If x is an integer, is 2^x a factor of 12!?

Regarding your first equation, why can't x be 7+5 =12?

Because the prime factorisation of 12! only includes 10 2's and therefore X cannot contain more than this. The answer is B.

Hello,
Can you please explain how are we getting 10 factors for 12!?

Aprajita760

\(12! = 12*11*10*9*8*7*6*5*4*3*2*1\)

For \(2^x\) to be a factor of 12!, 12! needs to be fully divisible by \(2^x\) (i.e. no remainder).

If we break down the 12! sequence into prime factors (e.g. \(12=2^2*3^1\)) we would get: \((2^2*3^1)(11^1)(2^1*5^1)(3^2)(2^3)(7^1)(2^1*3^1)(5^1)(2^2)(3^1)(2^1)(2^0)\), which simplifies to: 2^10*\(3^5*5^2*7^1*11^1\).

Now, we can see that 2^10 is the largest power of 2 that is a factor of 12!. If we know whether x is more or less than 10 we will know whether \(2^x\) is a factor of 12!.

St-1: x is the sum of two distinct prime factors: if the prime factors are 2 and 3 then x=5 and \(2^5\) is a factor of 12!, but if the prime factors are 5 and 7 then x=12 and 2^12 is NOT a factor of 12!. Insufficient

St-2: 0 < x < 11: As x is less than 11 and an integer (question stem) we know that the max value of x can be 10 and based on the above we know that 2^10 is a factor of 12!. Sufficient

Answer is therefore B

Hope this helps

aliakberza

How to find maximum power of 2 in 12!

We have to divide 12 by every \(2^x\), where x is a positive integer and \(2^x<12\), ignore the remainder and keep the quotient

12/\(2^1\)=6
12/\(2^2\)=3
12/\(2^3\)=1

6+3+1=10

So the maximum value of x can be 10

Hope, it's clear now

12! = 2*4*6*8*10*12*k, where k is an odd integer
—> 12! = 2*2^2*2*2^3*2*2^2*m where m is any odd integer
—> 12! = 2^10*m

So maximum value of x for which 2^x can be a factor of 12! is “10”

(1) x is the sum of two distinct single-digit prime numbers.

Maximum value of x possible = 5 + 7 = 12
—> x > 10

Insufficient

(2) 0 < x < 11
—> Maximum value of x possible = 10

Sufficient

IMO Option B

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