The Name's Donder, not Donner!

Algebra Level 4

Donder is sick of people calling him Donner all the time, so he has decided to start a tour around the world on Christmas (2015) to see how many people were calling him the wrong name.

He will spend $n$ days in 1 place, then $n-1$ days in 2 places, then $n-2$ days in 3 places, $\dots$ , 2 days in $n-1$ places then 1 day in $n$ places. Assume that his travelling does not take up time (as he factors that into his stay at each place). ( $n$ is a positive integer.)

His tour can be at most 365 days long as he must get back to Santa on Christmas Eve. (Yes, 2016 is a leap year)

What is the maximum length of his tour in days?

This was inspired from a discussion of the reindeer Donder's name from Day 9: The Reindeer Quiz . Thanks go to Colin Carmody for telling me this!

The answer is 364.

1 solution

Michael Ng
Dec 10, 2015

His tour is of length $\sum_{k=1}^{n} k(n+1-k)$ days.

Split the sum to give $\sum k(n+1) - \sum k^2 \\ = \frac{n(n+1)^2}{2} - \frac{n(n+1)(2n+1)}{6} \\ = \frac{n(n+1)(n+2)}{6}$

Now $\frac{n(n+1)(n+2)}{6} \leq 365$

By far the easiest method from here is to simply calculate some values. At $n=12$ we get $364$ and it is clear that for larger $n$ the expression will exceed $365$ , so the answer is $\boxed{364}$ days, as required.

Thanks for giving me credit! Nice problem too!

Colin Carmody - 5 years, 6 months ago

No problem; thank you!

Michael Ng - 5 years, 6 months ago

The Name's Donder, not Donner!

This was inspired from a discussion of the reindeer Donder's name from Day 9: The Reindeer Quiz . Thanks go to Colin Carmody for telling me this!

The answer is 364.

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