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Find all natural numbers with the property that when the first digit is moved to the end, the resulting number is $3.5$ times the original number.

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Will this work if the natural number you begin with is odd? – Dilip Sarwate Oct 30 '12 at 2:37
up vote 7 down vote accepted

Suppose that our original number is $k+1$ digits long. Let $a$ denote the first digit of the number. We represent the number as $$10^ka + b$$ where $b < 10^k$ is the remaining portion of our number. The condition is thus expressed as $$3.5(10^ka + b) = 10b + a$$ Clearing fractions and rearranging, we end up with $$(7\cdot 10^k - 2)a = 13b$$ Clearly $13\nmid a$ and we must therefore have $$7\cdot 10^k - 2 \equiv 0 \pmod{13} \implies 10^k \equiv 4 \pmod{13}$$ We find $k = 5$ as the smallest solution to the congruence. The order of $10$ modulo $13$ is $6$, so we have $k\equiv 5\pmod6$.

Getting back to our former equation, we note that $$a \ge 2 \implies \frac{14}{13}10^k - 2 > 10^k$$ which will contradict the restrictions on $b$. We then have $a = 1$. Therefore the numbers are of the form $$10^k + \frac{7\cdot 10^k - 2}{13} = \frac{2\cdot 10^{k+1} - 2}{13}$$ where $k \equiv 5 \pmod6$. The order of $10$ modulo $13$ is $6$ so the fraction $\frac{1}{13}$ repeats with period $6$. Denote the repeating block of $13^{-1}$ as $r = 076923$. Our number is of the form $$2\times \frac{10^{6n}- 1}{13} = 2\times\underbrace{rr\cdots r}_{n \text{times}}$$ where the latter expression denotes the number formed by concatenating $r$ with itself $n$ times for some natural number $n$. Putting it all together, our number is of the form $153846$ appended with itself an arbitrary number of times, $$\left\{153846,\ 153846153846,\ 153846153846153846,\ \cdots\right\}$$

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Besides EuYu's, another way to arrive at the answer is to say that the leading digit has to be $1$ or $2$, because if it were $3$ or greater there would be a carry and the product would have more digits than the original number. Let's try $1$. The multiplicand then starts with $1$ and the product ends with $1$. To have the product end with $1$, the multiplicand must end with $6$ as $6 \cdot \frac 72=21$. Then the product ends with $61$ and the multiplicand ends with $46$. We keep going until we have a $1$ at the front, giving $$ 153846 \\ \underline{\times \ \ \ \ 3.5} \\ 538461$$ Now we can see that multiplying by $2$ will carry, so will not work and this is the only primitive solution.

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