Question

given three consecutive integer a, a+1,a+2. show that one of them is divisible by 3.

Answer 1

What ideas have you tried thus far?

Answer 2

I'd try to prove that the product of the numbers is divisible by 3 using induction. Since the product is divisible by 3, one of the numbers must have had 3 as a factor.

Answer 3

@ joemath314159...I have no ideas

Answer 4

We know is that either a is even or odd. Thus a = 2k or a = 2k + 1 where k is an integer.

If a is even then the terms become: 2k, 2k + 1, 2k + 2
If a is odd then the terms become: 2k + 1, 2k + 2, 2k + 3

We notice that both sequences share 2k + 1 and 2k + 2. So we must prove that either
2k + 1 or 2k + 2 is a multiple of 3 for any value of k. First let's factor: 2k + 2 = 2(k + 1). So we know that 2k + 2 is always a multiple of 2, thus if we form a set A of all multiples of 2 given any integer k, the set becomes: A = (...-2, 0, 2...) <--Ellipses indicate positive and negative infinity.

We know that 2k + 1 is always odd so if we form a set B of all odd integers given any integer k, the set becomes: B = (....-1, 1, 3.....)

If we combine both sets A and B, i.e. A U B, we get: A U B = (...-3, -2, -1, 0, 1, 2, 3...)
This is the set of all integers from negative infinity to positive infinity.

Thus, we know that the union of the sets A and B of 2k + 1 and 2k + 2 contains all integers. Atleast one of them is divisible by 3 so we conclude that for any integer a, one of the terms of the sequence a, a + 1, a + 2 is always divisible by 3.

Answer 5

@Avoryibor