Question

show that exactly one of the number n,n+2,n+4 is divisible by 3?????

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Answer 1

Answer:

Hence Proved that  exactly one of the number n,n+2,n+4 is divisible by 3

Step-by-step explanation:

Three number are n , n+2 , n+4

n could be 3k , 3k+1 , 3k+2  (as 3k+3 = (3k+1) and so on)

where k = integer

Case 1     n = 3k

then n/3 = 3k/3 = k   (Divisible by 3)

(n+2)/3 =(3k+2)/3 = k + 2/3  (2/3 remainder is there)

(n+4)/3 = (3k +4)/3 = 3(k+1) + 1/3  (1/3 remainder is there)

so Only one number is divisible

Case 2    n = 3k+1

then n/3 = (3k+1)/3 = k +1/3   (1/3 remainder is there)

(n+2)/3 =(3k+1+2)/3 = k + 1   (Divisible by 3)

(n+4)/3 = (3k +1+4)/3 = 3(k+1) + 2/3  (2/3 remainder is there)

so Only one number is divisible

Case 3     n = 3k+2

then n/3 = (3k+2)/3 = k +2/3   (2/3 remainder is there)

(n+2)/3 =(3k+2+2)/3 = 3(k + 1) + 1/3   (1/3 remainder is there)

(n+4)/3 = (3k +2+4)/3 = 3(k+2)    (Divisible by 3)

so Only one number is divisible

So in all possible cases only number is divisble by 3

Hence Proved that  exactly one of the number n,n+2,n+4 is divisible by 3