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If rand s are positive integers, each greater than 1, and if [#permalink]
Expert's post 00:00

Question Stats: 50% (01:14) correct 50% (00:21) wrong based on 12 sessions
If rand s are positive integers, each greater than 1, and if $$11(s - 1) = 13(r - 1)$$, what is the least possible value of $$r + s$$?

(A) 2

(B) 11

(C) 22

(D) 24

(E) 26
[Reveal] Spoiler: OA

_________________ GRE Instructor Joined: 10 Apr 2015
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Re: If rand s are positive integers, each greater than 1, and if [#permalink]
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Expert's post
Carcass wrote:
If rand s are positive integers, each greater than 1, and if $$11(s - 1) = 13(r - 1)$$, what is the least possible value of $$r + s$$?

(A) 2

(B) 11

(C) 22

(D) 24

(E) 26

I have a feeling there's a better approach that mine, but here it goes....

GIVEN: $$11(s - 1) = 13(r - 1)$$

Expand: $$11s - 11 = 13r - 13$$

Add 13 to both sides: $$11s+2 = 13r$$

Subtract 11s from both sides: $$2 = 13r-11s$$

Subtract 2r from both sides: $$2 - 2r = 11r-11s$$

Factor both sides: $$2(1-r) = 11(r-s)$$

Since r and s are INTEGERS, we know that $$(r-s)$$ is an INTEGER, which means $$11(r-s)$$ is a multiple of 11

From this, we can conclude that $$2(1-r)$$ is a multiple of 11

What is the smallest value of r (given that r is a positive integer greater than 1) such that $$2(1-r)$$ is a multiple of 11??

If $$r = 12$$, then $$2(1-r)=2(1-12)=2(-11)=-22$$

So, $$r = 12$$ is the smallest value of r to meet the given conditions.

To find the corresponding value or s, take $$11(s - 1) = 13(r - 1)$$ and plug in $$r = 12$$ to get: $$11(s - 1) = 13(12 - 1)$$

Simplify : $$11(s - 1) = 13(11)$$

This tells us that $$s-1=13$$, which means $$s=14$$

So, the LEAST possible value of $$r+s =12+14=26$$

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails Re: If rand s are positive integers, each greater than 1, and if   [#permalink] 05 Sep 2019, 06:39
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