What's the smallest natural number that starts with $15$ that becomes $5$ times when these digits are moved to the end?

Obviously the original number must end in $3$ (so that $\times \, 5 = 15$). Also its third digit must be $7$ or $8$, since the third digit becomes the first digit of the altered number.

So if the intermediate part has a length of $n$ digits, the number must be of the form: $15XX\ldots X3$ where $XX\ldots X$ has $n$ digits.

Any ideas on how to continue?

  • $\begingroup$ Do you mean with first two digits 1 and 5? $\endgroup$ – Manish Kundu Jan 14 '18 at 10:13
  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Jan 14 '18 at 10:18
  • $\begingroup$ @ManishKundu yes! $\endgroup$ – Tom Galle Jan 14 '18 at 10:20
  • $\begingroup$ @JoséCarlosSantos: Thank you, I will try to understand the MathJax techniques by going through the documents. Sorry, I am 58 y.o. and not very familiar with technology :( $\endgroup$ – Tom Galle Jan 14 '18 at 10:22
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    $\begingroup$ $157894736842105263\cdot5=789473684210526315$ $\endgroup$ – Professor Vector Jan 14 '18 at 11:00

Let all the $m$ digits after "15" form the number $y$. Prepending the "15" means $15\cdot10^m+y$, while appending "15" means $100y+15$, so we have the equation $$5\,(15\cdot10^m+y)=100y+15,$$ i.e. $$y=\frac{15\cdot10^m-3}{19}.$$ Now we can solve this with a bit of modular arithmetic, or like this:

  • $\begingroup$ Professor Vector you are brilliant! $\endgroup$ – Tom Galle Jan 16 '18 at 11:35

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