Prove that one of any three consecutive positive integers must be divisible by 3

Solution:

Consider three consecutive positive integers as n, n +1 and n + 2

Now dividing n by 3, consider q as the quotient and r as the remainder.

Using Euclid’s division algorithm,

n = 3q + r, where 0 ≤ r < 3

n = 3q or n = 3q + 1 or n = 3q + 2.

Case I:

When n = 3q, divisible by 3

(n + 1) and (n + 2) are not divisible by 3.

Only n is divisible by 3.

Case II:

When n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1) is divisible by 3

n and (n + 1) are not divisible by 3.

Only (n + 2) is divisible by 3.

Case III:

When n - 3q + 2, then n + 1 = 3q + 3 = 3(q + 1) is divisible by 3

n and (n + 2) are not divisible by 3.

Only (n + 1) is divisible by 3.

Therefore, one of any three consecutive positive integers is divisible by 3.

✦ Try This: Prove that one of any three consecutive positive integers must be divisible by 6

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 1

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NCERT Exemplar Class 10 Maths Exercise 1.4 Problem 3

Prove that one of any three consecutive positive integers must be divisible by 3

Summary:

One of any three consecutive positive integers is divisible by 3. Hence proved

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☛ Related Questions:

• For any positive integer n, prove that n³ – n is divisible by 6
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