g(x) is defined as the product of all even integers k such t

Question

g(x) is defined as the product of all even integers k such that 0 < k ≤ x. For example, g(14) = 2 × 4 × 6 × 8 × 10 × 12 × 14. If g(y) is divisible by 4^11, what is the smallest possible value for y?

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

Please help.

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

Now, i just need help with the approach. One way is to manually count all the powers of 2 that would be sufficient.
However, i would like to use the formula for counting the power of a prime number as given in the gmat club maths book. 
i.e

For n! suppose we need to find out the powers of prime number p in n!:
n/p +n/p^2 +n/p^3 +......n/p^k where p^k <n.

I think it would make no difference even though it is not a factorial in the question, as we are anyway counting only powers of 2.

My question is, while using this formula, how many powers of p(2 in this question) should we take in this case to get the value of n(y in this question).

Can it be done this way?
Please help.

mdbharadwaj · Accepted Answer

g(x) is defined as the product of all even integers k such that 0 < k ≤ x. For example, g(14) = 2 × 4 × 6 × 8 × 10 × 12 × 14. If g(y) is divisible by 4^11, what is the smallest possible value for y?

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

Please help.

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44

4^11=2^22. So we have to find a product with atleast 22 2's in it.

in option 1 22 the total no of 2's =   +   +  +  = 11+5+2+1 = 19
in option 2 24 the total no of 2's =   +   +  +  = 12+6+3+1 = 22 . Hence B

mau5 · Answer

For the given function g(y), the generalized notation will be of the type : 2^(y/2)* !

We can see that g(14) = 2^7*7!. Now the number of times 2 appears in 7! is given by :  +  = 3+1 = 4; where   denotes the greatest integer less than or equal to x .Thus, the total number of times 2 appears in g(14) is 7+4 = 11 times.

Now, given that g(y) is divisible by 4^11 = 2^22. Thus the value of y should be such that 2 appears 22 times in g(y). Now g(22) = 2^11*11!. Just as above, we can see that the number

of 2's is 11+ + +  = 11+5+2+1 = 19. For g(24) = 2^12*12! . The total number of 2's = 12+ + +  = 12+6+3+1 = 22.

B.

ygdrasil24 · Answer

Method I : Counting Method. Method II : As you rightly pointed we can apply factorial theorem here , since presence of any odd factor will not affect our finding the # of 2s So you are supposed to take greatest integer of N/P + N/p2 + N/p3......till pn exceed N. So in our case, + + + = 12 + 6 + 3 + 1 = 22 Hope its clear.

AccipiterQ · Answer

How did you see that g(4)=2^7*7??? I'm always baffled how you guys look at these problems and instantly see that stuff. It takes most people several minutes to multiply the product of that one and you guys just see it right at it's exponent level instantly. I'm quite jealous

jasimuddin · Answer

g(24)=2.4.6……24=2^12*12!
Power of prime 2 in 12!
=12/2+12/4+12/8
=6+3+1
=10
2^(12+10)=2^22=4^11
Plug in method starting with middle man "C"
ANS: B

anairamitch1804 · Answer

Pasting official solution to this problem.

TestPrepUnlimited · Answer

We can deduce a closed formula for g(y). The idea is this product chain has terms increasing with a pattern. They are all even, thus if we divide every term by 2, we will end up with 1 * 2 * 3 * 4 .... 
For example,  g(14) = 2^7 * 1 * 2 * 3 * 4 * 5 * 6 * 7 = 2^7 * 7! . We can take out 2, 4, and 6 from the 7! and combined all 2's to get  2^{11} . 
We can start by plugging in some values from the answers now and check how close we are to  4^{11} = 2^{22} .
g(22) = 2^11 * 11!
 It is important to count the 2's in layers  (the formula is not needed for GMAT either, we can use this method every time we need to count multiples of a factor in a long chain)
We want to count the total number of multiples of 2's in 11!.
The even numbers 2, 4, 6, 8,10 are the first layer, that's 5 numbers with at least one multiple of 2.
The numbers 4 and 8 have another multiple of 2, that's 2 numbers with another "layer" of multiple of 2.
Finally, 8 has a 3rd layer, count an extra 2 from that.
Add up the layers now with the twos from 2^11, 11 + 5 + 2 + 1 = 19. We are missing a couple 2's but we can simply count the extra numbers now. 
g(24) gives an extra 24 in the product, 24 has 3 more multiples of two. 19 + 3 = 22, bingo.

Ans: B

Adarsh_24 · Answer

2*4*6*8*......y
Take 2 from each
2^y/2 (1*2*3*...y/2)

2^y/2 * y/2 !

Try numbers now
I will directly do 24( ideally go from 22)
We need to check 2^12 * 12! divisible by 2^22?
12!
12/2=6
12/4=3
12/8=1
12/16=0
Adding up 2^10

So 2^12 * 12! Becomes 2^12 * 2^10 = 2^22
Hence proved divisible.

Dooperman · Answer

We need 4^11 i.e. 2^22.
The presence of odd numbers wont contribute to the number of 2s.

So the problem essentially is that 2^22 is the number of 2s in N! for what value of N.

Test the Options. A will fail. B will work as follows:
24/2 =12 
24/4 = 6
24/8 = 3
24/16 = 1
Total = 22.

ablatt4 · Answer

for 4^11 to be a factor of the equation we must have 22 2s in the equation.

There is probably a better shortcut but I just wrote out the first 14 terms and counted until i got to 22 2s, which took like 90 seconds.

Whole questions boils down to understanding we are solving for a number with 22 2s.

Rishitkaul · Answer

I used the counting method for this. But what if we get options which are pretty big and counting would be impractical. Do we have a proper method for these types of questions.

g(x) is defined as the product of all even integers k such t

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

g(x) is defined as the product of all even integers k such t

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards