Same concept different way of thinkingHow do I ensure that a product of three numbers is divisible by 3 ?
Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3
Step 2 : I ensure that I chose my numbers such that each belongs to a different part
Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.
How do I achieve this ?
Step 1 : The division I choose is numbers of the form \(3\alpha, 3\alpha+1, 3\alpha+2\). The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3.
Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)
Important Note : If I shift any number by 3, its constituent set amongst \(\alpha, 3\alpha+1, 3\alpha+2\) does not change
So how do we solve this question ?
Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.
A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice
Hence, solution is A
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