Problem C. 1793. (January 2024)
C. 1793. Function \(\displaystyle f\) defined on the real numbers satisfies the following two conditions (for all \(\displaystyle x \in \mathbb{R}\)):
| \(\displaystyle (1)\) | \(\displaystyle f(x)=f(147-x),\) |
| \(\displaystyle (2)\) | \(\displaystyle f(x+100)=f(46-x).\) |
Find the value of \(\displaystyle f(200)+f(201)+f(202)\), if we also know that \(\displaystyle f(50)+f(51)+f(52)+f(53)=2024\).
Proposed by Katalin Abigél Kozma, Győr
(5 pont)
Deadline expired on February 12, 2024.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Az \(\displaystyle (1)\)-es feltétel miatt \(\displaystyle f(x+100)=f(147-(x+100))=f(47-x)\). Ezt \(\displaystyle (2)\)-vel összevetve \(\displaystyle f(47-x)=f(46-x)\) minden valós \(\displaystyle x\)-re, ami azt jelenti, hogy alkalmasan megválasztva \(\displaystyle x\)-et \(\displaystyle f(50)=f(51)=f(52)=f(53)= \ldots =f(200)=f(201)=f(202)\). Létezik is ilyen függvény, például a konstansfüggvény kielégíti az összes feltételt. Az előzőek alapján \(\displaystyle 4 \cdot f(50)=2024\), amiből \(\displaystyle f(50)=2024/4=506.\) Ekkor \(\displaystyle 506=f(200)=f(201)=f(202)\), azaz a megoldás
\(\displaystyle f(200)+f(201)+f(202)=3 \cdot 506=1518.\)
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Problems in Mathematics of KöMaL, January 2024