Question 13
Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
Theory
As per Euclid’s Division Lemma
If a and b are 2 positive integers, then
a = bq + r
where 0 ≤ r < b
If b = 3,
a = 3q + r
where 0 ≤ r < 3
So, r = 0, 1, 2
∴ Numbers = 3q + 0, 3q + 1, 3q + 2
Let’s assume
n = 3q, 3q + 1, 3q + 2
Now, we check whether n, n + 2, n + 4 is divisible by 3
If n = 3q
n = 3q
Since n can be divided by 3
It is divisible by 3
n + 2 = 3q + 2
Putting q = 1
n + 2 = 3(1) + 2 = 5
Since 5 is not divisible by 3
n + 2 is not divisible by 3
n + 4 = 3q + 4
Putting q = 1
n + 4 = 3(1) + 4 = 7
Since 7 is not divisible by 3
n + 4 is not divisible by 3
If n = 3q + 1
n = 3q + 1
Putting q = 1
n = 3(1) + 1 = 4
Since 4 is not divisible by 3
n is not divisible by 3
n + 2 = 3q + 1 + 2
n + 2 = 3q + 3
n + 2 = 3(q + 1)
Since n + 2 can be divided by 3
It is divisible by 3
n + 4 = 3q + 1 + 4
n + 4 = 3q + 5
Putting q = 1
n + 4 = 3(1) + 5 = 8
Since 8 is not divisible by 3
n + 4 is not divisible by 3
n + 4 = 3q + 1 + 4
n + 4 = 3q + 5
Putting q = 1
n + 4 = 3(1) + 5 = 8
Since 8 is not divisible by 3
n + 4 is not divisible by 3
If n = 3q + 2
n = 3q + 2
Putting q = 1
n = 3(1) + 2 = 5
Since 5 is not divisible by 3
n is not divisible by 3
If n = 3q + 1
n = 3q + 1
Putting q = 1
n = 3(1) + 1 = 4
Since 4 is not divisible by 3
n + 2 is not divisible by 3
n + 4 = 3q + 2 + 4
n + 4 = 3q + 6
n + 4 = 3(q + 2)
Since n + 4 can be divided by 3
It is divisible by 3
We see that in all 3 cases,
Exactly one of the numbers n, n + 2, n + 4 is divisible by 3
Hence proved

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.