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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
j, k, m, x, y and z are positive integers. When j is divided
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22 May 2019, 05:20
22 May 2019, 05:20
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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
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
j, k, m, x, y and z are positive integers. When j is divided
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22 May 2019, 08:54
22 May 2019, 08:54
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Bookmarks
GreenlightTestPrep 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
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
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
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Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
j, k, m, x, y and z are positive integers. When j is divided
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26 Aug 2021, 11:57
26 Aug 2021, 11:57
1
given ky=75, let k=5 and y=15
j,5,m,x,15,z
j/5 with remainder m and m={1,2,3,4}, all numbers must be positive
x/15 with remainder z and z={1,2,3...14}
possible values of mz must be the product of two set numbers, set m and set z
49=7*7 not possible
50 not possible too
56=4*14 is possible
Answer is I & II, and it's B
j,5,m,x,15,z
j/5 with remainder m and m={1,2,3,4}, all numbers must be positive
x/15 with remainder z and z={1,2,3...14}
possible values of mz must be the product of two set numbers, set m and set z
49=7*7 not possible
50 not possible too
56=4*14 is possible
Answer is I & II, and it's B
GreenlightTestPrep 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
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
