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Bunuel
N is a positive integer. When N + 1 is divided by 5, the remainder is 4. What is the remainder when N is divided by 5?

A. 6
B. 5
C. 4
D. 3
E. 2

N+1 = 5a+4

i.e. N+1 = 9, 14, 19, 24, ... etc.

i.e. N = 8, 13, 18, 23, ... etc.

When N is divided by 5 Remainder is always 3

Answer: Option D
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Quote:
When N + 1 is divided by 5, the remainder is 4
Smallest numbet possible is 9 and the possible value of N = 8
Quote:
What is the remainder when N is divided by 5?
8/5 = Reminder 3

Hence answer is (D)
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N+1=5q+4

=> N=5q+4-1 => N=5q+3, so the answer is: the remainder is 3 when N is divided by 5.
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easiest way => take N+1=4
=> N=3
hence Remainder => 3
hence D

peace out :)
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Another way:

N+1 =5a+4 ---->N=5a+3
and
N=5a+ x (x is what we looking for)
therefore: 5a+3=5a+x ---> x=3
OS -->D
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Abhishek009
Quote:
When N + 1 is divided by 5, the remainder is 4
Smallest numbet possible is 9 and the possible value of N = 8

Can someone correct me if wrong? Isn't the smallest possible number 4?
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Abhishek009
Quote:
When N + 1 is divided by 5, the remainder is 4
Smallest numbet possible is 9 and the possible value of N = 8

Can someone correct me if wrong? Isn't the smallest possible number 4?

Can not be 4 as with N=4, N+1 = 5 will not give you a remainder of 4 when divided by 5. But if N=8, N+1=9 will give you a remainder of 4 when divided by 5.
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Engr2012
Can not be 4 as with N=4, N+1 = 5 will not give you a remainder of 4 when divided by 5. But if N=8, N+1=9 will give you a remainder of 4 when divided by 5.

Aha! That is, indeed, when it is in relation with the second part of the question (What is the remainder when N is divided by 5?). Thank you. Makes sense now.
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Bunuel
N is a positive integer. When N + 1 is divided by 5, the remainder is 4. What is the remainder when N is divided by 5?

A. 6
B. 5
C. 4
D. 3
E. 2

This can easily be solved by assuming values.
N+1, when divided by 5 leaves remainder 4
Assume N+1 = 9

Therefore, N = 8
N divided y 5 leaves remainder 3
Option D

nice work
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