12
$\begingroup$

Consider $f(x)=x\sqrt{8}+\frac{1}{x\sqrt{8}}-\sqrt{8}$. The function has 2 real roots, say, $x_1$ and $x_2$. If the $1994$th digit in the decimal expansion of $x_1$ is $6$, what is the $1994$th digit in the decimal expansion of $x_2$?

This is question 1 of the 34th Swedish Olympiad (1994). Either by explicitly finding the roots, or by Vieta's relations, we get that $x_1+x_2=1$. Now, if we write $1$ as $0.999\cdots$, we see that the corresponding digits in the decimal place add up to $9$. That is, the $m$th digit of $x_1$ plus the $m$th digit of $x_2=9$, for $m\geq2$. This means that the $1994$th digit of $x_2$ is $3$.

Is this correct? For some reason, I can't find the answer/solution anywhere.

Also, if we have $a,b\in\mathbb{R}$, and $(a+b)\in\mathbb{Z}$, then what can be said about the decimal expansion of $a$ and $b$? For example, $\pi=3.1415\cdots$, and $1-\pi=-2.1415\cdots$. We can see that the digits in the decimal expansion of $\pi$ and $1-\pi$ are equal (except for the first digit, of course). Can this be generalised?

Edit: I thought it would be helpful to include a screenshot that demonstrates that corresponding digits in the decimal expansions of $x_1$ and $x_2$ add up to $9$: enter image description here

$\endgroup$
5
  • $\begingroup$ Your answer looks correct to me. As to generalization, I think the answer would be that either the digits of $a,b$ are the same, or the digits of $b, d_b$, are $9-d_a$. $\endgroup$ Commented Dec 26, 2022 at 5:25
  • 5
    $\begingroup$ An exception must be made for terminating decimals, e.g. $0.63 + 0.37 = 1$. $\endgroup$ Commented Dec 26, 2022 at 5:33
  • 1
    $\begingroup$ but since the roots are non-terminating here, this exception can be overlooked here. $\endgroup$
    – D S
    Commented Dec 26, 2022 at 12:46
  • 1
    $\begingroup$ Also, note that this is negative for negative $x$ (all terms are negative). Zero clearly isn't a root. Thus, both roots are positive. Combined with the observation that the roots sum to $1$, this implies they're in the range $(0,1)$. $\endgroup$ Commented Dec 26, 2022 at 13:48
  • 2
    $\begingroup$ @RobertIsrael Technically, even if the 1994th digit onwards is $0.\dots600\dots$, the 1994th digit of the other root will be ambiguous, since it could be written either as $0.\dots400\dots$ or $0.\dots399\dots$. So answering with $3$ would be a valid solution regardless. $\endgroup$ Commented Dec 26, 2022 at 13:52

1 Answer 1

0
$\begingroup$

$9- $nth digit of $x_{1}=$ nth digit of $x_{2}$. "An exception must be made for terminating decimals, e.g. 0.63+0.37=1 . – Robert Israel Dec 26, 2022 at 5:33" As this is not a terminating decimal, but an irrational number, with $x_{1}+x_{2}=0.9999999...$ , Robert Israel's statement will not work here. So the other 1994th digit of the other root should be $9-3=6$. For your example of $\pi$ and $1-\pi$, as the numbers of both roots are positive, that case does not work here. (Try $4-\pi$)

$\endgroup$

You must log in to answer this question.