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?

  • $\begingroup$ 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

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 $5,10,20$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.