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GRE 1: Q160 V152
There are four positive integers x, y ...
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30 Jul 2021, 14:50
30 Jul 2021, 14:50
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There are four positive integers x, y, z and w. If x/y leaves a remainder 3 and w/z leaves a remainder 8, what is the smallest possible value for y+z ?
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
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Re: There are four positive integers x, y ...
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30 Jul 2021, 19:34
30 Jul 2021, 19:34
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motion2020 wrote:
There are four positive integers x, y, z and w. If x/y leaves a remainder 3 and w/z leaves a remainder 8, what is the smallest possible value for y+z ?
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
Remember: Whenever we divide a number with another one (\(\frac{N}{D}\)), where N < D, N becomes the remainder
Now, if \(\frac{x}{y}\) leaves remainder of 3
So, lets take x = 3 and y has to be greater than 3
Also, if \(\frac{w}{z}\) leaves remainder of 8
So, lets take w = 8 and z has to be greater than 8
Min. values of y and z are 4 and 9
So, y + z = 13
Hence, option C
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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Re: There are four positive integers x, y ...
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11 Nov 2021, 12:37
11 Nov 2021, 12:37
motion2020 wrote:
There are four positive integers x, y, z and w. If x/y leaves a remainder 3 and w/z leaves a remainder 8, what is the smallest possible value for y+z ?
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
Key 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
If x divided y leaves a remainder of 3, then we know that 0 ≤ 3 < y, which means 4 is the smallest possible value of y
Similarly, if w divided z leaves a remainder of 8, then we know that 0 ≤ 8 < z, which means 9 is the smallest possible value of z
So, the smallest possible value of y + z = 4 + 9 = 13
Answer: C
