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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
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@P@ is defined as the product of all even integers such r such that
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24 Aug 2021, 05:16
24 Aug 2021, 05:16
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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
(A) 22
(B) 24
(C) 28
(D) 32
(E) 44
Re: @P@ is defined as the product of all even integers such r such that
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27 Aug 2021, 10:19
27 Aug 2021, 10:19
Expert Reply
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.
Answer: B
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.
Answer: B
Re: @P@ is defined as the product of all even integers such r such that
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11 Sep 2021, 09:02
11 Sep 2021, 09:02
Can you explain in another way pls? or in a simpler way
