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

Question

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

KSBGC · Accepted Answer

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.

Archit3110 · Answer

13*11= 143
so at s= 14 and r =12 
we get 11(s - 1) = 13(r - 1)
IMO E ; 26

BrentGMATPrepNow · Answer

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

rsrighosh · Answer

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

MathRevolution · Answer

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

WiziusCareers1 · Answer

1. Analyze the Given Equation

The initial equation is:
11(s−1) = 13(r−1)
Since r and s are positive integers, the terms (s−1) and (r−1) must also be integers.

2. Apply the Concept of Least Common Multiple (LCM)

The equation shows that the expression 11(s−1) is equal to 13(r−1). This means that the product must be a common multiple of 11 and 13.
Since 11 and 13 are prime numbers, they have no common factors other than 1.
Therefore, the least common multiple (LCM) of 11 and 13 is their product: 11×13=143.

To find the least possible value for r + s, we must find the least possible non-zero value for the expression 11(s−1) = 13(r−1). This least value is the LCM, 143.

3. Solve for s and r

Set the expression equal to the LCM, 143:
A. Solve for s:
11(s−1) = 143
s − 1= 13
s = 14

B. Solve for r:
13(r−1) = 143
r − 1 = 11 
r = 12

4. Check the Constraints and Calculate r+s

The problem states that r and s must be positive integers, each greater than 1.
Our calculated values are r = 12 and s = 14. Both are positive integers and greater than 1. This is the least set of values that satisfies the equation and the constraints.

So, r + s = 12 + 14 = 26
The least possible value of r + s is 26.

The correct answer is E.

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

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