If positive integers x and y are NOT multiples of 11, do they leave th

Question

If positive integers x and y are NOT multiples of 11, do they leave the same remainder when divided by 11?

(1) x - y = 11k, where k is an integer
(2) x = 10^52 + 1 and y = 24

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BrentGMATPrepNow · Accepted Answer

Given : Positive integers x and y are NOT multiples of 11  
Let's say that, when we divide y by 11, we get remainder r
This means that  y is r greater than some multiple of 11 
In other words,   y = 11j + r  , for some integer j

Target question :    Do x and y leave the same remainder when divided by 11?

Statement 1 : x - y = 11k, where k is an integer  
Add y to both sides of the equation to get: x = 11k + y 
Substitute to get: x = 11k +   11j + r  
Factor the first two terms to get:  x = 11(k + j) + r 
Since  11(k + j)  is clearly a multiple of 11, we can conclude that  x is r greater than some multiple of 11 
So when we divide x by 11, we can be certain that the remainder will be r
So, the answer to the target question is  YES, x and y leave the same remainder (r) when divided by 11 
Since we can answer the  target question  with certainty, statement 1 is SUFFICIENT

Statement 2 : x = 10^52 + 1 and y = 24 
No need to perform any calculations here. 
Since we are given the   actual values   of x and y, we COULD divide each value by 11 to determine the remainder in each case, which means we COULD answer the  target question  with certainty (although we'd never waste valuable time on test day doing so)
Statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

szubham · Answer

x = 11a + r1
y = 11b + r2
where a and b are integers

(1)
x-y = 11(a-b) + (r1-r2) = 11k
r1 and r2 can range from 1 to 10, because
remainders can't be zero since x and y are not divisible by 11, remainders are obviously above zero and remainders have to be less than 11 => 1 to 10
first term on the RHS is divisible by 11, second term has to be divisible by 11, which can only happen when r1=r2 given the constraints mentioned in the sentence right before this one
so the remainders are same

Sufficient

(2)
r2 = 2
r1 = whatever, answer is fixed and hence the conditions are

Sufficient

In (2) by the way, if someone really wants to solve
{10^52 + 1}/11 can be written as
{(11-1)^52/11 + 1/11}
and the remainder of the complete expression is sum of remainders of individual terms just like always
so remainder = 0 + remainder of {(-1)^52/11} + 1 (this follows from some crazy-a** binomial expansion)
= 0 + 1 + 1 = 2
Again, sufficient
Remainder of this complete expression is

carouselambra · Answer

IMO D Took whooping 3 mins

If positive integers x and y are NOT multiples of 11, do they leave the same remainder when divided by 11? (1) x - y = 11k, where k is an integer Simplest, let k=1 x-y=11 19-8=11 / 36-25=11 Each will yield the same remainder(s) when divided by 11. (A) is sufficient (2) x = 10^52 + 1 and y = 24 For this, I solved it in a different way. x can be written as : (2*5)^52 +1 I am just trying to find out the last digits here, to get a fair idea- we will be able to resolve the remainder. Since y=24, remainder is 2. (B) is sufficient.

CEdward · Answer

Can someone explain statement 1? 
If k = 0 then they will always have the same remainder since x = y
If k = 1 then they will never have the same remainder
e.g. k = 1, y = 1 then x = 12 and y = 1 NO

carouselambra · Answer

12/11 -> remainder is  1 
1/11 -> remainder is -10
In the scenario of negative remainder, add +11(because 11*dividend) to -10, hence remainder is  1.

Hence, the remainders are the same.

MHIKER · Answer

(1) I tried values for x and y as follows; Consider first one as x and the second as y

12-1=11k; the remainder is 1 for both x and y

13-2=11k; the remainder is 2 for both x and y

15-4=11k; the remainder is 4 for both x and y

Sufficient.

(2) If  \frac{10^2+1}{11}  the remainder will be 2; the same thing will be for  10^{52} . And  \frac{24}{11}  leaves the remainder 2; Sufficient.

The answer is D

szubham · Answer

64 and 53,
128 and 150, or
71 and 115

That is, difference of 2 numbers is a multiple of 11
In such a scenario, are the remainders of 2 numbers same?

Posted from my mobile device

MHIKER · Answer

k can't be "0"
Because it is told that x-y=11*k, you see 11 is a positive number. k can take any value without "0".

When  k=1  and with your example  \frac{x}{11} \ and \ \frac{y}{11}  will leave the same remainder  1 .

Vickydagger · Answer

I am posting a solution for the first time. I saw people complicating this question, that's why I am posting this. Please correct me if I am wrong.

Option 1.

X-Y=11K => X= 11K + Y (Quotient & remainder form) i.e when X is divided by 11 it leaves Y as remainder. so X/11 gives Y as remainder.

as the max remainder that's possible when divisor is 11 is '10' => Y is less than or equal to 10, so when you divide Y/11 the remainder will be Y. so this option is sufficient to check the remainders. in fact the remainders are equal in this case.

Option 2  
Absolute values of X and Y are given so we can get the remainders, so this data is sufficient.

Answer: D

Fdambro294 · Answer

Although you obtained the correct answer, the reasoning is a bit faulty.

Y does not necessarily have to be less than the divisor of 11.

Just because the rebuilding the dividend equation is written in that form does not necessarily follow that any equation written in that form will have to involve a remainder that is less than the divisor.

So even though —->. X = 11k + Y

It does not necessarily follow that Y is 10 or less.

Posted from my mobile device

If positive integers x and y are NOT multiples of 11, do they leave th

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If positive integers x and y are NOT multiples of 11, do they leave th

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