BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

What is the lowest positive integer that is divisible by

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2276
Joined: Sat Oct 14, 2017 6:10 am
Followed by:3 members

What is the lowest positive integer that is divisible by

by VJesus12 » Mon Apr 08, 2019 5:35 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

What is the lowest positive integer that is divisible by each of the odd integers between 15 and 21, inclusive?

A) 3×17×19×21
B) 5×17×19×23
C) 7×15×17×19
D) 7×15×19×21
E) 15×17×19×21

[spoiler]OA=C[/spoiler]

Source: Economist GMAT Tutor
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8085
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Wed Apr 10, 2019 4:35 pm
VJesus12 wrote:What is the lowest positive integer that is divisible by each of the odd integers between 15 and 21, inclusive?

A) 3×17×19×21
B) 5×17×19×23
C) 7×15×17×19
D) 7×15×19×21
E) 15×17×19×21

[spoiler]OA=C[/spoiler]

Source: Economist GMAT Tutor

The odd integers from 15 to 21 inclusive are:

15, 17, 19, 21

The LCM of these numbers is 3 x 5 x 17 x 19 x 7.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pm
Followed by:5 members

by deloitte247 » Thu Apr 25, 2019 10:03 pm
Odd integers between 15 and 21 = 15, 17,19 and 21
Lowest positive integers that is divisible by each of these numbers will be its L.C.M
15 = 3 * 5
17 and 19 are prime numbers
21 = 3 * 7

L.C.M of 15 and 21 = 3*5*7
L.C.M of 15,17,19,21
$$=\left(3\cdot5\right)\cdot7\cdot17\cdot19$$
$$=15\cdot7\cdot17\cdot19$$
$$7\cdot15\cdot17\cdot19$$

$$Answer\ is\ Option\ C$$
Top
Legendary Member
Posts: 2499
Joined: Sun Oct 29, 2017 2:04 pm
Followed by:6 members

by swerve » Fri Apr 26, 2019 7:18 am
Ok, so we need to make sure that the number is divisible by 15, 17, 19, and 21.

It is mandatory to have 17 and 19.
\(15 = 3*5\)
\(21 = 3*7\)
so we need at least one factor of 3, then one of 5, and one of 7.

overall:
we need
1 factor of 3, 1 factor of 5, 1 factor of 7, 17, and 19.

Therefore, __C__
Top
Post new topic Post Reply
• Page 1 of 1