An Odd Sum

Number Theory Level 1

The sum of several odd terms is odd. What can we say about the number of terms?

It's even It's prime It's odd It's a perfect square

2 solutions

Hana Wehbi
Apr 9, 2016

We can proceed by contradiction.

Suppose that we have $2k$ terms, each of the form $2 n_i + 1$ .
Then, the sum is $2 ( n_1 + n_2 + \ldots + n_{2k} ) + 2k$ which is even. Hence, the number of terms cannot be even.

Suppose that we have $2k+1$ terms, each of the form $2 n_i + 1$ .
Then, the sum is $2 ( n_1 + n_2 + \ldots + n_{2k} + n_{2k+1} ) + (2k+1)$ which is odd. Hence, we know that the number of terms must be odd.

Note that the number of terms need not be prime.

Yes, that's an simple specific example. How do we know that it must be always odd?

Calvin Lin Staff - 5 years, 2 months ago

We can prove by mathematical induction that: 3+5+7+...+(2n+1)= n(n+2) which is odd.

Hana Wehbi - 5 years, 2 months ago

We do not necessarily add the consecutive odd numbers.

For example, we could have $1 + 5 + 11 = 17$ .

Calvin Lin Staff - 5 years, 2 months ago

@Calvin Lin – Yes, and we are concerned about the number of terms. We can proceed by contradiction, that suppose it is true that the sum of " i " ( where i is even) terms of odd numbers is odd, say (2n+1)+ (2k+1)= 2(k+n)+2(1)= 2( k+n+1), which is even. We have reached a contradiction. Thus "i" should be odd.

Hana Wehbi - 5 years, 2 months ago

Ashish Menon
May 28, 2016

Odd + Odd is even and Evn + Odd is Odd. So, to have an odd aum, ww must add $\color{#69047E}{\boxed{\text{odd}}}$ number of odd numbers.

An Odd Sum

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