Question

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

n = bq + r where b = 3

n = 3q + r, where r = 0, 1, 2

Check for n, n + 1, n + 2]​

Answer 1

ANSWER:

• One out of n, n+1, n+2 is divisible by 3.

GIVEN:

• A positive integer and which is divided by 3.

TO PROVE:

• One Out of n, n+1, n+2 is divisible by 3.

SOLUTION:

Let n be any positive integer which is divided by 3 we get some quotient 'q' and some remainder 'r'.

n = 3q+r

Let r = 0

=> n = 3q (divisible by 3)

Let r = 1

=> n = 3q+1

=> n+2 = 3q+1+2. (adding 2 both sides)

=> n +2 = 3q+3

=> n+2 = 3(q+1) (divisible by 3)

Let r = 2

=> n = 3q+2

=> n +1 = 3q+2+1. (adding 1 both sides)

=> n +1 = 3q+3

=> n+1 = 3(q+1). ( divisible by 3)

So one out of n, n+1, n+2 is divisible by 3.

• In the similar way we can solve : one out of n, n+4, n+8 , n+6 and n+12 is divisible by 5, where n is any positive integer.
• This is by Euclid division algorithm.

Answer 2

Answer:

Step-by-step explanation:

To Prove :-

Any three consecutive positive integers must be divisible by 3.

Solution :-

Let the three consecutive positive integers be n, n + 1 and n + 2.

Reminders, we get by dividing any number by 3 are 0 or or 2.

So, r = 0, 1 and 2.

n = 3q, 3q + 1, 3q + 2. where p is the integer

When, n = 3q

Then, n is divisible by 3.

When, n = 3q + 1,

= n + 2 = 3q + 1 + 2

= 3q + 3

= 3(q + 1)

Then n + 1 is divisible by 3.

= n + 1 = 3q + 2 + 1

= 3q + 3

= 3(q + 1)

Then, n + 2 is divisible by 3.

Hence,  any three consecutive positive integers must be divisible by 3.