0

Here is the full problem : Non-zero positive integers, not necessarily distinct, are written on the squares of an $8$ × $8$ chessboard (one number per square). At the beginning, five grasshoppers are on five differentsquares and hide the numbers. Gabriel calculates the sum of all the visible numbers and he obtains $100$. Simultaneously, each grasshopper jumps onto an adjacent square (it crosses a side shared by two squares). Gabriel calculates the sum of all the visible numbers and he gets $1000$. And so on, until he can no longer obtain a sum ten times greater than the previous one (when Gabriel calculates a sum, two grasshoppers are never on the same square). The total of the sixty-four numbers written on the chessboard is divisible by $35$ and it is the largest possible. What is this total?

This was the only problem on a math competition that I couldn't figure out and I still don't know how to solve it (I don't even know if I'm supposed to use number theory or combinatorics...) . The detailed solution isn't available on their website, the only thing that is said is that the answer is $11110785$. If anyone could help me figure out how to solve it, it would be appreciated!

John Omielan
  • 47,976
David G.
  • 95
  • “Non-zero positive integers” is redundant. – Thomas Andrews Dec 31 '21 at 03:27
  • 1
    No, for example, in Quebec, we consider that positive integers include zero. Plus, it's as it was written in the competition so I can't do much about it... – David G. Dec 31 '21 at 03:33
  • @ThomasAndrews FYI, in general you're right, but note that according to Willie Wong's answer, "In typical French mathematical usage, zero is both positive and negative. Or rather, in mathematical French "$x$ est positif" (literally "$x$ is positive") allows the case $x = 0$, while "$x$ est positif strictement" (literally "$x$ is strictly positive") does not." I've also seen the same thing mentioned about France's handling of positive/negative in a comment, but this was at least a few months ago and I don't recall with which post I saw it. – John Omielan Dec 31 '21 at 03:34

1 Answers1

2

Hint: each time the grasshoppers jump a new large value must be exposed so the sum increases so much. There were only five squares hidden at the start, so there are at most five steps upward. You can think of the grasshoppers all being next to each other in a row and jumping one square in the same direction. They can turn the corner if they are going to leave the board. So each time one new square is revealed and one deleted. Now use the factor $35$ to refine the problem.

Ross Millikan
  • 374,822
  • I named $a,b,c,d,e$ the $5$ initial squares (in order of deletion) and $m,n,o,p,q$ the $5$ squares revealed (in order). This is where I think I go wrong : $a-m = 900$, $b-n = 9000$, $c-o = 90000$, $d-p = 90000$ $e-q = 900000$ – David G. Dec 31 '21 at 03:45
  • Also, let $S$ be the sum of the remaining squares. $S \geq 54$ s.t each square's value is at least $1$ $\implies$ $m+n+o+p+q \leq 46$ s.t the starting sum is $100$ – David G. Dec 31 '21 at 03:47
  • Add those all together to get a constraint on the final sum. Round down to a multiple of $35$. – Ross Millikan Dec 31 '21 at 03:48
  • I only get $10000145$ as the maximum sum, which is lower than the answer. (In my first message, there was a typo, I meant $d−p=900000$ and $e−q=9000000$) – David G. Dec 31 '21 at 03:50
  • 1
    This is assuming that all five of the initially hidden numbers are visible at the end. I believe the solution has only one of those numbers visible at any given time. – Daniel Mathias Dec 31 '21 at 04:02
  • In my head, those $5$ hidden numbers are the largest, so they have to be all visible at the end. But I might be wrong... – David G. Dec 31 '21 at 04:06
  • 1
    @DavidG.: But they might not be visible at intermediate steps, which can increase the next number. – Ross Millikan Dec 31 '21 at 04:07
  • @DavidG. Confine the grasshoppers to a 2x3 region of the board. – Daniel Mathias Dec 31 '21 at 04:10
  • 2
    If I do this, the sum of the remaining squares is at least $58$ and let $a,b,c,d,e,f$ be the $6$ squares. WLOG $f = 42$ as it is the unhidden square. The rest are easily $942, 9942, 99942, 999942, 9999942$ and the sum is $11110810$. The closest smaller multiple of $35$ is indeed $11110785$. Thank you! – David G. Dec 31 '21 at 04:17
  • Being pedantic with the hint, the "new large value" will need to be demonstrated. For example, if we have 11 grasshoppers, we can get to 6 steps upward. – Calvin Lin Dec 31 '21 at 15:33