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Manager  G
Joined: 24 Feb 2020
Posts: 165
Location: Italy
WE: Analyst (Investment Banking)
@P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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10 00:00

Difficulty:   65% (hard)

Question Stats: 57% (02:15) correct 43% (02:32) wrong based on 96 sessions

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@P@ is defined as the product of all even integers such r such that $$0<r\leq{P}$$. For example $$@P@= 2 * 4 * 6 * 8 * 10 * 12 * 14$$. If @K@ is divisible by $$4^{11}$$, what is the smallest possible value for k?

(A) 22
(B) 24
(C) 28
(D) 32
(E) 44
Intern  B
Joined: 21 Aug 2019
Posts: 22
Re: @P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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Here's the trick, the number is not factorial thus we cannot use 44!/2 = 22 2's

here it will go by quick cal
2 * 1 = 2.... till.........12*2 = 24

then the sum of 2s will be 24
Target Test Prep Representative V
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Re: @P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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Ravixxx wrote:
@P@ is defined as the product of all even integers such r such that $$0<r\leq{P}$$. For example $$@P@= 2 * 4 * 6 * 8 * 10 * 12 * 14$$. If @K@ is divisible by $$4^{11}$$, what is the smallest possible value for k?

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

Since 4^11 = 2^22, we need to find the smallest value of k such that @k@ has 22 factors of 2.

Let’s now test the answer choices; but we will start with choice C. If it has exactly 22 factors of 2, we have found our answer. If it has fewer than 22 factors of 2, we can proceed to choices D and E. If it has more than 22 factors of 2, we can backtrack to choices B and A.

If k = 28, we see that each of the factors 2, 6, 10, 14, 18, 22, and 26 have 1 factor of 2; each of the factors 4, 12, 20, and 28 have 2 factors of 2; each of the factors 8 and 24 have 3 factors of 2; finally, 16 has 4 factors of 2. So the total number of factors @28@ has is 7 x 1 + 4 x 2 + 2 x 3 + 1 x 4 = 7 + 8 + 6 + 4 = 25 factors of 2. This is more than 22 factors of 2. So let’s backtrack to 24. If k = 24, we will remove the factors 26 and 28, which means we are removing 3 factors of 2 since 26 and 28 have 1 and 2 factors of 2, respectively. Therefore, we see that @24@ will have exactly 22 factors of 2.

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Intern  B
Joined: 06 Aug 2019
Posts: 2
Re: @P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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1
@k@ is divisible by 4^11

4^11 = 2^22

minimum valve of k can be the number which has the minimum 2^22 factor.

maximum power of 2 which divide the number is:

24!/2=12/2=6/2=3/2=1/2

maximum power of 2 which divide the 24! is 12+6+3+1=22
which satsifies the condition
hence B is correct

Posted from my mobile device
Intern  B
Joined: 01 Jan 2019
Posts: 22
Re: @P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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Ayushiawasthi wrote:
Here's the trick, the number is not factorial thus we cannot use 44!/2 = 22 2's

44! contains many more than 22 twos... in fact it contains:

22+11+5+2+1=41 twos.
Intern  B
Joined: 01 Jan 2019
Posts: 22
Re: @P@ is defined as the product of all even integers such r such that 0<  [#permalink]

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@K@ will contain the same number of twos as K! because the odd numbers in the factorial don't affect the number of twos.

Thus we can use the usual technique where we divide repeatedly by 2.

If we start with the middle number, 28: @28@ contains 14+7+3+1=25 twos. We only need 22 twos, by removing the 28 and the 26 we remove 3 twos, so 24 is the correct answer. Re: @P@ is defined as the product of all even integers such r such that 0<   [#permalink] 28 May 2020, 11:08

# @P@ is defined as the product of all even integers such r such that 0<   