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M22-27

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M22-27  [#permalink]

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New post 16 Sep 2014, 01:17
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (01:47) correct 29% (01:44) wrong based on 116 sessions

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Re M22-27  [#permalink]

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New post 16 Sep 2014, 01:17
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Official Solution:

If \(f(x)\) is defined as the largest integer \(N\) such that \(x\) is divisible by \(2^N\), which of the following numbers is the greatest?

A. \(f(24)\)
B. \(f(42)\)
C. \(f(62)\)
D. \(f(76)\)
E. \(f(84)\)


Often the hardest part is rewording a question to understand what it's really asking.

So, we have an integer \(x\). It has some power of 2 in its prime factorization (\(2^n\)) and \(f(x)\) is the value of that \(n\). Basically \(f(x)\) is the power of 2 in prime factorization of \(x\).

For example, if \(x\) is say 40, then \(f(x)=3\). Why? Because the largest integer \(n\) such that 40 is divisible by \(2^n\) is 3: \(\frac{40}{2^3}=5\), or \(40=2^3*5\) - the power of 2 in prime factorization of 40 is 3.

Hence all we need to do to answer the question is to factorize all options and see which one has 2 in highest power.

A. \(f(24)\): factorize 24: \(24 = 2^3*3\), thus \(f(24) = 3\)

B. \(f(42)\): factorize 42: \(42 = 2*21\), thus \(f(42) = 1\)

C. \(f(62)\): factorize 62: \(62 = 2*31\), thus \(f(62) = 1\)

D. \(f(76)\): factorize 76: \(76 = 2^2*19\), thus \(f(76) = 2\)

E. \(f(84)\): factorize 84: \(84 = 2^2*21\), thus \(f(84) = 2\)


Answer: A
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Re: M22-27  [#permalink]

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New post 02 Jul 2015, 05:32
Hi,

I find this answer hard to do the algebraic translation, does some body has a shortcut or an easier way to de the algebraic translate of question stem?

Thanks a lot.

Regards.

Luis Navarro
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Re: M22-27  [#permalink]

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New post 02 Jul 2015, 05:35
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luisnavarro wrote:
Hi,

I find this answer hard to do the algebraic translation, does some body has a shortcut or an easier way to de the algebraic translate of question stem?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Often the hardest part is rewording the question to understand what it's really asking.

So, we have an integer x. It has some power of 2 in its prime factorization (2^n) and f(x) is the value of that n. Basically f(x) is the power of 2 in prime factorization of x.

For example, if x is say 40, then f(x)=3. Why? Because the largest integer n such that 40 is divisible by 2^n is 3: 40/2^3=5, or 40=2^3*5 --> the power of 2 in prime factorization of 40 is 3.

Hence all we need to do to answer the question is to factorize all options and see which one has 2 in highest power.

A. f(24) --> 24 = 2^3*3
B. f(42) --> 42 = 2*21
C. f(62) --> 62 = 2*31
D. f(76) --> 76 = 2^2*19
E. f(84) --> 84 = 2^2*21

Answer: A.

Hope it's clear.
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Re: M22-27  [#permalink]

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New post 02 Jul 2015, 06:10
Bunuel wrote:
luisnavarro wrote:
Hi,

I find this answer hard to do the algebraic translation, does some body has a shortcut or an easier way to de the algebraic translate of question stem?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Often the hardest part is rewording the question to understand what it's really asking.

So, we have an integer x. It has some power of 2 in its prime factorization (2^n) and f(x) is the value of that n. Basically f(x) is the power of 2 in prime factorization of x.

For example, if x is say 40, then f(x)=3. Why? Because the largest integer n such that 40 is divisible by 2^n is 3: 40/2^3=5, or 40=2^3*5 --> the power of 2 in prime factorization of 40 is 3.

Hence all we need to do to answer the question is to factorize all options and see which one has 2 in highest power.

A. f(24) --> 24 = 2^3*3
B. f(42) --> 42 = 2*21
C. f(62) --> 62 = 2*31
D. f(76) --> 76 = 2^2*19
E. f(84) --> 84 = 2^2*21

Answer: A.

Hope it's clear.


Thanks, now it is very clear to me¡¡¡

Regards

Luis Navarro
Looking for 700
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Re: M22-27  [#permalink]

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New post 30 Jan 2018, 10:10
+1 for A
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Re: M22-27  [#permalink]

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New post 14 Jun 2019, 20:15
I reworded to find the largest power of 2 I could find in each function. Then I tested the answer choices.

F(24): 24 = 2^3 *3^1

All the others have 2 or less powers of 2.
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Re: M22-27   [#permalink] 14 Jun 2019, 20:15
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