GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 19 Feb 2020, 20:54

### 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

# The product of the first seven positive multiples of three is closest

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61302
The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

22 Jan 2020, 23:11
00:00

Difficulty:

75% (hard)

Question Stats:

47% (01:35) correct 53% (02:18) wrong based on 66 sessions

### HideShow timer Statistics

Competition Mode Question

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

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

Are You Up For the Challenge: 700 Level Questions

_________________
Director
Joined: 30 Sep 2017
Posts: 660
GMAT 1: 720 Q49 V40
GPA: 3.8
The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

Updated on: 25 Jan 2020, 03:36
2
Roughly approximate
= 3*(6*9*12*15)*(18*21)
= 3*~(10*10*10*10)*~(20*20)
= 3*10000*400
= 12,000,000
Thus, it is closest to 10^7.

Hit kudo if you like this explanation

Originally posted by chondro48 on 22 Jan 2020, 23:35.
Last edited by chondro48 on 25 Jan 2020, 03:36, edited 2 times in total.
VP
Joined: 20 Jul 2017
Posts: 1322
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

22 Jan 2020, 23:36
1
First seven positive multiples of 3 = {3, 6, 9, 12, 15, 18, 21}
--> Product = $$3*6*9*12*15*18*21 = 3*(2*3)*(3^2)*(3*2^2)*(3*5)*(3^2*2)*(3*7)$$
= $$2^4*3^9*5*7$$
= $$10*56*19683$$
= $$10*56*20000$$ (approx)
= $$112*10^5$$
= $$1.12*10^7$$

Option C
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5940
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 00:53
1
first 7 multiples of 3; 3,6,9,12,15,18,21 ; its product shall be closet to ~ 10^7
IMO C

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 105
Senior Manager
Joined: 20 Mar 2018
Posts: 465
Location: Ghana
Concentration: Finance, Real Estate
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 02:31
1
The product of the first seven positive multiples of three is closest to which of the following powers of 10?

Multiple of 3 is in the form 3x where x= any positive integer
First seven = 3,6,9,12,15,18,21
Product = 3•(2•3)•(3^2)•(2^2•3)•(3•5)•(2•3^2)•(3•7) = 2^4•3^9•5•7
= 2^3•3^9•7•10= (2^3•70)•3^9 = 560•19,683= 11,022,480
10,000,000 < 11,022,480 < 12,000,000
10^7 < 11,022,480 < 12•10^6 ~ 10^8
10^7 < 11,022,480 < 10^8

.: 11,022,480 is close to 10^7 since it difference is smaller than to 10^8

Hit that C

Posted from my mobile device
CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 385
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 03:40
1
This is an excellent question which tests you on your ability to do quick approximations. Of course, you also need to know some specific numbers here if you want to be good with your approximation.

The first seven positive multiples of three are 3,6,9,12, 15, 18 and 21. When we take the product of these numbers, we get 3*6*9*12*15*18*21.

We see that 3 is common in all of them. Therefore, we already have a 3^7 in the product. 3^7 = 2187. Once 3^7 is taken out, the product also has 1*2*3*4*5*6*7 = 7!.
7! = 5040.

Therefore, the product of the first seven positive multiples of 3 = 2187 * 5040. Here’s where you need to bring in your approximation skills instead of trying to calculate the exact answer.
I would round up 2187 to 2200 and round down 5040 to 5000. The product of 5000 and 2200 is 11,000,000. The closest power of 10 to this number is 10^7.

Therefore, the correct answer option is C.

Hope that helps!
_________________
VP
Joined: 24 Nov 2016
Posts: 1218
Location: United States
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 05:02
1
Quote:
The product of the first seven positive multiples of three 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

prod(3,6,9,12,15,18,21)=3^7(7!)
3^7=2187~2(10^3)
7!=5040~5(10^3)
prod(…)=10(10^6)=10^7

Ans (C)
Director
Joined: 07 Mar 2019
Posts: 693
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 09:21
1
The product of the first seven positive multiples of three is closest to which of the following powers of 10?

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

(3*1)*(3*2)*(3*3)*(3*4)*(3*5)*(3*6)*(3*7) = 3^7*1*2*3*4*5*6*7 = 2187*7! = 2187*5040
~ 2200 * 5000 = 11,000,000
~$$10^7$$

_________________
Ephemeral Epiphany..!

GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
GMATPREPSoft1 680(Q48,V35) June 26, 2019
Director
Joined: 25 Jul 2018
Posts: 549
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 15:18
1
The product of the first seven positive multiples of three is closest to which of the following powers of 10?

$$3*6*9*12*15*18*21= 3^{7}* 7!= (3*3^{6})*(720*7)= 21*729*720$$

--> $$21*729*720≈ 2*10*700*700$$

--> $$2*10*700*700= 2*10*490000= 980000*10 = 9800000≈ 10,000,000= 10^{7}$$

SVP
Joined: 03 Jun 2019
Posts: 2005
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 17:42
1
Product of first seven positive multiples of 3 = 3*6*9*12*15*18*21
= 3^7*7! = 2187*5040 = 11022480 which is closest to 10^7.

IMO C

Posted from my mobile device
CR Forum Moderator
Joined: 18 May 2019
Posts: 706
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 20:14
1
3*6*9*12*15*18*21
We can approximate 18 to 20, and 21 to 20. In addition, we can approximate 12 to 10 and 9 to 10.
So, 3*6*9*12*15*18*21≈3*6*10*10*15*20*20 = 3*6*15*4*10^4 = 3*6*60*10^4 = 3*36*10^5 = 108*10^5, but we can approximate 108 to 100, hence 108*10^5 ≈ 10^7.

Senior Manager
Joined: 01 Mar 2019
Posts: 438
Location: India
Concentration: Strategy, Social Entrepreneurship
Schools: Ross '22, ISB '20, NUS '20
GPA: 4
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 21:42
1
3*6*9*12*15*18*21
=162*180*378
~3*10^4*4*10^2
~12*10^6
~10^7

OA:C

Posted from my mobile device
Manager
Joined: 30 Jul 2019
Posts: 101
Location: Viet Nam
Concentration: Technology, Entrepreneurship
GPA: 2.79
WE: Education (Non-Profit and Government)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 22:49
1
$$3*6*9*12*15*18*21 = 11022480 = 1.1 * 10^7$$
=> Closest to $$10^7$$

=> Choice C
Math Expert
Joined: 02 Sep 2009
Posts: 61302
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

23 Jan 2020, 23:26
1
Bunuel wrote:

Competition Mode Question

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

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

Are You Up For the Challenge: 700 Level Questions

Similar questions to practice:
https://gmatclub.com/forum/the-product- ... 35192.html (OG)
https://gmatclub.com/forum/the-product- ... 28135.html
https://gmatclub.com/forum/the-product- ... 49050.html
http://gmatclub.com/forum/the-product-o ... 37076.html

Hope it helps.
_________________
GMAT Tutor
Status: Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator
Joined: 26 Jun 2014
Posts: 107
GMAT 1: 740 Q51 V39
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

24 Jan 2020, 12:32
Question:
The product of the first seven positive multiples of three is closest to which of the following powers of 10?

Solution:

The product of the first seven positive multiples of three
= (3*1) * (3*2) * (3*3) * (3*4) * (3*5) * (3*6) * (3*7)
= $$3^7 * (1 * 2 * 3 * 4 * 5 * 6 * 7)$$
= $$(3^4 * 3^3 )* (2 * 5) * (3 * 4) * (6 * 7)$$
= $$(81 * 27 )* (10) * (12) * (42)$$
=~ $$(80 * 30 )* (10) * (10) * (40)$$
= $$(96 * 10^5)$$
=~ $$(100 * 10^5)$$
= $$10^7$$

_________________
Sujoy Kumar Datta
Director - CUBIX Educational Institute Pvt. Ltd. (https://www.cubixprep.com)
IIT Kharagpur, TU Dresden Germany
GMAT - Q51 & CAT (MBA @ IIM) 99.98 Overall with 99.99 QA
_________
Feel free to talk to me about GMAT & GRE | Ask me any question on QA (PS / DS)
Let's converse!
Skype: sk_datta
Alt. Email: sujoy.datta@gmail.com
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9426
Location: United States (CA)
Re: The product of the first seven positive multiples of three is closest  [#permalink]

### Show Tags

28 Jan 2020, 06:22
1
Bunuel wrote:

Competition Mode Question

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

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

Are You Up For the Challenge: 700 Level Questions

We need the approximate value of:

3 x 6 x 9 x 12 x 15 x 18 x 21

We can group the numbers as follows:

9 x 12 = 108 ≈ 100 = 10^2

6 x 18 = 108 ≈ 100 = 10^2

3 x 15 x 21 = 45 x 21 ≈ 50 x 20 = 1000 = 10^3

Therefore, the product of the first seven positive multiples of 3 is approximately:

10^2 x 10^2 x 10^3 = 10^7

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
181 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: The product of the first seven positive multiples of three is closest   [#permalink] 28 Jan 2020, 06:22
Display posts from previous: Sort by