of 5's will be less than number of 3's , so knowing the number of 5's will not give the actual number of 3's. Hence insufficient.

GMATist1 · Answer

P!= 1*2*3*4*5*6*.....*(p-1)*p

6 is made up of two factors = 2 and 3
Factors of 6 = minimum of occurence(2,3)
Since 2 will always have more multiple than 3, it is not possible to increase any further factors
Therefore, highest power of 6 = highest power of 3

Similarly,
15 = 3*5
But multiples of 5 would have been less than 3 in the factorial
Therefore the highest power of 15 just gives the highest possible power of 5

Hence A

Kinshook · Answer

Asked: What is the highest integer power of 3 in P!, where P is a positive integer ?

(1) The highest integer power of 6 in P! is 9.
Since there will be lesser highest power of 3 available in P! as compared to highest power of 2 available in P!
Highest power of 3 in P! = Highest power of 6 in P! = 9
SUFFICIENT

(2) The highest integer power of 15 in P! is 4.
Since there will be lesser highest power of 5 available in P! as compared to highest power of 3 available in P!
Highest power of 5 in P! = Highest power of 15 in P! = 4
P may take any value in {20,21,22,23,24}
If P = 20; Highest power of 3 in P! = 6 + 2 = 8
If P = 21; Highest power of 3 in P! = 7 + 2 = 9
If P = 22; Highest power of 3 in P! = 7 + 2 = 9
If P = 23; Highest power of 3 in P! = 7 + 2 = 9
If P = 24; Highest power of 3 in P! = 8 + 2 = 10
Highest power of 3 in P! can not be found out.
NOT SUFFICIENT

IMO A

gmatophobia · Answer

The highest power of a number in a factorial can be found by getting the power of the highest prime factor.

Statement 1

6 can be prime factorized as 2 *3. Therefore the highest power of 6 in P! is the highest power of 3 in P!.

Hence the statement is sufficient.

The highest power of 3 in P! is 9.

Statement 2

15 can be prime factorized as 5 *3. Therefore the highest power of 15 in P! is the highest power of 5.

Hence we cannot conclude what the highest power of 3 would be.

For example is P = 15! or 18! the largest power of 5 will be same, however the largest power of 3 would differ.

Hence the statement is not sufficient.

IMO A

ElninoEffect · Answer

(1) The highest integer power of 6 in P! is 9.
Highest power of 6 in P! means highest power of 3 in P! as 6 is prime factorized into 3*2 now to get the highest power we need to take the bigger integer hence 3.
So it means highest power of 3 in P! is 9. 
  Sufficient.

(2) The highest integer power of 15 in P! is 4.
Using the same logic as above 15 can be prime factorized into 5*3, so to get highest power we need to take 5 in here.
So highest power of 5 in P! is 4.
  Not sufficient.

A is the answer.

behlmanmeet · Answer

Question Stem: 
What is the highest integer power of 3 in P!, where P is a positive integer?

(1) The highest integer power of 6 in P! is 9.
(2) The highest integer power of 15 in P! is 4.

Solution:

To find the highest power of 3 in P!, first, we need a value for P and then divide it by 3 continuously till we can get a quotient.

Statement 1: 
 The highest integer power of 6 in P! is 9.

Now 6=2*3.
Since 3 is greater than 2, the power is decided by 3.
We can guestimate the value of P to correlate to a power of 9 when divided by 3.
If P=27 then the highest power of 3 in P!:
27/3=9
9/3=3
3/3=1. The highest power of 3 in 27!=9+3+1= 13, so we move a bit lower.
If P=21, then similarly highest power of 3 in 21!= 9. But here P can be 22 or 23 and we will still get the highest power of 3 to be 9.
So, P can be 21,22, or 23. Still, we can get a unique value for the highest power of 3 in P!. Thus, statement 1 is sufficient.

Statement 2: 
 The highest integer power of 15 in P! is 4.

Now, 15=3*5. This can be solved or at least assessed very simply if we understand how we are able to get a unique value of 3 in P! in Statement 1 that is because the power depended on 3 in that case but now it depends on 5 since 5 is greater than 3. We will not be able to get a unique value of 3 in this case. To illustrate, if P=20 then we can get the highest value of 15 in P!=4 with a unique value for 3 and 5. But if P! is 21! then also we can get the highest power of 15 in 21! as 4 but now the value of 3 will change. Thus, statement 2 is not sufficient.

Thus, our answer is A

bumpbot · Answer

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