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

It is currently 22 Nov 2019, 02:45

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

The product of the first eight positive even integers is closest to

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

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 439
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post Updated on: 26 Dec 2017, 09:37
5
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (02:03) correct 45% (01:59) wrong based on 328 sessions

HideShow timer Statistics

The product of the first eight positive even integers is closest to which of the following powers of 10?

(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?

_________________
Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Originally posted by enigma123 on 25 Feb 2012, 17:01.
Last edited by Bunuel on 26 Dec 2017, 09:37, edited 2 times in total.
Edited.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59245
Re: The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post 25 Feb 2012, 17:16
3
1
enigma123 wrote:
The product of the fir…st eight positive even integers is closest to which of the following powers of 10?
(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?


We should find the approximate value of 2*4*6*8*10*12*14*16 to some power of 10.

2*4*6=~50 (actually less than 50);
8*12=~100 (actually less than 100);
10;
14*16=~200 (actually more than 200);

2*4*6*8*10*12*14*16=~50*100*10*200=10^7 (6 zeros plus one zero produced by 2*5).

Answer: C.

Similar question to practice: the-product-of-all-the-prime-numbers-less-than-20-is-closest-126587.html

Hope it helps.
_________________
Manager
Manager
avatar
Joined: 07 Apr 2012
Posts: 87
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Re: The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post 22 Nov 2013, 07:29
Bunuel wrote:
enigma123 wrote:
The product of the fir…st eight positive even integers is closest to which of the following powers of 10?
(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?


We should find the approximate value of 2*4*6*8*10*12*14*16 to some power of 10.

2*4*6=~50 (actually less than 50);
8*12=~100 (actually less than 100);
10;
14*16=~200 (actually more than 200);

2*4*6*8*10*12*14*16=~50*100*10*200=10^7 (6 zeros plus one zero produced by 2*5).

Answer: C.

Similar question to practice: the-product-of-all-the-prime-numbers-less-than-20-is-closest-126587.html

Hope it helps.



Is there a more logical way of solving?
Manager
Manager
avatar
Joined: 25 Oct 2013
Posts: 141
Re: The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post 22 Nov 2013, 07:49
1
I solved it this way... since this is approximation question, we can approximate as below

\(2*4*6*8*10*12*14*16\) rounding up or down each of the numbers to the nearest 10.

\(2*4*10*10*10*10*10*20\)
= \(2*4*10^5*2*10\)
= \(16*10^6\)
=\(20*10^6\)
=\(2*10^7\)

it is closer to \(10^7\) than \(10^8\). so its C.
_________________
Click on Kudos if you liked the post!

Practice makes Perfect.
Intern
Intern
avatar
S
Joined: 13 Dec 2016
Posts: 41
Re: The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post 28 Jan 2017, 02:37
1
I solved it the following way:
2*4*6*8*10*12*14*16
= 1*5*5*10*10*10*15*20 (approximately)
=25*1000*300
=75 *10^5
=100*10^5 (approximately)
=10^7
Senior Manager
Senior Manager
User avatar
P
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 440
Premium Member
Re: The product of the first eight positive even integers is closest to  [#permalink]

Show Tags

New post 09 Jan 2019, 07:37
The goal is to pick from the positive even ints 2,4,6,8,10,12,14,16 to get products close to powers of 10.

10
6*16=96 =~ 100
8*12=96 =~ 100
4*14*2 = 56*2 = 112 =~ 100

So, we get something close to 10^7
GMAT Club Bot
Re: The product of the first eight positive even integers is closest to   [#permalink] 09 Jan 2019, 07:37
Display posts from previous: Sort by

The product of the first eight positive even integers is closest to

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





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