Question

Las sucesiones de enteros positivos $1, a_2, a_3, \ldots$ y $1, b_2, b_3, \ldots$ son una sucesión aritmética creciente y una sucesión geométrica creciente, respectivamente. Sea $c_n = a_n + b_n.$ Existe un entero $k$ tal que $c_{k-1} = 100$ y $c_{k+1} = 1000.$ Halla $c_k.$

The sequences of positive integers $1, a_2, a_3, \ldots$ and $1, b_2, b_3, \ldots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n = a_n + b_n.$ There is an integer $k$ such that $c_{k-1} = 100$ and $c_{k+1} = 1000.$ Find $c_k.$

Respuesta

Solución:

Escribe $a_n = 1 + (n-1)d$ y $b_n = r^{n-1}$ con enteros $d \ge 1$ y $r \ge 2.$ Como $c_1 = 2 \lt 100,$ tenemos $k \ge 3,$ y las dos condiciones dicen $(k-2)d + r^{k-2} = 99,$ $kd + r^k = 999.$

Restando, $2d + r^{k-3}(r-1)r(r+1) = 900.$ El producto de tres enteros consecutivos es divisible por $3,$ así que $3 \mid 2d,$ por tanto $3 \mid d.$ Entonces $(k-2)d + r^{k-2} = 99$ obliga a $3 \mid r^{k-2},$ así que $3 \mid r.$ Las cotas $r^{k-2} \le 98$ y $r^k \le 998$ dejan solo $(r, k) = (3, 3),$ $(3, 4),$ $(3, 5),$ $(3, 6),$ $(6, 3),$ $(9, 3).$

Probando cada uno contra $(k-2)d = 99 - r^{k-2}$ y $kd = 999 - r^k,$ solo $(r, k) = (9, 3)$ da un valor consistente, $d = 90.$ Entonces $c_3 = 1 + 2 \cdot 90 + 9^2 = 262.$

Write $a_n = 1 + (n-1)d$ and $b_n = r^{n-1}$ with integers $d \ge 1$ and $r \ge 2.$ Since $c_1 = 2 \lt 100,$ we have $k \ge 3,$ and the two conditions read $(k-2)d + r^{k-2} = 99,$ $kd + r^k = 999.$

Subtracting, $2d + r^{k-3}(r-1)r(r+1) = 900.$ The product of three consecutive integers is divisible by $3,$ so $3 \mid 2d,$ hence $3 \mid d.$ Then $(k-2)d + r^{k-2} = 99$ forces $3 \mid r^{k-2},$ so $3 \mid r.$ The bounds $r^{k-2} \le 98$ and $r^k \le 998$ leave only $(r, k) = (3, 3),$ $(3, 4),$ $(3, 5),$ $(3, 6),$ $(6, 3),$ $(9, 3).$

Testing each against $(k-2)d = 99 - r^{k-2}$ and $kd = 999 - r^k,$ only $(r, k) = (9, 3)$ gives a consistent value, $d = 90.$ Then $c_3 = 1 + 2 \cdot 90 + 9^2 = 262.$

2016 AIME II Problema 9

Solución:

El Problema 9 en otros años