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

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

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Updated on: 01 Apr 2013, 04:34
4
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65% (hard)

Question Stats:

63% (02:31) correct 37% (02:29) wrong based on 353 sessions

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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

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?

Originally posted by 12bhang on 01 Apr 2013, 04:31.
Last edited by Bunuel on 01 Apr 2013, 04:34, edited 1 time in total.
Renamed the topic and edited the question.
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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01 Apr 2013, 10:34
4
4
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
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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01 Apr 2013, 05:59
3
2
12bhang wrote:
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

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.
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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11 Nov 2013, 07:24
1
12bhang wrote:
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

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?

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.
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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11 Nov 2013, 18:32
mau5 wrote:
12bhang wrote:
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

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.

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
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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08 Nov 2015, 00:04
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
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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26 Feb 2017, 17:49
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Re: g(x) is defined as the product of all even integers k such t  [#permalink]

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16 Oct 2018, 01:47
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Re: g(x) is defined as the product of all even integers k such t   [#permalink] 16 Oct 2018, 01:47
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