Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer

Solution:

Using the Euclid’s division Lemma,

Consider a positive integer = n , b = 3

n = 3q + r,

where q is the quotient and

r is the remainder

0 < r < 3 means that remainders can be 0, 1 and 2

n can be in the form of 3q, 3q + 1, 3q + 2

When n = 3q

n + 2 = 3q + 2

n + 4 = 3q + 4

n is only divisible by 3

When n = 3q + 1

n + 2 = 3q = 3

n + 4 = 3q + 5

n + 2 is divisible by 3

When n = 3q + 2

n + 2 = 3q + 4

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

n + 4 is divisible by 3.

Therefore, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3

✦ Try This: Prove that one and only one out of q, q + 2 and q + 4 is divisible by 3, where q is any positive integer

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

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NCERT Exemplar Class 10 Maths Exercise 1.4 Problem 2

Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer

Summary:

One and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer. Hence Prove

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☛ Related Questions:

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