If r and s are positive integers, each greater than 1, and if 11(s - 1

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 \(r+s =12+14=26\)

Answer: E

Cheers,
Brent

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

This means

\(s-1 = 13k => s= 13k+1\)
\(r-1 = 11p => r = 11p+1\)

\(r+s = 13k + 11p + 2\)

Mininmum value can be found of k=p=1

Hence \(r+s = 13+11+2 = 26\)

If s - 1 = 13 and r - 1 = 11 then we will get 11(s - 1) = 13(r - 1).

=> s - 1 = 13. Therefore, s = 14
=> r - 1 = 11. Therefore, r = 12

=> Least possible value of r + s = 14 + 12 = 26

=> We can also add equations directly as: r + s - 2 = 24 giving us r + s = 26.

Answer E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.