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