show that exactly one of the no. N,n+2, n+4 is devisible by 3
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You need to consider the cases-
1) When 'n' is divisible by 3.
If this is true, then no.s "n-3, n+3, n+6,...." are divisible by 3 but not n-2 or n-4
2) When 'n' is not divisible by 3.
If this is true, further two subclasses can be made, i.e, either n-1 is divisible by 3 or n+1. This is true for all no.s existing.
For example:-5,6,7 or 12,13,14 or 125,126,127 or 554,555,556..
What I mean to say is if you take three consecutive numbers, one of them has to be divisible by 3
Now,
If n-1 is divisible then
"n-1, n+2, n+5, n+8..." are divisible by 3 but not n or n+4
If n+1 is div..
Ah, you know well enough how to go about it now. Good luck (^_-)
1) When 'n' is divisible by 3.
If this is true, then no.s "n-3, n+3, n+6,...." are divisible by 3 but not n-2 or n-4
2) When 'n' is not divisible by 3.
If this is true, further two subclasses can be made, i.e, either n-1 is divisible by 3 or n+1. This is true for all no.s existing.
For example:-5,6,7 or 12,13,14 or 125,126,127 or 554,555,556..
What I mean to say is if you take three consecutive numbers, one of them has to be divisible by 3
Now,
If n-1 is divisible then
"n-1, n+2, n+5, n+8..." are divisible by 3 but not n or n+4
If n+1 is div..
Ah, you know well enough how to go about it now. Good luck (^_-)
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