Question

Dado que $\begin{aligned} &\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} \\ &\quad {}+ \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} \\ &\quad {}+ \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}, \end{aligned}$ halla el mayor entero que es menor que $\frac{N}{100}$ .

Given that $\begin{aligned} &\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} \\ &\quad {}+ \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} \\ &\quad {}+ \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}, \end{aligned}$ find the greatest integer that is less than $\frac{N}{100}.$

Respuesta

Solución:

Multiplica ambos lados por $19!$ . Cada término de la izquierda se convierte en $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ para $k = 2, \ldots, 9$ , mientras que el lado derecho se convierte en $19N$ .

Como $\binom{19}{k} = \binom{19}{19 - k}$ , la primera mitad de la fila binomial suma $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18}$ , así que $\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}$

Por tanto $N = \frac{262124}{19} = 13796$ , así que $\frac{N}{100} = 137.96$ , y el mayor entero menor que esto es $137$ .

Multiply both sides by $19!.$ Each term on the left becomes $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ for $k = 2, \ldots, 9,$ while the right side becomes $19N.$

Since $\binom{19}{k} = \binom{19}{19 - k},$ the first half of the binomial row sums to $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18},$ so $\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}$

Hence $N = \frac{262124}{19} = 13796,$ so $\frac{N}{100} = 137.96,$ and the greatest integer less than this is $137.$

2000 AIME II Problema 7

Solución:

El Problema 7 en otros años