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What are the bases $b_{1},b_{2}$ in $443642_{b_1}=53818_{b_2}$

What would be a general approach to this type of problems that would reduce the tedious manual calculations?

(This problem is important, because we could have two unknown sources that have different initial conversion and we must make an educated guess about the bases.)

For two bases, what is the range of numbers/bases that is undecidable, meaning you may miss the correct base? What are the restrictions needed?

Is there an example of two numbers with different base that are not within this range that may have two or more interpretations, apart from trivial ones?

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    Well, from the numbers themselves, we can tell the first must be larger than base $6$ and the second larger than base $8$ and you can likely assume max of base $16$. This certainly reduces the range. in this case, first number is in base $7$, second is base $11$ with the equivalent number equal to $78185$. – Moo Jan 02 '16 at 21:31
  • If $b_1$ is odd, the number is odd, so $b_2$ is odd, and vice versa. If $b_1$ is even, the number is $\equiv2\pmod8$, so $b_2\equiv2\pmod8$ – Empy2 Jan 02 '16 at 21:38

4 Answers4

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An obvious observation here is that $b_1 \ge 7$ and $b_2 \ge 9$ (because the first number contains the digit 6, and the second one contains the digit 8).

One possible solution is $b_1 = 7$ and $b_2 = 11$. I wrote a small program to find this.

As far as I know there's no general method for solving such problems (manually, without a computer). You just have to make some observations (most probably number theoretic ones), and come up with some ingenious findings which lead you to the solution. There may be some general method though. I am just not aware of one.

peter.petrov
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Actually your problem can be in general very hard to solve because, in your particular case, one has the following diophantine equation of degree 5 with two unknowns:

$$4x^5+4x^4+3x^3+6x^2+4x+2=5y^4+3y^3+8y^2+y+8$$

whose solution is $$(x,y)=(b_1,b_2)=(7,11)$$ (in both sides you have the number $78 185$).

Piquito
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  • @AlexPeter: It is not difficult to prove that the solution is unique because equality between two polynomials occurs when one being smaller than the other to some extent and after of this point (just one when your equation is produced) remains higher (first less than, second equal to and third largest than). Sorry for bad English – Piquito Jan 02 '16 at 23:17
  • Be careful because there is an intersection point but not always its coordinates are integer. – Piquito Jan 03 '16 at 00:10
  • I advise you to read about "local" and "global" (if you do not know this, of course. Let allow myself to guess that no much). – Piquito Jan 03 '16 at 00:30
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You can write, $443642_{b_1} = 4*b_1^5+4*b_1^4+3*b_1^3+6*b_1^2+4*b_1+2$ and $53818_{b_2} = 5*b_2^4+3*b_2^3+8*b_2^2+b_2+8$. Now set the two expressions equal. You may need another restriction on $b_1$ and $b_2$.

Sanwar
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If you know that a there is an integer solution, why not just use the rational zeros theorem to find the zeros of both polynomials on both sides of the equation? Your list would be fairly short and could quickly find an answer

j215c228
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