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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
(A) 22
(B) 24
(C) 28
(D) 32
(E) 44
Show SpoilerMy question
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.
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
Joined: 26 Feb 2013
Last visit: 10 Oct 2017
Posts: 22
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30
WE:Consulting (Telecommunications)
GMAT 1: 660 Q50 V30
Posts: 22
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
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 = [22/2] + [22/4] +[22/8] +[22/16] = 11+5+2+1 = 19
in option 2 24 the total no of 2's = [24/2] + [24/4] +[24/8] +[24/16] = 12+6+3+1 = 22 . Hence B
(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 = [22/2] + [22/4] +[22/8] +[22/16] = 11+5+2+1 = 19
in option 2 24 the total no of 2's = [24/2] + [24/4] +[24/8] +[24/16] = 12+6+3+1 = 22 . Hence B
12bhang
For the given function g(y), the generalized notation will be of the type : 2^(y/2)*[y/2]!
We can see that g(14) = 2^7*7!. Now the number of times 2 appears in 7! is given by : [7/2]+[7/4] = 3+1 = 4; where [x] 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+[11/2]+[11/4]+[11/8] = 11+5+2+1 = 19. For g(24) = 2^12*12! . The total number of 2's = 12+[12/2]+[12/4]+[12/8] = 12+6+3+1 = 22.
B.
_________________
General Discussion
12bhang
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, [24/2] + [24/4] + [24/8] + [24/16]
= 12 + 6 + 3 + 1 = 22
Hope its clear.
AccipiterQ
Joined: 26 Sep 2013
Last visit: 03 Sep 2020
Posts: 144
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
GMAT 2: 730 Q49 V41
Posts: 144
mau5
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
Joined: 16 Oct 2012
Last visit: 02 Nov 2016
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Status:My heart can feel, my brain can grasp, I'm indomitable.
Affiliations: Educator
Location: Bangladesh
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Posts: 34
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
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
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TestPrepUnlimited
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Tutor
Joined: 16 Sep 2014
Last visit: 30 Jun 2022
Posts: 1,221
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Expert reply
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
_________________
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
12bhang
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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.
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.
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.
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
Joined: 18 Dec 2024
Last visit: 24 Sep 2025
Posts: 87
Location: United States (FL)
Concentration: Finance
Schools: Duke '26 Darden '26 Johnson '26 UNC '26
Schools: Duke '26 Darden '26 Johnson '26 UNC '26
Posts: 87
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.
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.
12bhang
Posts: 7
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.
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