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

multiple of 11What 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 \(r+s =12+14=26\)

Answer: E

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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