x and y are positive integers. When x is divided by 11, the remainder

Question

x and y are positive integers. When x is divided by 11, the remainder is 5, and when x is divided by 34, the remainder is 27. When y is divided by 17, the remainder is 11, and when y is divided by 3, the remainder is 2. What is the least possible value of x + y?

A) 7
B) 19
C) 38
D) 53
E) 89

Source: GMATPrepNopw.

Kudos for a correct solution.

A) 7
B) 19
C) 38
D) 53
E) 89

Source: GMATPrepNopw.

Bunuel · Accepted Answer

GMAT Prep Now OFFICIAL SOLUTION:

When x is divided by 11, the remainder is 5: So, the possible values of x are: 5, 16, 27, 38, etc.
 
When x is divided by 34, the remainder is 27: So, the possible values of x are: 27... STOP. Since both lists include 27, the smallest possible value of x is 27.
 
When y is divided by 17, the remainder is 11: So, the possible values of y are: 11, 28, 45, etc.
 
When y is divided by 3, the remainder is 2: So, the possible values of y are: 2, 5, 8, 11...STOP. Since both lists include 11, the smallest possible value of y is 11
 
Since the smallest possible values of x and y are 27 and 11 respectively, the smallest possible value of x + y is 38. So,  C is the correct answer  to the original question.
 
 The Big Takeaway:

When solving remainder questions on the GMAT, you can sometimes save yourself a lot of work by listing possible values and, more importantly, by beginning with the smallest possible value.

Skywalker18 · Answer

x/11= remainder is 5 x/34= remainder is 27 x=27 y/17= remainder is 11 y/3 = remainder is 2 y=11 x+y=38 Updated Answer to C , Thanks for correcting Engr2012 Cheers!! :thanks

ENGRTOMBA2018 · Answer

Your solution is not correct as shown below:

x= 11p+5 ---> x = 5 , 16, 27 ....
x= 34q+27 = 27, 61,...

Min common value of x =27

y=17r+11= 11,28,45...
y=3s+2=2,5,8,11...
Min y = 11

Thus least x+y value = 11+27=38

C is the correct answer.

thembaseeker · Answer

x=27 when divided by 11 leaves remainder 5, when divided by 34 leaves remainder 27.
y=11 when divided by 17leaves remainder 11, when divided by 3 leaves remainder 2

Hence x+y=38

Rule for finding the numbers are

Let x=34k+27 on division by 34 will leave remainder as 27

34k+27-5=34k+22 will be divisible by 11. For k=0.

Hence x=27.

Similarly Y can be found.

Answer : C

HarrishGowtham · Answer

For the least possible sum, lets assume that the quotient of all divisions are zero and that the given remainders are the actual numbers.
When x is divided by 11, the remainder is 5, and when x is divided by 34, the remainder is 27.
So the least value of X is 27.

When y is divided by 17, the remainder is 11, and when y is divided by 3, the remainder is 2.
So the least value of Y is 11.

Sum is 38

preeti20166 · Answer

Hope my approach is correct

x= 11m+5 .... (1)
x=34n+27......(2)

From 1 and 2;
11m+5=34n+27
so, 11m-34n=22 ... (3)
for n=0, we get m=2, putting that in eq 1, we get x=22
for m=0, we get n=-22/7, we get -ve value of x, but we know x is a positive integer.

similarly, for y= 17p+11...(4)
and y= 3q+2 ....(5)
so, 17p-3q= -9 .....(6)
for, p=0, we get q=3 and y= 11 (from eq 5)
for q=0, we get p=-19/7 and a negative value of y.
but we know y is a positive integer.

so, x+y= 38
option c

this is my first ever post on gmat club

Do give kudos if you feel i have helped in some way !
Cheers

Do give kudos if you feel i have helped in some way ! Cheers

franz711 · Answer

Hi,

As it is mentioned  when x is divided by 11, the remainder is 5 , and when x is divided by 34, the remainder is 27.  When y is divided by 17, the remainder is 11 , and when y is divided by 3, the remainder is 2

Hence, I concluded when X is even (11+5) and Y is even (17+11); Even + Even = Even. Since there is just one option with an Even number. I picked C. Is this approach correct?

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x and y are positive integers. When x is divided by 11, the remainder

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