show that only one of the …

Case-I: When n = 3q In this case, we have n= 3q, which is divisible by 3 Now, n = 3q n+2 = 3q+2 n+2 leaves remainder 2 when divided by 3 Again, n = 3q n+4 = 3q+4=3(q+1)+1 n+4 leaves remainder 1 when divided by 3 n+4 is not divisible by 3. Thus, n is divisible by 3 but n+2 and n+4 are not divisible by 3. Case-II: when n = 3q+1 In this case, we have n= 3q+1, n leaves remainder 1 when divided by 3. n is divisible by 3 Now, n = 3q+1 n+2 = (3q+1)+2=3(q+1) n+2 is divisible by 3. Again, n = 3q+1 n+4 = 3q+1+4=3q+5=3(q+1)+2 n+4 leaves remainder 2 when divided by 3 n+4 is not divisible by 3. Thus, n+2 is divisible by 3 but n and n+4 are not divisible by 3. Case-III: When n + 3q+2 In this case, we have n= 3q+2 n leaves remainder 2 when divided by 3. n is not divisible by 3. Now, n = 3q+2 n+2 = 3q+2+2=3(q+1)+1 n+2 leaves remainder 1 when divided by 3 n+2 is not divisible by 3. Again, n = 3q+2 n+4 = 3q+2+4=3(q+2) n+4 is divisible by 3. Thus, n+4 is divisible by 3 but n and n+2 are not divisible 3.