Difference between revisions of "2023 WSMO Team Round Problems/Problem 8"
(Created page with "==Problem== Let <math>f(x)=\sqrt{x-\sqrt{x-\sqrt{x-\ldots}}}.</math> Find the modulo 1000 on the minimum integer <math>a</math> such <math>f(f(f(f(f(a)))))</math> is a positi...") |
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==Solution== | ==Solution== | ||
| + | Note that | ||
| + | <cmath>\begin{align*} | ||
| + | f(x) &= \sqrt{x-f(x)}\implies\\ | ||
| + | [f(x)]^2 &= x-f(x)\implies\\ | ||
| + | x &= [f(x)]^2+f(x). | ||
| + | \end{align*}</cmath> | ||
| + | Since <math>[f(x)]^2+f(x)</math> is a strictly increasing function, the minimum value of <math>a</math> occurs when | ||
| + | <cmath>\begin{align*} | ||
| + | f(f(f(f(f(a))))) &= 1\implies\\ | ||
| + | f(f(f(f(a)))) &= 1^2+1 = 2\implies\\ | ||
| + | f(f(f(a))) &= 2^2+2 = 6\implies\\ | ||
| + | f(f(a)) &= 6^2+6 = 42\implies\\ | ||
| + | f(a) &= 42^2+42 = 1806\implies\\ | ||
| + | a &= 1806^2+1806\implies\\ | ||
| + | a&\equiv \boxed{442}\pmod{1000}. | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | ~pinkpig | ||
Latest revision as of 14:28, 13 September 2025
Problem
Let 
Find the modulo 1000 on the minimum integer 
such 
is a positive integer.
Solution
Note that
![\begin{align*} f(x) &= \sqrt{x-f(x)}\implies\\ [f(x)]^2 &= x-f(x)\implies\\ x &= [f(x)]^2+f(x). \end{align*}](http://latex.artofproblemsolving.com/d/c/b/dcb01b16aea6e713b02d099e7484e61e87861675.png)
Since ![$[f(x)]^2+f(x)$](http://latex.artofproblemsolving.com/1/a/6/1a6b8bd61ec2bbf4bab8360e8a2c26e7af2178f5.png)
is a strictly increasing function, the minimum value of occurs when
~pinkpig
