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Question

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

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Solution

## THE QUESTION MUST BE WRONG IT SHOULD BE LIKE n, n+2 , n+4 is divisible by 3 let n be any positive integer and b=3 n =3q+r where q is the quotient and r is the remainder 0_ <r<3 so the remainders may be 0,1 and 2 so n may be in the form of 3q, 3q=1,3q+2 CASE-1 IF N=3q n+4=3q+4 n+2=3q+2 here n is only divisible by 3 CASE 2 if n = 3q+1 n+4=3q+5 n+2=3q=3 here only n+2 is divisible by 3 CASE 3 IF n=3q+2 n+2=3q+4 n+4=3q+2+4 =3q+6 here only n+4 is divisible by 3 HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE METHODE 2 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.

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