If x is an integer, is 3^x a factor of 15! ?

IMO : B

Highest power of 3 in 15! = 6

St 1 : x is the sum of two distinct single-digit prime numbers.

Prime = {2,3,5,7}
for sum(2,3) = 2+3 = 5 ... Satisfies
for sum(5,7) = 5+7 = 12 ... Doesn't satisfy
Hence not suff

St 2 : 0 < x < 6
Since highest power of 3 in 15! can be 6.
Max value of x = 6
Hence given condition is well in the range.

Hence suff

For 3^x to be a factor of 15! => 15! should be divisible by 3^x, leaving no remainder

Power of 3 in 15! = 15/3 + 15/9 = 5 + 1 = 6 => x <= 6.
Note that x can be negative too, and for all negative integer values of x, 15! / 3^x will result in an integer. The only constraint is that x <= 6

St. 1 => X is sum of two distinct single-digit prime numbers4
=> x = 2+3 = 5 => Answer is Yes & x = 2+5=7 => Answer is No => Not sufficient

St. 2 => 0 < x < 6 => Answer is always Yes as discussed above => Sufficient

Answer B

Should not it be D? The first statement may work even if X is 5+7. If X = 12, 36 can be a factor of 15!. Am I wrong?
15! = 15*.... * 12 * .. 3 * 2 * 1.

No. Statement 1 is not sufficient. Look below.

if x = 12, then \(3^{12}\)should be a factor of 15!

For maximum power of \(A^n\)in M! , we use the formula: M/A + M/\(A^2\) + M/\(A^3\) .... M/\(A^n\) where \(A^n\)< M

Thus, maximum p9ower of 3 in 15! = 15/3+ 15/3^2 = 5+1 = 6. So in other words, 3,\(3^2\),\(3^3\), \(3^4\), \(3^5\), \(3^6\) are the only powers of 3 that will give you as a factor of 15!. Anything greater than \(3^6\) (\(3^7\) or \(3^{12}\)) will not give you an integer when you calculate 15!/\(3^{12}\).

I believe you are confusing '^' symbol with the symbol '*'.

"^" is 'raised to the power of' and NOT the multiplication symbol. We write multiplication symbol as either 'x' or '*'

3^12 = \(3^{12}\)= 3*3*3*3*3*3*3*3*3*3*3*3 \(\neq\) 36

3*12 = 36

Hope this helps.

Ahh yes, that makes sense. Confused the symbols. Thank you!

If x is an integer, is 3^x a factor of 15! ?
reworded,
is x<=6?
this is because 15! has six 3's aka 3^6

(1) x is the sum of two distinct single-digit prime numbers.
x=2+3=5 yes
x=3+5=8 no
Insufficient

(2) 0 < x < 6
x is less than 6
Sufficient

Answer: B

15! Has 6 3s so any number from 3^0 to 3^6 will be a factor of 15!..hence b


Sent from my iPhone using Tapatalk

For this we need to find the highest power of 3 in 15!

(15/3)+(15/9)=5+1=6
(1) x is the sum of two distinct single-digit prime numbers. 2,3,5,7 are the four possible prime nos. 2+3=5 looks good but 3+7=10 exceeds 6.Insufficient
(2) 0 < x < 6. Now, if xis more than 0 but less than 6, this is what we want since highest power of 3 in 15! is 6. Sufficient.

Answer B

As per my normal approach for solving problems, where possible I try to avoid actual calculations in favour of more intuitive methods (don't have a strong math background).

If x is an integer, is 3^x a factor of 15! ? (this is a Yes or No question)

How many times does the number 3 turn up in 15!
15! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15
# of 3's =....1...........1...........2..............1...............1 = 6 x 3's

From this we can determine that as long as x <= 6 it will be divisible by 15! ----> if x <= 6 Answer will be YES (Sufficient), if x > 6 answer will be NO (Sufficient)

On to the statements...

(1) x is the sum of two distinct single-digit prime numbers... single digit primes are 2, 3, 5 & 7 ----> x is between 2 + 3 = 5 and 5 + 7 = 12 ----> INSUFFICIENT
(2) 0 < x < 6.... Based of the previous information we can definitively answer this question as YES ----> SUFFICIENT

Answer = B

P.S. as I have posted on a few questions now trying to build my own confidence (teaching helps me understand) but if you have any feedback on the way I am presenting it please let me know. Thanks in advance

PRINCETON REVIEW OFFICIAL SOLUTION:

This is a much more complicated question than the first one, and we really have to do some thinking up front before we worry about the statements. First, this is a Yes/No question, so we need to be clear that it doesn’t matter whether the answer is yes or no, whether 3^x is a factor of 15! or not. All that matters is that we know for certain one way or the other. So how will we know?

3^x describes a certain number of 3s multiplied together. If 3^x is a factor of 15! then all those 3s divide evenly into 15! with nothing left over (if z is a factor of some number then that number is divisible by z). The question becomes, “How many factors of 3 are there in 15! ?” Write out 15! and count the factors of 3.

15! = 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

There’s one in the 15 (3 x 5), one in the 12 (3 x 4), two in the 9 (3 x 3), one in the 6 (3 x 2) and one in the 3 (3 x 1). That’s a total of six 3s. So now we can finally say what the missing piece of the puzzle is. Since 15! contains six factors of 3, if x = 6 or anything less than 6 then 3^x will be a factor of 15! and the answer will be Yes. If x is anything larger than 6 then 3^x will not be a factor of 15! because it will have too many 3s. The answer will be No. So the key question we’re concerned with as we turn to the statements is do we know whether x<=6 or x>6 so we have answered our question definitively. Statement 2 is sufficient and the answer to the question is (B).

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