Question

Show that of the numbers n , n + 2 , n + 4 , only one of them is divisible by 3​

Answer 1

Given:

The numbers n, (n+2), (n+4), where n is a positive integer.

Now by dividing n by 3,

Let q be the quotient and r be the remainder

so,

n = 3q+r

where 0 ≤ r < 3

[By EDL- Euclid's Division Lemma]

So, n = 3q+r

Remainders (r) = 0, 1, 2

• Case 1, when r = 0

n = 3q+0

n+2 = 3q + 2

n+4 = 3q + 4

[Here n = 3q is only divisible by 3]

• Case 2, when r = 1

n = 3q+1

n+2 = 3q 2 + 1 = 3q + 3

n+4 = 3q+ 4 + 1 = 3q+5

[Here (n+2) = (3q+3) is only divisible by 3]

• Case 3, when r = 2

n = 3q+2

n+2 = 3q+2 + 2 = 3q + 4

n+4 = 3q+ 4 + 2 = 3q + 6

[Here n+4 = 3q+6 is only divisible by 3]

Hence,

We conclude that from the numbers n, (n+2), (n+4), only one of them is divisible by 3​

Answer 2

Answer- The above question is from the chapter 'Real Numbers'.

Concept used: 1) Any positive number is of the form 3q, 3q + 1 or 3q + 3.

2) Euclid's Division Lemma which states that given two positive integers, a and b such that a > b, there exist unique integers q and r satisfying a = bq + r,

$0 \leq b < r$ .

Given question: Show that one and only one out of n, n + 2, n + 4 is divisible by 3.

Solution: (Please see the attachments.)