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

Expert replies
by Brent@GMATPrepNow » Wed May 22, 2019 3:01 pm

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A

B

C

D

E

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Difficulty

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

Answer: B
Source: www.gmatprepnow.com
Difficulty level: 650-700
Brent Hanneson - Creator of GMATPrepNow.com
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Source: — Problem Solving |

by Brent@GMATPrepNow » Thu May 23, 2019 8:14 am
Brent@GMATPrepNow wrote: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

Answer: B
Source: www.gmatprepnow.com
Difficulty level: 650-700
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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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