# Show that exactly

As per Euclid’s lemma,

If a and b are two positive integers, then they can be represented in the form,

a = bq + r

where 0 ≤ r < b

If b = 3, this can be written as

a = 3q + r

where 0 ≤ r < 3

So, r = 0, 1, 2

a = 3q, 3q + 1, 3q + 2

Let n = 3q, 3q + 1, 3q + 2

First taking n = 3q …(i)

This clearly implies n is divisible by 3.

Adding 2 on both sides of equation (i),

n + 2 = 3q + 2

Putting q = 1, we get

n + 2 = 3(1) + 2

n + 2 = 5

Now since 5 is not divisible by 3, this implies (n + 2) is not divisible by 3.

Adding 4 on both sides of equation (ii),

n + 4 = 3q + 4

Putting q = 1, we get

n + 4 = 3(1) + 4 = 7

Since 7 is not divisible by 3, this implies (n + 4) is not divisible by 3.

Now taking n = 3q + 1 …(ii)

Putting q =1, we get

n = 3(1) + 1 = 4

Since 4 is not divisible by 3, n is also not divisible by 3.

Adding 2 on both sides of equation (ii),

n + 2 = 3q + 1 + 2
= 3q + 3
= 3(q + 3)

It’s clear that 3(q + 3), which is a multiple of 3, is divisible by 3.

(n + 2) is divisible by 3.

Adding 4 on both sides of equation (ii),

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

Putting q = 1, we get

n + 4 = 3(1) + 5
= 3 + 5
= 8

8 is not divisible by 3.

Since (3q + 5) is not divisible by 3, this implies (n + 4) is also not divisible by 3.

Taking n = 3q + 2 …(iii)

Putting q = 1, we get

n = 3(1) + 2
= 5

5 is not divisible by 3.

(3q + 2) is not divisible by 3

n is not divisible by 3.

Adding 2 on both sides of equation (iii),

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

Putting q = 1, we get

n + 2 = 3(1) + 4
= 7

Since 7 is not divisible by 3, this means (3q + 4) is not divisible by 3.

(n + 2) is not divisible by 3.

Adding 4 on both sides of equation (iii),

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

Since 3(q + 2) is a multiple of 3, this means it is divisible by 3.

(n + 4) is divisible by 3.

Thus, we have seen exactly one of the numbers n, n + 2, n + 4 is divisible by 3.

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