show that exactly one of the number n ,n+2 or n+4 is divisible by 3
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We know that,
n= qm +r
Here q= 3
Therefore, n= 3m+r ( Euclid's division lemma)
Hence the possible remainder are 0,1,2
so,
1) n= 3m
Which is divisible by 3
2) n= 3m+1
n+2 = 3m+3= 3(m+1)
which is also divisible by 3
3) ) n= 3m+2
n+4 = 3m+6= 3(m+2)
which is also divisible by 3
Hence proved!
Cheers
n= qm +r
Here q= 3
Therefore, n= 3m+r ( Euclid's division lemma)
Hence the possible remainder are 0,1,2
so,
1) n= 3m
Which is divisible by 3
2) n= 3m+1
n+2 = 3m+3= 3(m+1)
which is also divisible by 3
3) ) n= 3m+2
n+4 = 3m+6= 3(m+2)
which is also divisible by 3
Hence proved!
Cheers
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