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

Question

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.

Siva2021 · Accepted Answer

n is an even integer and n leaves 0 or 1 when divided by3.
When n is 10 it leaves a remainder 1 when divided by 3.n(n+2)=10*12=120. 120 when divided by 24 leaves no remainder.
Next case n=12 which is evenly divided by 3. 12*14 =168. 168 when divided by 24 leaves 0 as remainder
Take other cases n=18,22. In all cases n(n+2) is perfectly divided by 24.So the answer is C Both statements are sufficient.
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MathRevolution · Answer

Condition 1)
There are two kinds of even integers, which are n = 4k or n = 4k + 2 for some integer k. That means n could have 0 or 2 as a remainder when it is divided by n. If n = 4k, n(n+2) = 4k(4k+2) = 8k(2k+1) and n(n+1) is a multiple of 8. If n= 4k+2, n(n+2) = (4k+2)(4k+2+2) = 2(2k+1)*4(k+1) = 8(2k+1)(k+1) and n(n+1) is a multiple of 8. For both cases, n(n+1) is a multiple of 8. However, we can’t identify the remainder when it is divided by 3 from the condition 1).

Condition 2)
n = 3k or n = 3k +1. If n = 3k, n(n+2) = 3k(3k+2) is a multiple of 3. If n = 3k + 1, n(n+2) = (3k+1)(3k+3) = 3(3k+1)(k+1) is a multiple of 3. Thus, for both cases, n(n+1) is a multiple of 3. However, we can’t identify the remainder when it is divided by 8 from the condition 2).

Condition 1) & 2)
From the condition 1), n(n+1) is a multiple of 8. And n(n+1) is a multiple of 3 from the condition 2). Therefore, n(n+1) is a multiple of 24 from the both conditions 1) & 2) together.

Ans: C

DharLog · Answer

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.

Kinshook · Answer

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

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