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14 Aug 2017, 01:28
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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 is an even integer
2) n has remainder 0 or 1 when it is divided by 3.
14 Aug 2017, 01:40
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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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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.
Please Give me kudos . I need them badly
General Discussion
16 Aug 2017, 02:39
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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
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
