show that only one n , n+1 , n+2 is divisible by 3
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Hey
Here is your answer,
We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0
i.e n = 3q -------- (1),
n = 3q +1 --------- (2),
n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.
Hope it helps you!
Here is your answer,
We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0
i.e n = 3q -------- (1),
n = 3q +1 --------- (2),
n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.
Hope it helps you!
harshit91960:
thnkx bhau
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