Problem C. 1837. (December 2024)
C. 1837. The ancient Greeks were already familiar with the fact that \(\displaystyle \pi\approx 3.1416\) can be approximated with the fraction \(\displaystyle 22/7 \approx 3.1429\). How many pairs of positive integers \(\displaystyle (a;b)\) satisfy the following properties: \(\displaystyle 1<b<100\) and the decimal form of \(\displaystyle \frac{a}{b}\) starts as \(\displaystyle 3.14\)?
Proposed by: Katalin Abigél Kozma, Győr
(5 pont)
Deadline expired on January 10, 2025.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Azokat az \(\displaystyle 0<a\) és \(\displaystyle 0<b<100\) egész számokat keressük, amelyekre
\(\displaystyle \frac{314}{100} \le \frac{a}{b} < \frac{315}{100}.\)
Kivonva \(\displaystyle 3\)-at:
\(\displaystyle \frac{14}{100} \le \frac{a-3b}{b} < \frac{15}{100}.\)
Legyen \(\displaystyle c=a-3b\) pozitív egész és innentől külön-külön vizsgáljuk az
\(\displaystyle (1)~~\frac{14}{100} \le \frac{c}{b} \quad \text{ és a } \quad (2)~~\frac{c}{b} < \frac{15}{100} \text{ egyenlőtlenséget}.\)
\(\displaystyle (2)\)-ből következően: \(\displaystyle c<\frac{15b}{100}<15\), azaz \(\displaystyle c \in\{1;2;3;4;5;6;7;8;9;10;11;12;13;14 \}.\) Vesszük \(\displaystyle (2)\) mindkét oldalának reciprokát, majd kivonunk \(\displaystyle 7\)-et:
\(\displaystyle \frac{b}{c}>\frac{100}{15} \Rightarrow \frac{b-7c}{c}>\frac{-5}{15}=-\frac{1}{3}.\)
Teljesen hasonlóan \(\displaystyle (1)\)-ből kapjuk, hogy
\(\displaystyle \frac{b}{c} \le \frac{100}{14} \Rightarrow \frac{b-7c}{c} \le \frac{2}{14}=\frac{1}{7}.\)
Mivel \(\displaystyle b\) és \(\displaystyle c\) egész, ezért \(\displaystyle d=b-7c\) is az. Ekkor
\(\displaystyle \frac{-1}{3}<\frac{d}{c} \le \frac{1}{7},\)
\(\displaystyle -\frac{c}{3}< d \le \frac{c}{7}.\)
Tehát ahogy \(\displaystyle c\) végigfut \(\displaystyle 1\)-től \(\displaystyle 14\)-ig, meg kell számolnunk, hogy hány egész szám van a \(\displaystyle \displaystyle{\Big]-\frac{c}{3}; \frac{c}{7}\Big]}\) intervallumban. Ezt táblázat segítségével tesszük meg:
| begincenter \(\displaystyle c\) értéke | \(\displaystyle ]-\frac{c}{3}; \frac{c}{7}]\) | az intervallum egész elemei | \(\displaystyle d\) lehetséges értékeinek száma |
| \(\displaystyle 1\) | \(\displaystyle ]-\frac{1}{3}; \frac{1}{7}] \) | \(\displaystyle 0\) | \(\displaystyle 1\) |
| \(\displaystyle 2\) | \(\displaystyle ]-\frac{2}{3}; \frac{2}{7}]\) | \(\displaystyle 0\) | \(\displaystyle 1 \) |
| \(\displaystyle 3\) | \(\displaystyle ]-1; \frac{3}{7}] \) | \(\displaystyle 0 \) | \(\displaystyle 1 \) |
| \(\displaystyle 4\) | \(\displaystyle ]-\frac{4}{3}; \frac{4}{7}] \) | \(\displaystyle -1; 0 \) | \(\displaystyle 2 \) |
| \(\displaystyle 5\) | \(\displaystyle ]-\frac{5}{3}; \frac{5}{7}] \) | \(\displaystyle -1; 0\) | \(\displaystyle 2\) |
| \(\displaystyle 6\) | \(\displaystyle ]-2; \frac{6}{7}]\) | \(\displaystyle -1; 0\) | \(\displaystyle 2 \) |
| \(\displaystyle 7\) | \(\displaystyle ]-\frac{7}{3}; 1]\) | \(\displaystyle -2;-1; 0;1 \) | \(\displaystyle 4 \) |
| \(\displaystyle 8\) | \(\displaystyle ]-\frac{8}{3}; \frac{8}{7}] \) | \(\displaystyle -2;-1; 0;1 \) | \(\displaystyle 4 \) |
| \(\displaystyle 9\) | \(\displaystyle ]-3; \frac{9}{7}]\) | \(\displaystyle -2;-1; 0;1 \) | \(\displaystyle 4\) |
| \(\displaystyle 10\) | \(\displaystyle ]-\frac{10}{3}; \frac{10}{7}] \) | \(\displaystyle -3;-2;-1; 0;1 \) | \(\displaystyle 5 \) |
| \(\displaystyle 11\) | \(\displaystyle ]-\frac{11}{3}; \frac{11}{7}] \) | \(\displaystyle -3;-2;-1; 0;1 \) | \(\displaystyle 5 \) |
| \(\displaystyle 12\) | \(\displaystyle ]-4; \frac{12}{7}]\) | \(\displaystyle -3;-2;-1; 0;1 \) | \(\displaystyle 5\) |
| \(\displaystyle 13\) | \(\displaystyle ]-\frac{13}{3}; \frac{13}{7}] \) | \(\displaystyle -4;-3;-2;-1; 0;1 \) | \(\displaystyle 6 \) |
| \(\displaystyle 14 \) | \(\displaystyle ]-\frac{14}{3}; 2] \) | \(\displaystyle -4;-3;-2;-1; 0;1;2 \) | \(\displaystyle 7\) |
A feladat feltétele, hogy \(\displaystyle b<100\), ezért \(\displaystyle c=14\) esetén (lásd utolsó sor) \(\displaystyle d=b-7c=b-7 \cdot 14=b-98<100-98=2\), azaz \(\displaystyle d<2\). Így \(\displaystyle c=14\)-re a \(\displaystyle d=2\) nem jó, ezért csak \(\displaystyle 7-1=6\) megoldás van.
Összesen: \(\displaystyle 3 \cdot (1+2+4+5)+2 \cdot 6=48\) megfelelő \(\displaystyle (a;b)\) számpár van.
Megjegyzés. A Pick-tétel alkalmazásával is megoldható a feladat.
Statistics:
42 students sent a solution. 5 points: Farkas András, Iván Máté Domonkos, Kókai Ákos, Pink István. 4 points: Fercsák Flórián. 3 points: 9 students. 2 points: 6 students. 1 point: 10 students. 0 point: 8 students.
Problems in Mathematics of KöMaL, December 2024