When a positive integer \(m\) is divided by a positive integer \(x,\) the reminder is 7 and when a positive integer

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 am; Followed by:2 members

When a positive integer \(m\) is divided by a positive integer \(x,\) the reminder is 7 and when a positive integer

by Gmat_mission » Sun May 03, 2020 1:32 pm

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When a positive integer \(m\) is divided by a positive integer \(x,\) the reminder is 7 and when a positive integer \(n\) is divided by a positive integer \(y,\) the reminder is 11. Which of the following is a possible value for \(x + y?\)

I. 18
II. 19
III. 20

A. I only
B. II only
C. III only
D. II and III only
E. None

[spoiler]OA=C[/spoiler]

Source: GMAT Club Tests

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Source: Beat The GMAT — Problem Solving | Sun May 03, 2020 1:32 pm

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

Re: When a positive integer \(m\) is divided by a positive integer \(x,\) the reminder is 7 and when a positive integer

by Scott@TargetTestPrep » Sat May 09, 2020 5:47 am

Gmat_mission wrote: ↑
Sun May 03, 2020 1:32 pm
When a positive integer \(m\) is divided by a positive integer \(x,\) the reminder is 7 and when a positive integer \(n\) is divided by a positive integer \(y,\) the reminder is 11. Which of the following is a possible value for \(x + y?\)

I. 18
II. 19
III. 20

A. I only
B. II only
C. III only
D. II and III only
E. None

[spoiler]OA=C[/spoiler]

Source: GMAT Club Tests

Let’s review a simple divisibility rule, using division by 3 as an example: The possible remainders when a number is divided by 3 are 0, 1, or 2. Note that the largest possible remainder when division by 3 is performed is 1 less than 3, which is 2. In other words, if we have a remainder 2 from a division problem, we know that the minimum value of the divisor is 2 + 1 = 3.

So, in general, when division by w is performed and a remainder of r is obtained, we know that the minimum value of the divisor w is r + 1. Thus, when division by x yields a remainder of 7, then the minimum value of x is 7 + 1 = 8. Similarly, when division by y yields a remainder of 11, then the minimum value of y is 11 + 1 = 12. Thus, the smallest value of x + y is 8 + 12 = 20, so only statement III can be true.

Answer: C

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