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How can one find the base $r$ in which $1065 = 13 \cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.

$$r^{3}-5r^{2}-13r-7=0$$ I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.

Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance

Jyrki Lahtonen
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Theorem
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    In what base does $3$ times $4$ have a units digit of $5$? There are not that many. – Michael Burr Jul 21 '18 at 12:25
  • @MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong? – Theorem Jul 21 '18 at 12:28
  • Multiplication actually works the same way, it's only when you're changing bases do all the digits matter. – Michael Burr Jul 21 '18 at 12:29
  • @MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3\cdot 4) \text{mod} r=5$? – Theorem Jul 21 '18 at 12:32
  • If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial. – Mark Bennet Jul 21 '18 at 12:40

4 Answers4

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Since $3\cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).

To give some details, if $r$ is the base, then, in base $10$, $13_r=1\cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law, $$ 13_r\cdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12\equiv12\pmod{r}. $$ On the other hand, in base $10$, $$ 1065_r=1\cdot r^3+0\cdot r^2+6r+5\equiv 5\pmod{r}. $$ Therefore, for equality, it must be that $5\equiv 12\pmod{r}.$.

Michael Burr
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    But the base is not $1$, because then it cannot have a $6$ as digit. – wythagoras Jul 21 '18 at 12:31
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    Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2\cdot 2$ – Joffan Jul 21 '18 at 14:11
  • @Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1\cdot 1^3+0\cdot 1^2+6\cdot 1+5=12$. Therefore, the entire statement becomes $4\cdot 9=12$ (in base $10$), which is also silly. – Michael Burr Jul 21 '18 at 14:21
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The base is clearly less than $11$, since $13\cdot54\lt20\cdot54=T80\lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.

*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.

Barry Cipra
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  • I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $\lt 12$). ++1 – Mark Bennet Jul 21 '18 at 16:25
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\begin{align} [1,3]_B \cdot [5,4]_B &= [1,0,6,5]_B \\ \end{align}

In base $B$:

\begin{array}{r} & & 1 & 3 \\ & \times & 5 & 4 \\ -- & --& -- & -- \\ & & 4 & 12 \\ & 5 & 15 & 0 \\ -- & --& -- & -- \\ & 5 & 19 & 12 \\ \end{array}

And, so $[5, 19,12]_B = [1,0,6,5]_B$

If we are to believe that $[1,0,6,5]_B$ is in normalized notation, then $B \ge 7$

It seems then that $12$ must be equal to $[1,5]_B$; which implies that $B = 7$. This is easily verified:

$[5,19,12]_7 = [5,20,5]_7 = [7,6,5]_7 = [1,0,6,5]_7$

Steven Alexis Gregory
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Let $b$ be the base then

$$ (b+3)(5b+4)=b^3+6b+5\Rightarrow 5b^2+15b+4b+12 = b^3+6b+5 $$

hence $b = 7$ then

$$ 5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5 $$

Cesareo
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  • @Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right. – Cesareo Jul 21 '18 at 13:17
  • it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^{3}-5b^{2}-13b-7=0$, for which s/he wants an alternative... – farruhota Jul 21 '18 at 13:49
  • @farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer. – Cesareo Jul 21 '18 at 14:01