# Cycling Digits puzzle

I'm trying to answer the following:

"I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by $2$. Am I telling the truth?"

I think the answer to that is no. It's easy to prove that it's false for numbers with two digits: Let $N = d_0 + 10 \cdot d_1$. Then $2N = 2 d_0 + 20 d_1$ and the "swapped" number is $N^\prime = d_1 + 10 d_0$. We would like to have $2d_0 + 20 d_1 = d_1 + 10d_0$ which amounts to $8d_0 = 19d_1$. The smallest value for which this equality is fulfilled is $d_0 = 19, d_1 = 8$ but $19$ is not $\leq 9$ that is, is not a digit, hence there is no solution.

Using the same argument I can show that the claim is false for $3$-digit numbers.

I conjecture that it's false for all numbers. How can I show that? Is there a more general argument than mine, for all numbers? Thanks for helps.

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Zero respects the constraint, but it's a bit a dull solution – Jean-Sébastien Sep 28 '12 at 17:21
Does the number have to be in base 10? Does it matter? – David Conrad Feb 18 '14 at 20:02
@DavidConrad Yes the number is assumed to be in base $10$. – Rudy the Reindeer Feb 18 '14 at 20:28

If you follow your argument but let $N=a+10b$ where $a$ is a single digit but let $b$ have $n$ digits, then $2N=10^na+b$ and you get $b=\frac {10^n-2}{19}a$ If $n=17$, this is integral. Then $a$ has to be at least $2$ to make $b$ have $17$ digits. The smallest solution is $105263157894736842$

Another way to get there is to just start multiplying. If you guess that the ones digit is $2$, you double it the ones digit of the product will be $4$, which will be the tens digit of the first number, and so on. Stop when the product starts with a $2$ and doesn't carry. You get $$105263157894736842 \\ \underline {\times\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2} \\ 210526315789473684$$ If you had started with a $1$, you would miss the leading zero.

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Excellent, thank you! – Rudy the Reindeer Sep 28 '12 at 18:19

The number has to be divisible by 9

The remainder left by a number when divided by $9$ is equal to the sum of its digits. Now, here, you are not changing the digits when transferring the digits, so the remainder does not change. However, multiplying by two should double the remainder module nine and this is a contradiction unless the number is divisible by $9$.

Solution

Assume that the number $u = 9k$ and the representation of $u$ is $$\overline{a_{n}a_{n-1}\ldots a_{1} a_{0}} = u = 9k$$. Then $$\overline{a_{0}a_{n}\ldots a_{2} a_{1}} = 2u = 18k$$.

Hence multiplying by 10 we get, $$\overline{a_{0}a_{n}\ldots a_{2} a_{1}} * 10 + a_{0} = 180k + a_{0}$$.

Regrouping the digits and then writing the original number as $9k$, we get

$$a_{0}*10^{n+1} + \overline{a_{n}a_{n-1}\ldots a_{1} a_{0}}= 180k + a_{0}$$

which implies that

\begin{align} &a_{0}*10^{n+1} + 9k = 180k + a_{0}\\ \\ &\left(10^{n+1} -1 \right) a_{0} = 171k\\ \\ &\underbrace{\overline{99\ldots999}}_{n+1}a_{0} = 171k \end{align}

that is $$171k = 9*\left( \underbrace{\overline{11\ldots111}}_{n+1}\right) a_{0}$$

hence $$19k = \left( \overline{11\ldots 111}\right) a_{0}$$

So, the problem comes down to finding the solutions of the above equation.

Combinining this with Ross Millikan's Analysis, we get the equation

$$10^{n+1} = 171k + a_{0}$$ and $$9k \equiv a_{0} \, \mathrm{ mod } \, 10$$

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But you have the solution $2\cdot 0=0$, so this argument only shows that the sum of digits must be divisible by 9. – Per Manne Sep 28 '12 at 17:29
Unless the number was a multiple of 9 to begin with – Ross Millikan Sep 28 '12 at 17:30
@PerManne Yup. Sorry about that, I was going to update it. – Jayesh Badwaik Sep 28 '12 at 17:30
Thanks. I don't understand what you do below the line (in particular not how you conclude that all the digits are ones) but arguing using remainders hadn't occurred to me and it's very interesting to see a new piece of information. – Rudy the Reindeer Sep 28 '12 at 18:17
Added some more explanation. I hope it is more clear now. – Jayesh Badwaik Sep 28 '12 at 18:25