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The last digit of some number 𝑋 in base 𝑘 is 2. Last number of $12_{10}*X$ in the same system is 4. How many systems that are suitable for these conditions for any X.

I m really confused for example I take $3_{10}$ as X, so for any system last digit will be 3, so there are no such systems?

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  • $\begingroup$ I don't understand the first sentence. Did you mean to say, "The last digit of some number $X$ in base $k$ is $2?$" $\endgroup$ – saulspatz Feb 9 at 17:22
  • $\begingroup$ Correct, I'll edit that... $\endgroup$ – Hmmman Feb 9 at 17:23
  • $\begingroup$ I think you misunderstand the question. What they are asking is, for what values of $k$ is it true that if the last digit of a number $X$ in base $k$ is $2,$ then the last last digit of of $12_{10}X$ in base $k$ is $4.$ For example, this is true if $k=10.$ $\endgroup$ – saulspatz Feb 9 at 17:41
  • $\begingroup$ It seems like you are right... but how can I determine all such systems? $\endgroup$ – Hmmman Feb 9 at 17:56
  • $\begingroup$ And to be sure that it's correct for all digits, for example k = 10: 2, 12*2 = 24 how can i be sure that for ...2, 12*...2 = ....4 $\endgroup$ – Hmmman Feb 9 at 17:57
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Let me get you started. First of all, we must have $k>4$ or we couldn't have the digit $4$. Then the last digit of $X$ in base $k$ is $2$ if and only if $$X\equiv2\pmod{k}$$ Then we know that $$12X\equiv24\pmod{k}$$ and we need $$12X\equiv4\pmod{k}$$ in order for the last digit to be $4$.

So the question becomes, for what values of $k>4$ is it true that $$Y\equiv24\pmod{k}\implies Y\equiv4\pmod{k}$$

Well, if $k|(-4)$ and $k|(Y-24)$ then $k$ divides their difference, so $k|20$. The admissible $k$ are $5,10,20$.

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  • $\begingroup$ I understand that and feeling guilty that I can't make further conclusions... $\endgroup$ – Hmmman Feb 9 at 18:26
  • $\begingroup$ Big thanks!!!!!! $\endgroup$ – Hmmman Feb 9 at 18:35
  • $\begingroup$ @Hmmman It was my pleasure. $\endgroup$ – saulspatz Feb 9 at 18:37

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