j, k, m, x, y and z are positive integers. When j is divided by k,

Question

j, k, m, x, y and z are positive integers. When j is divided by k, the remainder is m. When x is divided by y, the remainder is z. If ky = 75, which of the following CANNOT be the value of mz?

i) 49
ii) 50
iii) 56

A) iii only
B) i and ii
C) i and iii
D) ii and iii
E) i, ii and iii

Archit3110 · Answer

my take on below question
k*y
75= 3*5^2
and given
j=k*a+m and x=y*b+z
a& b are multiple values
we can solve this question by using answer options

49 . ; 7*7 = so m & n have to be both 7 & 7 , we can have 7 as divisior only in case when either of k & y is 15 but other 7 wont be possible as the other digit will be 3 
50; 5^2*2 ; either of m or n can be 2 ; if we have value of k as 3 but other remainder value of 25 will not be possible as other digit will be 15
56 ; 2^3*7 ; 14*4 possible with 15*5 So possible
Option 1&2 not possible
IMO B

bidskamikaze · Answer

ky=75= 5*5*3
So, the values of k, y can be 15 and 5 (5*3=15 and other number 5: 15*5 = 75),
OR the values of k, y can be 25 and 3 (5*5=25 and other number 3: 25*3 = 75).
(We can ignore the case 75 * 1 = 75, since dividing positive integer by 1 won't give any remainder)
Let us assume k is the larger number and y is the smaller number.

J can be any positive integer, but when divided by k (k is either 15 or 25 ), the remainder (which is m) can be between 0 to 14 if k=15 or 0 to 24 if k=25.
Similarly, x can be any positive integer, but when divided by y (y is either 3 or 5) the remainder (which is z) can be between 0 to 2 if y=3 or 0 to 4 if y=5.

Let us now consider the options:

i) 49: only possible factors are 1 and 7 (as 7 is a prime number and 7*7=49) so m*z cannot be 49.

ii) 50: 10*5 = 50 but 5 cannot be the remainder so this is not possible
25*2= 50 but 25 as a remainder is not possible so this not possible as well.

iii) 56: 14*4= 56. If m and y are 15 and 5, then remainders 14 and 4 ARE possible. So 56 is a possible value of m*y.

Answer B

BrentGMATPrepNow · Answer

Useful remainder property: 
   When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D   
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

When j is divided by k, the remainder is m  
The above   property   tells us that  0 ≤ m < k

When x is divided by y, the remainder is z  
The above   property   tells us that  0 ≤ z < y

If ky = 75, which of the following CANNOT be the value of mz?  
There are 3 ways in which ky = 75
1) k and y are 1 and 75
2) k and y are 3 and 25
3) k and y are 5 and 15

We can ELIMINATE case 1, because dividing by 1 will give us a remainder of 0, and we're told that all values are POSITIVE integers
So, k and y are EITHER 3 & 25 OR 5 & 15 
So, we have two possible cases:
case a)  m < 3  and  k < 25 
case b)  m < 5  and  k < 15

Now let's check the statements. 
iii) 56
Are there values of m and z such that mz = 56, AND one of the two cases (above) are met? 
YES!
If m = 4 and k = 14, then mk = (4)(14) = 56
This meets the conditions in case b. That is,  m < 5  and  k < 15 
Since mz CAN equal 56, we can ELIMINATE answer choices A, C, D and E (since they say that mz CANNOT equal 56)

By the process of elimination, the correct answer is B

LeoN88 · Answer

All given variables are +ve integers.

j = Q1K + m
x= Q2y + z,

ky=75, so 
Case 1, (K,Y) can be (5,15) in any order
Case 2 (K,Y) can be (25,3) in any order.

We have to find m*z which will also give +ve Integer.

For case 1.
0<=m<j and 0<=z<x
We can get 56 by 4*14 but can't get 49 and 50.

For case 2.
0<=m<j and 0<=z<x
The max m*z we get is 24*2= 48.

So, m*z can't be 49 and 50.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

j, k, m, x, y and z are positive integers. When j is divided by k,

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

j, k, m, x, y and z are positive integers. When j is divided by k,

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards