GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Oct 2019, 23:28 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  M22-27

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58398

Show Tags 00:00

Difficulty:   35% (medium)

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

HideShow timer Statistics

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. $$f(84)$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58398

Show Tags

1
5
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 $$f(84) = 2$$

_________________
Intern  Joined: 24 Jun 2015
Posts: 45

Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 58398

Show Tags

1
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

Hope it's clear.
_________________
Intern  Joined: 24 Jun 2015
Posts: 45

Show Tags

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

Hope it's clear.

Thanks, now it is very clear to me¡¡¡

Regards

Luis Navarro
Looking for 700
Senior Manager  S
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33

Show Tags

+1 for A
_________________
" The few , the fearless "
VP  P
Joined: 14 Feb 2017
Posts: 1213
Location: Australia
Concentration: Technology, Strategy
Schools: LBS '22
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 WE: Management Consulting (Consulting)

Show Tags

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.
_________________
Goal: Q49, V41

+1 Kudos if I have helped you Re: M22-27   [#permalink] 14 Jun 2019, 20:15
Display posts from previous: Sort by

M22-27

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  