Question

show that one and only one out of n, n+1, n+2, n+4 is divisible by 3, where n is any positive integer.​

Answer 1

Answer:

We applied Euclid Division algorithm on n and 3. a = bq +r on putting a = n and b = 3 n = 3q +r , 0<r<3 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.

This may be

Answer 2

Step-by-step explanation:

On applying the euclid's division lemma,

a=bq+r , where 0≤r<b

here a=n and b=3

1) On putting r= 0

n = 3q (it is divisible by 3)

2) On putting r= 1

n = 3q+1 (it is divisible by 3)

3) On putting r= 2

n = 3q+2 (it is divisible by 3)

4) On putting r= 3

n = 3q+2 (it is divisible by 3)

∴ One and only out of n, n+1, n+2, n+3 is divisible by 3.

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