Show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.
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Answered by
3
let n be 2
so n=2
n+2=4
n+4=6
so n + 4 is divisible by 3
so n=2
n+2=4
n+4=6
so n + 4 is divisible by 3
Answered by
3
Checking the argument :- Only one of n, n+2 ,n+4 would be divisible by 3 when n is +ve integer.
Case 1 :-
n is divisible by 3 .
then n+2/3 = n/3 + 2/3 this shows a remainder 2 .
then n+4/3 =1+ n/3 + 1/3 this shows a remainder 1.
Case 2 :-
n+2 is divisible by 3 .
n/3 = n+2-2/3 = n+2/3 - 2/3 this leaves a remainder (3-2) = 1 .
n+4/3 =n+2+2/3 = n+2/3 + 2/3 leaves a remainder 2 .
Case 3 :-
n+4 is divisible by 3 .
n/3 = n+4-4/3 = (n+4)/3 -1 - 1/3 this leaves a remainder (3-1) = 2.
n+2/3 =n+4-2/3 = n+4/3 - 2/3 leaves a remainder (3-2)= 1 .
Hence proved that; one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer
Case 1 :-
n is divisible by 3 .
then n+2/3 = n/3 + 2/3 this shows a remainder 2 .
then n+4/3 =1+ n/3 + 1/3 this shows a remainder 1.
Case 2 :-
n+2 is divisible by 3 .
n/3 = n+2-2/3 = n+2/3 - 2/3 this leaves a remainder (3-2) = 1 .
n+4/3 =n+2+2/3 = n+2/3 + 2/3 leaves a remainder 2 .
Case 3 :-
n+4 is divisible by 3 .
n/3 = n+4-4/3 = (n+4)/3 -1 - 1/3 this leaves a remainder (3-1) = 2.
n+2/3 =n+4-2/3 = n+4/3 - 2/3 leaves a remainder (3-2)= 1 .
Hence proved that; one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer
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