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10 Nov 2022, 02:55
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Difficulty:
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Question Stats:
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32%
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If n is a positive integer, which of the following must be divisible by 3?
(A) n(n + 4)(n − 4)
(B) n(n + 5)(n − 1)
(C) n(n + 3)(n − 2)
(D) n(n + 1)(n − 2)
(E) n(n + 2)(n − 3)
(A) n(n + 4)(n − 4)
(B) n(n + 5)(n − 1)
(C) n(n + 3)(n − 2)
(D) n(n + 1)(n − 2)
(E) n(n + 2)(n − 3)
When divided by 3,
we have to understand that (n+4) will have the same effect as (n+1) and (n-4) will have the same effect as (n-1)
Similarly, (n+5) will correspond to (n+2)
Now, a product of 3 consecutive numbers is always divisible by 3 as one of the numbers will definitely be divisible by 3. (cycle)
Rewriting -
(A) n(n + 4)(n − 4) -> n(n+1)(n-1) -> (n-1)n(n+1) -> consecutive numbers
(B) n(n + 5)(n − 1) -> n(n+2)(n-1) -> (n-1)n(n+2)
(C) n(n + 3)(n − 2) -> (n-2)n(n+3)
(D) n(n + 1)(n − 2) -> (n-2)n(n+1)
(E) n(n + 2)(n − 3) -> (n-3)n(n+2)
Only option A has 3 consecutive numbers. Hence, it'll be divisible by 3.
Alternate
Try 3 consecutive values of n (positive integer, as question says). Say n=4, 5, 6
This will cover all cases.
Only option A will be divisible by 3 when n is any of 4, 5 or 6.
we have to understand that (n+4) will have the same effect as (n+1) and (n-4) will have the same effect as (n-1)
Similarly, (n+5) will correspond to (n+2)
Now, a product of 3 consecutive numbers is always divisible by 3 as one of the numbers will definitely be divisible by 3. (cycle)
Rewriting -
(A) n(n + 4)(n − 4) -> n(n+1)(n-1) -> (n-1)n(n+1) -> consecutive numbers
(B) n(n + 5)(n − 1) -> n(n+2)(n-1) -> (n-1)n(n+2)
(C) n(n + 3)(n − 2) -> (n-2)n(n+3)
(D) n(n + 1)(n − 2) -> (n-2)n(n+1)
(E) n(n + 2)(n − 3) -> (n-3)n(n+2)
Only option A has 3 consecutive numbers. Hence, it'll be divisible by 3.
Alternate
Try 3 consecutive values of n (positive integer, as question says). Say n=4, 5, 6
This will cover all cases.
Only option A will be divisible by 3 when n is any of 4, 5 or 6.
10 Nov 2022, 10:37
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kapil1995
We know that the product of three consecutive integers is always divisible by 3. So, let’s say n = 1, and look at each answer choice:
A) -3, 1, and 5
Using answer choice A, we see that we have a set evenly spaced set since -3 + 4 = 1 and 1 + 4 = 5.
Thus, we have a set of three consecutive numbers, so it’s divisible by 3.
Answer: A
18 Nov 2022, 10:38
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kapil1995
When we substitute n = 1 and get -3 * 1 * 5, not only is the product a multiple of 3, but also the factors are spaced evenly. I think it is already clear that a product of three consecutive integers is always divisible by 3, but more is true: if we have three evenly spaced integers, the product of those integers will be divisible by 3 as long as the difference between consecutive terms is not a multiple of 3.
Let's say we have three integers which are evenly spaced. We can represent those integers by x, x + k, and x + 2k. If k is not divisibe by 3, then the remainder when k is divided by 3 is either 1 or 2. If it is 1, then the remainder when 2k is divided by 3 is 2. If it is 2, then the remainder when 2k is divided by 3 is 1. So we know exactly one of k and 2k produce a remainder of 1 when divided by 3, and exactly one of those numbers produce a remainder of 2 when divided by 3.
If x is divisible by 3, then the product x * (x + k) * (x + 2k) is divisible by 3 because x is divisible by 3.
If x produces a remainder of 1 when divided by 3, then adding either k or 2k (the one producing a remainder of 2) will produce a remainder of 0 when divided by 3; i.e. it will be a multiple of 3.
If x produces a remainder of 2 when divided by 3, then again adding either k or 2k (this time, the one producing a remainder of 1) will produce a remainder of 0 when divided by 3.
We see that as long as k is not a multiple of 3, one of x, x + k or x + 2k will be a multiple of 3, which means that the product x(x + k)(x + 2k) will be divisible by 3.
We can actually apply this principle directly to the numbers n - 4, n, and n + 4 by letting x = n - 4 and k = 4 to conclude that the product (n - 4)(n)(n + 4) must be divisible by 3.
