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Math Expert V
Joined: 02 Sep 2009
Posts: 58464
If r and s are positive integers, each greater than 1, and if 11(s - 1  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 62% (02:00) correct 38% (01:53) wrong based on 76 sessions

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If r and 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

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Re: If r and s are positive integers, each greater than 1, and if 11(s - 1  [#permalink]

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1
Bunuel wrote:
If r and 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

13*11= 143
so at s= 14 and r =12
we get 11(s - 1) = 13(r - 1)
IMO E ; 26
VP  D
Joined: 31 Oct 2013
Posts: 1468
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If r and s are positive integers, each greater than 1, and if 11(s - 1  [#permalink]

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3
Bunuel wrote:
If r and 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

11(s - 1) = 13( r - 1 )

13 is not divisible by 11. So, r - 1 is divisible by 13. Least value of r = 14. Therefore (14 - 1) becomes multiple of 13.

The same way as above, 11 is not divisible by 13 .Therefore, s - 1 is a multiple of 11. Least value of s =12.

r = =14

s = 12

r + s = 14 + 12 = 26.

E is the correct answer.
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4019
Re: If r and s are positive integers, each greater than 1, and if 11(s - 1  [#permalink]

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Top Contributor
Bunuel wrote:
If r and 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

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
_________________ Re: If r and s are positive integers, each greater than 1, and if 11(s - 1   [#permalink] 05 Sep 2019, 07:40
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