Sum is odd, sum is even

Number Theory Level 3

$\begin{aligned} 495 &= 51+52+\cdots+59\\ &=45+46+\cdots+54\\ &=40+41+\cdots+50 \end{aligned}$

As shown above, 495 can be expressed as the sum of 9, 10, or 11 consecutive integers.

Can we find an integer that can be expressed as the sum of $X, X+1, X+2,$ or $X+3$ consecutive integers?

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1 solution

Otto Bretscher
Oct 29, 2018

Sadly perhaps, this is impossible to do, comrade! Taking the heading as a hint, we observe that the sum of $4n$ consecutive integers is even, while the sum of $4n+2$ consecutive integers is odd (in both cases, the average value of the summands is half an odd integer).

Another great problem! I never thought about this issue before. Thank you for posting!

Yay! Luckily this problem does not have an issue (unlike my other recent ones).

The way I formulated this setup is based on "The sum of (2n+1) consecutive integers must be divisible by (2n+1)".

Pi Han Goh - 2 years, 7 months ago

But how does "The sum of (2n+1) consecutive integers must be divisible by (2n+1)" show that this problem has no solution? (It's late at night, and I'm probably missing something obvious.)

Otto Bretscher - 2 years, 7 months ago

To clarify: I know that the sum of an odd number of consecutive integers is divisible by that same odd number, so I'm wondering if it works for even numbers as well. Turns out no! So I turn this observation into a question like this.

Pi Han Goh - 2 years, 7 months ago

@Pi Han Goh – OK, I get it now; I'm a little slow sometimes ;)

Otto Bretscher - 2 years, 7 months ago

Sum is odd, sum is even

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