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

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

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New post 08 Mar 2019, 01:35
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E

Difficulty:

  65% (hard)

Question Stats:

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

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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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New post 08 Mar 2019, 03:59
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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
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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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New post 08 Mar 2019, 05:31
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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.
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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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New post 05 Sep 2019, 07:40
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 \(r+s =12+14=26\)

Answer: E

Cheers,
Brent
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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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