What is the lowest positive integer that is divisible by

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

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

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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

Source: Beat The GMAT — Problem Solving | Mon Apr 08, 2019 5:35 am

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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]

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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed 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
$$=\left(3\cdot5\right)\cdot7\cdot17\cdot19$$
$$=15\cdot7\cdot17\cdot19$$
$$7\cdot15\cdot17\cdot19$$

$$Answer\ is\ Option\ C$$

swerve: Legendary Member; Posts: 2604; Joined: Sun Oct 29, 2017 2:04 pmFollowed 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__

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