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This quirky equation was presented to me by a fellow teacher, along with the instruction "solve it both ways up".

upside

The equation is simple enough to solve, with one integer root. What's more interesting is that the page can be rotated $180$° and a new equation can be read.

downside

This second equation has two roots. I found it fairly satisfying that both of its roots are integers, but I was hoping that it would share a root with the first equation.

I set about looking for an equation that makes sense when read upside-down, where the two equations share a root.

My students immediately came up with trivial examples like "$x=1+1$" and "$1=\frac{1}{x}$". I haven't found anything more interesting yet. I haven't even figured out an approach better than trial and error, with different permutations of the digits $1$, $6$, $8$ and $9$.

Can anybody help me to find such an equation, ideally one that is similarly "difficult"/"interesting" as the equation presented to me?

Edit: credit is due to Rob Eastaway who (I have learned since posting this question) originally posted the first image on Twitter.

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    $\begingroup$ This is simpler than you’re asking for, but the solution to $\mathsf{66-6x=0}$ is $\mathsf{x=l\,l}$ . If you turn the equation and the solution upside down, you get $\mathsf{0=x9-99}$ and $\,\mathsf{l\,l=x}$ , which is still correct. Maybe there’s a way to hide this in something more interesting. $\endgroup$ – Steve Kass Dec 19 '18 at 20:33
  • $\begingroup$ By 'upside-down' do youy mean rotated by 180 degrees, or reflected in a horizontal axis? This makes a difference; the one allows you to use sixes and nines, the other allows you to use threes. $\endgroup$ – Servaes Dec 20 '18 at 10:55
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    $\begingroup$ @Servaes have a read of the sentence between the two images. The problem is similar if the equation is reflection, though: I set this as an extension problem! $\endgroup$ – Malkin Dec 20 '18 at 18:17
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    $\begingroup$ Here is an Aperiodical article on this exact topic: aperiodical.com/2018/12/…. $\endgroup$ – Peter Kagey Dec 21 '18 at 17:36
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If I may be allowed to use $2$ and $5$ as well (their digital representations can be flipped ... and result in themselves):

$$\frac{82-x}{59+61-x+51}=\frac{x-28}{65+x-19+15}$$

Both original and flipped version have $x=55$ as a solution

Here is a digitized version:

enter image description here

and its flipped version:

enter image description here

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  • $\begingroup$ Thanks a lot for taking the time to write out the "digitized" version of your example. I will definitely be presenting it as a problem to some of my high school students. $\endgroup$ – Malkin Dec 20 '18 at 19:04
  • $\begingroup$ @Malkin You're welcome! I do wish Mathjax just had a font for this :) $\endgroup$ – Bram28 Dec 20 '18 at 19:07
  • $\begingroup$ Your example is nicely generalisable to equations of the form $\frac{x-a}{x-b}=\frac{a'-x}{b'-x}$ where $a'$ and $b'$ are the rotated forms of $a$ and $b$ (which may be sums; in your case $b=59+61+51$). A quick search on a spreadsheet for examples with integer solutions has yielded lots of such equations! For example, $\frac{x-81}{98-x}=\frac{18-x}{x-86}$ $\endgroup$ – Malkin Dec 20 '18 at 19:11
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    $\begingroup$ @Malkin Ah, nice! I didn't have access to a spread sheet while sitting in my comfy chair by the fire last night, so had to rely on pencil and paper to find something that worked. So I started with $82-x=x-28$ with solution $x=55$. I then took $61-x=x-19$ with solution $x=40$. So, I needed to increase the latter solution by $15$ by adding pairs of $a$ and $a'$. Using $59$ and $65$ the solution would increase (or decrease) by $3$. Using $15$ and $51$ it increases or decreases by $18$ ... bingo! (well, it took more work that that: I looked at all pairs, and noted their solutions and differences) $\endgroup$ – Bram28 Dec 20 '18 at 19:50
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There's of course

$$ \frac{1 + x}{1} = \frac{1}{x} ~~~\mbox{and}~~~ \frac{x}{1} = \frac{1}{x + 1} $$

These two don't have any integer roots, but they have the golden ratio as one

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  • $\begingroup$ OP is "looking for an equation that makes sense when read upside-down, where the two equations share a root." Upside-down, we get nothing. $\endgroup$ – Mohammad Zuhair Khan Dec 19 '18 at 19:33
  • $\begingroup$ @MohammadZuhairKhan Upside-down reads the same $\endgroup$ – caverac Dec 19 '18 at 19:39
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    $\begingroup$ Well, not before you edited it. +1 $\endgroup$ – Mohammad Zuhair Khan Dec 19 '18 at 19:44
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    $\begingroup$ @MohammadZuhairKhan Fair enough :) $\endgroup$ – caverac Dec 19 '18 at 23:26
  • $\begingroup$ Thanks for this. Your idea of starting with the golden ratio is nice, given its connection with its reciprocal. I will play with this starting point and try to shoehorn in some 8s and 6s, to make the equation a little more "interesting". $\endgroup$ – Malkin Dec 20 '18 at 19:12
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I think $0=(x-1)(x-1)$ does it.

Maybe $0=(x-6)(x-1)$ shares only one root.

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The answers by @caverac and @Bram28 very much helped me to come to some answers of my own.

My method has been to guess at the form of an equation, such as $ \frac{x+a}{b} + c = \frac{x+1}{d}$, which can be rotated to give $ \frac{d'}{1+x}=c'+\frac{b'}{a'+x} $, where $a$, $b$, $c$ and $d$ are constants that can be read as $a'$, $b'$, $c'$ and $d'$ when rotated.

I then used a fairly simple Excel spreadsheet to search for equations of this form that share solutions with their rotated forms, with the constants being selected from the set of all one or two digit numbers that can be read upside down:

$$S=\{1,8,6,9,11,18,81,16,61,19,91,88,86,68,89,98,66,99,69,96\}$$

(I have not included $5$ because my own written $5$s don't look like $5$s when upside down.)

The above form delivered no results. Neither did the forms $\frac{x}{a} =b- \frac{1-x}{c}$ or $\frac{a-x}{b} +c= \frac{1+x}{d}$.

But the form $ \frac{x+a}{b} =c- \frac{x-1}{d}$ delivered the equation:

$$ \frac{x+61}{6} =1- \frac{x-1}{8}$$

which shares an integer root with its rotated form, and:

$$ \frac{x+61}{8} =6- \frac{x-1}{16}$$

which shares a rational root with its rotated form.

I guess that more examples could be found by trying other equation forms; by including three digit numbers in $S$; or by allowing the constants to be sums of numbers from $S$, for example $a=14=8+6$ with $a'=17=9+8$.

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  • $\begingroup$ Nice! Thanks for sharing! $\endgroup$ – Bram28 Dec 29 '18 at 19:00

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