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What is the remainder, when n(n+2) is divided by 24 for a positive integer x?

1) n is an even integer
2) n has remainder 0 or 1 when it is divided by 3.

1) n = 2k
2k(2k+2)=4k(k+1)
If k = 1, the remainder is 8, if k = 2, the remainder is 0 ---> A is not the answer.

2) n = 3p or n=3m+1
3p(3p+1) or (3m+1)(3m+3)=3(3m+1)(m+1)
If p = 1, the remainder is 12, if p = 2, ther remainder is 18 ---> B is not the answer.

Lets combine them. 1)+2)
n(n+2)=4k(k+1)=3p(3p+1)
or
n(n+2)=4k(k+1)=3(3m+1)(m+1)
From (1) we have n(n+2) is divisible by 8 (because k and k+1 are 2 consecutive numbers, so one of them is even)
From (2) we have n(n+2) is divisible by 3 (both cases)
----->n(n+2) is divisible by 24, the raminder is always equal to 0.
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MathRevolution
What is the remainder, when n(n+2) is divided by 24 for a positive integer x?

1) n is an even integer
2) n has remainder 0 or 1 when it is divided by 3.

Asked: What is the remainder, when n(n+2) is divided by 24 for a positive integer x?

1) n is an even integer
n = 2k
n(n+2) = 2k(2k+2) = 4k(k+1)
The remainder, when n(n+2) is divided by 24 = {8,0}
NOT SUFFICIENT

2) n has remainder 0 or 1 when it is divided by 3.
n = 3k or n = 3k+1
If n = 3k ; n(n+2) = 3k(3k+2) = 9k^2 + 6k
The remainder, when n(n+2) is divided by 24 = {15,0,3}
If n=3k+1; n(n+2) = (3k+1)(3k+3) = 3(k+1)(3k+1)
The remainder, when n(n+2) is divided by 24 = {0,15,12}
NOT SUFFICIENT

(1) + (2)
1) n is an even integer
n = 2k
n(n+2) = 2k(2k+2) = 4k(k+1)
The remainder, when n(n+2) is divided by 24 = {8,0}
2) n has remainder 0 or 1 when it is divided by 3.
n = 3k or n = 3k+1
If n = 3k ; n(n+2) = 3k(3k+2) = 9k^2 + 6k
The remainder, when n(n+2) is divided by 24 = {15,0,3}
If n=3k+1; n(n+2) = (3k+1)(3k+3) = 3(k+1)(3k+1)
The remainder, when n(n+2) is divided by 24 = {0,15,12}
Combining, we get
The remainder, when n(n+2) is divided by 24 = 0
SUFFICIENT

IMO C
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