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The problem is as follows:

$\textrm{Find the sum of a+b+c}$

$\overline{abc}=\overline{c000}_{(3)}$

What I have tried is to transform back to base $10$ the number on base $3$ by doing the following:

$\overline{c000}_{(3)}=3^{3}\times c+3^{2}\times 0+3^{1}\times 0 +3^{1}\times 0 + 3^{0}\times 0$

therefore:

$\overline{c000}_{(3)}=27\times c$

Then I came up with the idea that $c$ can only be (judging from the numeric system above) only $1$ or $2$ since a base $3$ number limits to those values. But if I do multiplication between $1$ or $2$ by $27$. It does not produce a number of three digits.

$27\times 1=27$

$27\times 2=54$

However I know that the sum of the numbers which results from multiplication of a number of divisibility by $9$ (in this case $27$) with any other number is equal to $9$.

Therefore let's say if

$27\times 4=108$

$1+0+8=9$

But the above procedure does not seem to be right as $4$ is off the limits from the base in the number.

I'm not sure if the sum of $a+b+c=9$, but if it is, is there any way to prove it? Did I made a mistake?

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    $\begingroup$ Yes, this is a badly formed question. There is no such $a$, $b$, and $c$. I think what was intended was that $c=5$, which gives the number 135, but, as you pointed out, we must have $c<3$, so this is not actually a solution. There is none. Where did this question come from? $\endgroup$ – Matthew Conroy Nov 8 '17 at 5:57
  • $\begingroup$ @Henno Brandsma I restored the tag that you removed. Sources belong [to this site] [1], [Book by Neal Kobitz (number theory)] [2], [An article in Journal of Computer Applications] [3], as all indicate as change of base in numbers are within the realm of cryptography and by the way as it is used in this problem its of right use. [1]: gpgtools.tenderapp.com/kb/how-to/introduction-to-cryptography [2]: books.google.co.nz/… [3]: research.ijcaonline.org/volume118/number14/pxc3903150.pdf $\endgroup$ – Chris Steinbeck Bell Nov 8 '17 at 20:44
  • $\begingroup$ @Matthew Conroy I transcribed this question from an old exam, the alternatives given for the answer (that is the sum of $a+b+c$) were $9,12,15,8,16$ as I mentioned my best guess was that it could be $9$ but not sure if that would apply. Can you explain how did you get to the number $c=5$ as what it was intended?. If that would be the case, wouldn't the base be $6$? therefore $6^{3}\times 5+6^{3}\times 0+6^{2}\times 0+6^{1}\times 0=1080$ making it a four digit number. $\endgroup$ – Chris Steinbeck Bell Nov 8 '17 at 21:31
  • $\begingroup$ @Matthew Conroy What i've found is $\overline{432}_{(10)}=\overline{2000}_{6}$ and $\overline{864}_{10}=\overline{4000}_{6}$ check with the way how it would had been intended provided that the base three is incorrect, pluggin other numbers by increasing the base $7,8,9$ exceed the 3 digit capacity in the base 10 number. Therefore I think the numbers from above could be the answer but their sum is different $4+3+2=9$ and $8+6+4=18$ unless you follow up the sum of the second option as $1+8=9$ but wouldn't it be forcing it to be $9$?, sorry I'm slow with these things, is my rationale correct?. $\endgroup$ – Chris Steinbeck Bell Nov 8 '17 at 21:32
  • $\begingroup$ You do know that last "paper" you quoted is total BS? $\endgroup$ – Henno Brandsma Nov 8 '17 at 21:51
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If the number $n=\overline{abc}=\overline{c000}_\beta$, then $$ c\beta^3 = 100a+10b+c $$ so $$c(\beta^3-1)=100a+10b$$ with $1\le c < \beta, 0 \le a,b \le 9$. Since $100a+10b<1000$, we conclude that $\beta < 10$. Checking all possibilities then yields $432=2\cdot 6^3$ and $864=4\cdot 6^3$ as the only solutions to the problem.

The base $3$ specification yields no solution. The problem is incorrectly written.

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  • $\begingroup$ It is a nice observation but I don't understand what happens with the left side of the equation I mean $c(\beta^{3}-1)$ How does this is used with the other inequalities to conclude $\beta<10$? In other words can you add more steps? this part I'm a bit lost. $\endgroup$ – Chris Steinbeck Bell Nov 9 '17 at 23:23
  • $\begingroup$ $100a+10b\le990$, so $\beta^3 \le 991$ so $\beta<10$. Cheers! $\endgroup$ – Matthew Conroy Nov 9 '17 at 23:35

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