Question

The sum of 7 consecutive whole numbers is odd. How many of these numbers must be odd?

Answer 1

lets try an example
\[1+2+3+4+5+6+7\] is it odd?

Answer 2

i get 28 which is even
now lets try
\[2+3+4+5+6+7+8\]

Answer 3

i get an odd number
namely 35

Answer 4

Yes it is

Answer 5

35

Answer 6

so it looks like the first and last are even, leaving 3 odd, 4 even

Answer 7

nothing like an experiment to see what is going on

Answer 8

I like that, thank you.

Answer 9

Note that in \(7\) numbers we either have \(3\) even and \(4\) odd or \(4\) even and \(3\) odd. In such a case, consider whether our sums are divisible by \(2\). Let every even number be written in the form \(2n_k\) and every odd \(2m_k+1\) and consider the case where we have \(3\) even and \(4\) odd:$$2n_1+2n_2+2n_3+2m_1+1+2m_2+1+2m_3+1+2m_4+1\\\ \ \ \ \ =2(n_1+n_2+n_3+m_1+m_2+m_4+2)$$... which is even. We clearly then need \(4\) even and \(3\) odd to yield an odd sum.

Answer 10

at the risk of repeating myself, nothing like an experiment ...

Answer 11

In terms of modular arithmetic, we know that every even number is congruent to \(0\mod2\) and every odd congruent to \(1\mod2\). We know that in 7 integers either 3 will be odd and 4 even or vice versa. For an odd sum, we require there to thus be an odd number of odd integers hence the answer is \(3\) odds.