# Finding numbers that double when you switch the first and last digit

So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.

I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:

$$2(10a + b) = 10b + a$$

This simplifies to:

$$19a = 8b$$

Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.

Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):

$$199a + 10b = 98c$$ And for four-digit numbers, $$1999a + 100b + 10c = 998d$$

Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.

I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.

I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.

Thank you!

We may represent integer $\ n>0\$ as $\ n\ =\ b\cdot 10^d+M\cdot 10 + a\$ (so that $\ a>0\$ or $\ M>0\$ or else we would have $a=M=b=0)\$ so that

$$2\cdot(a\!\cdot\! 10^d + M\!\cdot\! 10 + b) \ =\ b\!\cdot 10^d+M\!\cdot 10 + a$$

where $\ a\ b\$ are decimal digits (i.e. $\ 0\ldots 9),\$ and $\ M\$ is a non-negative integer such that $\ M<10^{d-1}.$ Equivalently,

$$(2\!\cdot\! 10^d-1)\!\cdot a + 10\!\cdot\!M \,\ =\,\ (10^d-2)\!\cdot\! b$$ or $$b\,\ =\,\ 2\!\cdot\!a\ +\ \frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2}$$

Thus, $$0\ <\ b-2\cdot a\ = \frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2}\ \le\ 1\ +\ \frac{3\cdot a-8}{10^d-2}$$

Also if we had $\ d\ge 2\$ then this very last fraction has an absolute value $\ < 1\$ so that

$$\frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2}\ =\ 1$$

while, knowing that $\ a<5\$ (of course!),

$$10\!\cdot\!M\ +\ 3\!\cdot\!a\,\ \not\equiv \,\ 10^d-2\quad \mod 10$$

This contradiction shows that $\ d\le 1.\$ Thus there are only 2-digit solutions (if any). In particular, $\ M=0.\$ This simplifies the equation:

$$19\cdot a\ \ =\ \ 8\cdot b$$

However, $19$ does not divide $\ 8\cdot b.\$ Thus, after all,

$\qquad\qquad$ THERE ARE NO SOLUTIONS.