Question

Diga que un entero positivo $n$ es $k$ -bonito si $n$ tiene exactamente $k$ divisores positivos y $n$ es divisible entre $k.$ Por ejemplo, $18$ es $6$ -bonito. Sea $S$ la suma de los enteros positivos menores que $2019$ que son $20$ -bonitos. Halle $\frac{S}{20}.$

Call a positive integer $n$ $k$ -pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k.$ For example, $18$ is $6$ -pretty. Let $S$ be the sum of the positive integers less than $2019$ that are $20$ -pretty. Find $\frac{S}{20}.$

Respuesta

Solución:

Necesitamos $20 \mid n$ y $\tau(n) = 20.$ Escriba $n = 2^a 5^b m$ con $\gcd(m, 10) = 1;$ entonces $a \ge 2,$ $b \ge 1,$ y $(a + 1)(b + 1)\tau(m) = 20$ con $a + 1 \ge 3$ y $b + 1 \ge 2.$ El factor $a + 1$ debe ser un divisor de $20$ que sea al menos $3:$ uno de $4,$ $5,$ $10,$ $20.$

Si $a + 1 = 4,$ entonces $(b + 1)\tau(m) = 5$ fuerza $b = 4,$ $m = 1,$ así que $n = 2^3 5^4 = 5000,$ demasiado grande. Si $a + 1 = 10,$ entonces $b = 1,$ $m = 1,$ y $n = 2^9 \cdot 5 = 2560,$ demasiado grande; $a + 1 = 20$ es aún mayor. Si $a + 1 = 5,$ entonces $(b + 1)\tau(m) = 4,$ dando o bien $b = 3,$ $m = 1,$ así que $n = 2^4 5^3 = 2000 \lt 2019,$ o bien $b = 1$ y $\tau(m) = 2,$ de modo que $m = p$ es un primo distinto de $2$ y $5$ y $n = 80p \lt 2019,$ es decir $p \le 25:$ $p \in \{3, 7, 11, 13, 17, 19, 23\}.$

Por lo tanto $\begin{aligned} S &= 2000 \\ &\quad {}+ 80\tiny(3 + 7 + 11 + 13 + 17 + 19 + 23) \\ &= 2000 + 80 \cdot 93 \\ &= 9440, \end{aligned}$ así que $\frac{S}{20} = 472.$

We need $20 \mid n$ and $\tau(n) = 20.$ Write $n = 2^a 5^b m$ with $\gcd(m, 10) = 1;$ then $a \ge 2,$ $b \ge 1,$ and $(a + 1)(b + 1)\tau(m) = 20$ with $a + 1 \ge 3$ and $b + 1 \ge 2.$ The factor $a + 1$ must be a divisor of $20$ that is at least $3:$ one of $4,$ $5,$ $10,$ $20.$

If $a + 1 = 4,$ then $(b + 1)\tau(m) = 5$ forces $b = 4,$ $m = 1,$ so $n = 2^3 5^4 = 5000,$ too large. If $a + 1 = 10,$ then $b = 1,$ $m = 1,$ and $n = 2^9 \cdot 5 = 2560,$ too large; $a + 1 = 20$ is larger still. If $a + 1 = 5,$ then $(b + 1)\tau(m) = 4,$ giving either $b = 3,$ $m = 1,$ so $n = 2^4 5^3 = 2000 \lt 2019,$ or $b = 1$ and $\tau(m) = 2,$ so $m = p$ is a prime other than $2$ and $5$ and $n = 80p \lt 2019,$ i.e. $p \le 25:$ $p \in \{3, 7, 11, 13, 17, 19, 23\}.$

Therefore $\begin{aligned} S &= 2000 \\ &\quad {}+ 80\tiny(3 + 7 + 11 + 13 + 17 + 19 + 23) \\ &= 2000 + 80 \cdot 93 \\ &= 9440, \end{aligned}$ and $\frac{S}{20} = 472.$

2019 AIME II Problema 9

Solución:

El Problema 9 en otros años