What is the highest power of 3 in the value of X! ?

(1) The highest power of 9 in the value of x! is 9.
(2) The highest power of 6 in the value of x! is 19.

Using statement 1
The value of 9 will account for all 3 that formed groups of 2 such as 3 *3
hence highest value will be 9*2 =18. But there can be some 3 which does not form a pair
hence it could b 19 also. Thus not sufficient.

Using statement 2
highest power of 6 = 2*3
for every 3 there will be a corresponding 2 hence the highest power of 3 will be same as highest power of 6 = 19
Thus B is sufficient

Answer B
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What is the highest power of 3 in the value of x!, where x is a positive integer?

(1) The highest power of 9 in the value of x! is 9.
(2) The highest power of 6 in the value of x! is 19.


From St 1 we just know that 9 is the highest power 9 in x!. So we can have either 18 or 19 units of 3 in x!. For example, I have tried 39! or 42!. In both cases, we have 9 in power of or 9^9. Thus not sufficient


We are told that highest power of 6 in the value of x! is 19. Since we get more 2s than 3s, power of 6 in x! is determined by number of 3s in x!. So, \(\frac{x!}{3}\)+\(\frac{x!}{3^2}\)+\(\frac{x!}{3^3}\)=19. Value 42! fits
So x! is 42!, from here we can count number of 3. Hence, St 2 is sufficient
Our answer B

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

(1) The highest power of 9 in the value of x! is 9. --> not correct: x! = 9^9 *k = 3^18*k(k is an integer, not multiple of 9, but can be or can't be multiple of 3) where k can be or can't be multiple of 3, so the highest power of 3 can be 18 or 19
(2) The highest power of 6 in the value of x! is 19. --> correct: x! = 6^19*m(m is an integer, not multiple of 3) = 3^19*n, so the highest power of 3 must be 19

Answer: B

imo option B,

condition a, signifies 9 ^9, which says 3^18 or 3^19 both can be answers

condtion b, says 6^19, which is equivalent to 3^19 as no of 2 will be definitely more than 19, to form 6^19

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

(1) The highest power of 9 in the value of x! is 9.
(2) The highest power of 6 in the value of x! is 19.

Answer is B
The power of 6 always indicates power of 3 as power of 2 will always be more than power of 3.

For eg: if x=9 then in 9! the highest power of 2 is 7 but highest power of 3 is 4. so highest power of 6 will be 4 which is same as highest power of 3.

we know that for the power 3 in x factorial we need to count all 3 present in X!

statement 1 : we know the power of 9 in x! but there could be a 3 that couldn't form a pair to be 9 and was not counted so insufficient
statement 2 : we know the power of 6 in x! therefore we have already accounted for all the 3 as for any factorial we will see the number of 2 in the factorial will be greater than the number 3 so all 3's have formed pair .therefore sufficient

answer B

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

Option 1: The highest power of 9 in the value of x! is 9.

Now in option 1 let us consider 2 possibilities,

1 - Can the highest power of 9 give us the highest power of 3s in x!? - Maybe. If yes, then 9^9 = 3^18 => highest power of 3 being 18 --> this is possible.
2 - Can the highest power of 9 not give us the highest power of 3s in x!? - Maybe. If yes, then we will have atleast 1 more 3 which will ensure highest power of 9 remains 9 and that highest power of 3 is not 18 but rather 19. Imagine this like 9^9 * 3 being a part of the x!.

Hence since both above considered cases are possible, option 1 is not sufficient.

Option 2: The highest power of 6 in the value of x! is 19.

6^19 = 2^19 x 3^19

Now in option 2 let us consider 2 possibilities,

1 - Can the highest power of 6 give us the highest power of 3s in x!? - Yes. Highest power of 6 will always contain the highest power of 3 as every 3 will havea corresponding 2 to form 6 (vice versa may not be true).
2 - Can the highest power of 6 not give us the highest power of 3s in x!? - No. There cannot be a case where we have highest power of 6 in x! and also an extra 3 which does not have a corresponding 2 available to form a 6.

Hence, since both above considered cases are in synch, option 2 is sufficient.

The highest power of 3 in x! depends on value of x

Statement 1: Power of 9 in x! is 9

but power of 3 in x! may be 18 or 19 hence
NOT SUFFICIENT

Statement 2: Power of 6 in x! = 19
but each 6 is made of one 2 and one 3
Powers of 2 will be higher than power of 3 in any factorial so power of 6 will be as many as the power of 3
hence, power of 3 in x! = 19
SUFFICIENT

Answer: Option B
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We are required to find the highest power of 3 in x!

(1) The highest power of 9 in the value of x! is 9.

Consider 9!
The highest power of 3 in 9! is = Quotient of (\(\frac{9}{3}\)) + Quotient of (\(\frac{9}{3^2}\)) = 3 + 1 = 4
Thus, the number of 9s i.e \(3^2\) that can be formed is 2
So the highest power of 9 in 9! is 2
Thus in this case, highest power of 3 = 2 * Highest power of 9

Consider 12!
The highest power of 3 in 12! Is = Quotient of (\(\frac{12}{3}\)) + Quotient of (\(\frac{12}{3^2}\)) = 4 + 1 = 5
Thus, the number of 9s i.e \(3^2\) that can be formed is 2 (using four 3s out of five)
So the highest power of 9 in 12! is 2
Thus in this case, highest power of 3 NOT EQUAL to 2 * Highest power of 9

Therefore, we cannot be sure of the highest power of 3 in a factorial by knowing highest power of 9.

Not Sufficient.

(2) The highest power of 6 in the value of x! is 19.
6 is a composite number so the highest power of 6 in a factorial will depend upon the number of pairs of 2 and 3 that can be formed in the factorial.

Consider 9!
The highest power of 2 in 9! Is = Quotient of (\(\frac{9}{2}\)) + Quotient of (\(\frac{9}{2^2}\)) + Quotient of (\(\frac{9}{2^3}\)) = 4 + 2 + 1 = 7
The highest power of 3 in 9! Is = Quotient of (\(\frac{9}{3}\)) + Quotient of (\(\frac{9}{3^2}\)) = 4 + 1 = 5
The maximum number of 6s (2 * 3)s that can be formed by taking Seven 2s and Five 3s is 5
Therefore, the highest power of 6 in 9! = 5 = Highest power of 3 in 9!

Consider 12!
The highest power of 2 in 12! Is = Quotient of (\(\frac{12}{2}\)) + Quotient of (\(\frac{12}{2^2}\)) + Quotient of (\(\frac{12}{2^3}\)) = 6 + 3 + 1 = 10
The highest power of 3 in 12! Is = Quotient of (\(\frac{12}{3}\)) + Quotient of (\(\frac{12}{3^2}\)) = 4 + 1 = 5
Thus, the number of 6s that can be formed = 5
Therefore, the highest power of 6 in 12! = 5 = Highest power of 3 in 12!

Hence, in any factorial, the number of 2s will ALWAYS BE MORE than the number of 3s. So, the 2s can always pair up with all the 3s to form 6s. So highest power of 3 will always be equal to the highest power of 6

Sufficient.

Answer B

Can you please explain how the power of 3 in x! can be 19. I was only able to obtain 18.

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