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Bunuel
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

When positive integer x is divided by 11, the remainder is m and when positive integer y is divided by 11, the remainder is n. What is the value of m + n ?

(1) x + y is a multiple of 11.
(2) x - y is a multiple of 11.


 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 


Let x=11a+m and y=11b+n

S1:- x+y = 11c
=> 11a+m+11b+n = 11c => 11(a+b) + (m+n) = 11c
For RHS to be a multiple of 11 LHS should also be a multiple of 11. Since 11(a+b) is already a multiple of 11, therefore m+n should also be a multiple of 11 and since m,n<11, their sum<22. So either m+n=0 or 11.
Not sufficient.

S2:- x-y = 11d
=> 11a+m-11b-n=11d => 11(a-b)+(m-n)=11d
m,n<11 therefore (m-n)max = 10 (m=10 and n=0). Only possible multiple of 11 is 0 for (m-n) when either m=n=0 or m=n=any number.
Not sufficient.

S1&S2:- We get m+n=0.

C is correct.
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Bunuel
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

When positive integer x is divided by 11, the remainder is m and when positive integer y is divided by 11, the remainder is n. What is the value of m + n ?

(1) x + y is a multiple of 11.
(2) x - y is a multiple of 11.


 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 


Let x = 11a+m and y = 11b+n , Since x and y are positive integers, m and n are non-negative integers and always less than divisor (11)
\(0\leq{m}<{11}\) and \(0\leq{n}<{11}\)

\(0\leq{m+n}<{22}\)

(1) x + y is a multiple of 11.
x+y = 11(a+b)+ (m+n)
For x+y to be divisible by 11, either m+n =0 or multiple of 11 i:e 11

Case 1: when x=22 and y=33, m=0, n=0 , m+n=0
m+n=0 ( in which both m and n are 0)

Case 2: when x= 18 and y =15, m =7 and n=4, m+n=11
m+n =11
Insufficient

(2) x - y is a multiple of 11.
x-y = 11(a-b)+ (m-n)
For x-y to be divisible by 11 either m-n =0 or multiple of 11
Case 1: When m-n =0 or m=n
A. x=14, y= 25, m=3, n=3, m-n=0
m+n=6

B. x=15, y= 26 m=4, n=4, m-n=0
m+n= 8
Multiple sum of values of m+n exists.

Insufficient

Combining 1 and 2
We get m-n=0 and m+n=0 or m+n=11. Since m and n are integer \(m+n\neq{11}\)
m+n=0 and m-n=0
This is only possible when m=n=0. m+n=0

IMO C.
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