10
$\begingroup$

There is a machine, which when you put a number card which has written $x$ on it, it will output a number card with $$x^2+10x+20$$ on it. John has a number card on his hand, then repeated the following procedure ten times:

Put the card on his hand in the machine, and get the output.

John found out that he had the number card $0$ on his hand after repeating the procedures.

What number could have been written on John’s card at first?

There is a clever solution with an 'aha' moment.


Problem by me

$\endgroup$
1
  • 3
    $\begingroup$ Why downvote? There is a beautiful solution. $\endgroup$ Jun 26 '20 at 12:47
9
$\begingroup$

This is probably similar to msh210's idea but if we write

$$p(x) = (x+5)^2 - 5 $$ Then, we notice that $$ p(p(x)) = (p(x) + 5)^2 - 5 =((x+5)^2 - 5 + 5)^2 - 5 = (x+5)^4 - 5 $$ and for $k$ iterations $$p^k(x) = p(p(\ldots p(x)\ldots)) = (x+5)^{2^k} - 5$$ so we just need to solve $$p^{10}(x) = (x+5)^{1024} - 5 = 0 $$ which has solutions $$x = -5 + 5^{1/1024}$$ where the $1024$th root can be any such root. This is what others have found, just expressed in a different way.

$\endgroup$
1
  • $\begingroup$ Yes, this is my solution! $\endgroup$ Jun 26 '20 at 14:13
3
$\begingroup$

First, note that the minimum of $x^2+10x+20$ is equal to $-5$, so equations like $x^2+10x+20=y$ are only solvable for $y \ge -5$

Now solve the general case equation: $x^2+10x+20=y$. The solutions are $-5-\sqrt{y+5}$ and $-5+\sqrt{y+5}$. Now we have to repeat the process 10 times, starting with $y = 0$. If we choose the new $y$ to be the first solution (except for the last iteration), the next equation won't be solvable.

The first solution is $-5+\sqrt{5}$, the second solution is $-5+\root^4\of{5}$, the third solution is $-5+\root^8\of{5}$ and so on. In general, the $n$th solution is $-5+5^{(1/2)^n}$. The solution on the ninth iteration is thus $-5+\root^{512}\of{5}$.

Therefore, the two possible (real) starting numbers are the solutions of $x^2+10x+20 = -5+\root^{512}\of{5}$. These solutions are (using the general case formula above) $-5-\root^{1024}\of{5}$ and $-5+\root^{1024}\of{5}$.

$\endgroup$
2
  • $\begingroup$ Can you compose an elegant proof for the general formula? $\endgroup$ Jun 26 '20 at 12:52
  • $\begingroup$ @CulverKwan The -5+5^(1/2)^n one? It can be seen that to get the next iteration of the formula, when calculating the next solution, we add 5, take the square root and subtract 5 again, so every time after the first iteration, the first $-5$ is unchanged, while the second $5$'s power halves every iteration. $\endgroup$ Jun 26 '20 at 12:54

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.