# How can I get to Mars with a polynomial?

In order to get to Mars you must win a video game. The video game chooses $10$ points $(a_i,b_i)$ where $a_i$ and $b_i$ are single-digit integers, and places a disk with radius $1/3$ on each of the points. You must find a polynomial $f$ such that the graph of $f$ hits all $10$ discs.

However, you must choose your polynomial before seeing where the disks are. Find a polynomial that guarantees you a trip to mars.

The only clue I have for this problem is that for some point $x=h$ I must have a polynomial which is quite "steep" at that point and resembles the graph of a vertical line.In this way I am guaranteed that any other lattice point $(a_i,b_i)$ above it is included.

The only way I can imagine a polynomial doing that is when I have a product of positive expressions, i.e for $x=h$ I would have something like $(x-h)(x+r)(x+d)$ where $r,d$ are some large numbers.

The main conceptual difficulty I am facing here is to produce this kind of behavior for every single digit integer on the number line...

• Ah, but you have natural bounds in the question: "$a,b$ are single-digit integers"... – abiessu Feb 2 '16 at 14:55
• Is there a maximum to the degree of $f$? Can't you create a polynomial passing through $(a-\tfrac 1{10},1)$ and $(a+\tfrac 1{10},9)$ for $a=1,2,\cdots,9$? Surely it will then be in range of every lattice point in $[1,1]\times [9,9]$ – vrugtehagel Feb 2 '16 at 14:55
• @vrugtehagel I don't think there's a maximum degree for $f$ (the problem asks just for some polynomial )and no is for the second question .I guess it's something pretty easy (for others...) but still nothing comes to my mind. – Mr. Y Feb 2 '16 at 14:58
• The title is rather misleading. Polynomials have no mass, so I'd say the problem reduces to just going to Mars (which is hard enough), since having achieved that, you can take the polynomial with you at no extra cost. But personally I would want to avoid going to Mars at all price. – Marc van Leeuwen Feb 2 '16 at 15:25

We know that $(a,b)\in[0,0]\times [9,9]$ (since $a,b$ are single-digit integers. If negative numbers are considered "single-digit integers", too, then this is generalizable for those, too). So, what about $$p(x)=100(x-0)(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)$$ A plot made with Wolfram Mathematica 10.0:
• I edited the answer so that it passes through $0$. I'll change the plot image in a second. – vrugtehagel Feb 2 '16 at 15:13
• @Mr.Y Now it would be a nice exercise to try to minimize the $100$ – N74 Feb 2 '16 at 15:16
• @N74, I don't even think the factor $100$ is needed. The plot looks the same if I just completely get rid of it (so changing it to $1$). But if you want to go to Mars really badly, then better make that factor as large as you can! – vrugtehagel Feb 2 '16 at 15:17
• So the $1/3$ thing was ....somewhat obsolete .However you did win that trip to Mars :P. – Mr. Y Feb 2 '16 at 15:21
• The smallest hump in the graph is between $x=4$ and $x=5$, so it's enough to make sure that the graph goes through the point $(3\frac23,9)$. Evaluating the monic version of the polynomial at $x=3\frac23$ gives a value $> 867.9$, so you can get away with a factor of $\frac{9}{867.9} \approx 0.0104$. – TonyK Feb 2 '16 at 15:29