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Let An and Bn be two sequences such that An = 11 + 8(n – 1) and Bn = 13 + 9(n – 1) for all natural numbers n ≤ 500. Then, find the largest integer common to both these sequences.

Question

Let An and Bn be two sequences such that An = 11 + 8(n – 1) and Bn = 13 + 9(n – 1) for all natural numbers n ≤ 500. Then, find the largest integer common to both these sequences.

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Solution

Para encontrar el mayor entero común a ambas secuencias AnA_n y BnB_n, primero expresamos las secuencias en términos de nn:

An=11+8(n1) A_n = 11 + 8(n - 1) Bn=13+9(n1) B_n = 13 + 9(n - 1)

Simplificamos las expresiones:

An=11+8n8=8n+3 A_n = 11 + 8n - 8 = 8n + 3 Bn=13+9n9=9n+4 B_n = 13 + 9n - 9 = 9n + 4

Buscamos el mayor entero común a ambas secuencias, es decir, buscamos nn y mm tales que:

8n+3=9m+4 8n + 3 = 9m + 4

Reorganizamos la ecuación:

8n9m=1 8n - 9m = 1

Esta es una ecuación diofántica lineal. Para resolverla, buscamos soluciones enteras nn y mm. Usamos el método de búsqueda de soluciones particulares y luego la forma general de la solución.

Primero, encontramos una solución particular. Probamos con valores pequeños de nn y mm:

Para n=1n = 1:

8(1)9m=1 8(1) - 9m = 1 89m=1 8 - 9m = 1 9m=7 9m = 7

No es una solución entera. Probamos con n=2n = 2:

8(2)9m=1 8(2) - 9m = 1 169m=1 16 - 9m = 1 9m=15 9m = 15

No es una solución entera. Probamos con n=3n = 3:

8(3)9m=1 8(3) - 9m = 1 249m=1 24 - 9m = 1 9m=23 9m = 23

No es una solución entera. Probamos con n=4n = 4:

8(4)9m=1 8(4) - 9m = 1 329m=1 32 - 9m = 1 9m=31 9m = 31

No es una solución entera. Probamos con n=5n = 5:

8(5)9m=1 8(5) - 9m = 1 409m=1 40 - 9m = 1 9m=39 9m = 39 m=39/9=4.333 m = 39 / 9 = 4.333

No es una solución entera. Probamos con n=7n = 7:

8(7)9m=1 8(7) - 9m = 1 569m=1 56 - 9m = 1 9m=55 9m = 55

No es una solución entera. Probamos con n=8n = 8:

8(8)9m=1 8(8) - 9m = 1 649m=1 64 - 9m = 1 9m=63 9m = 63 m=7 m = 7

Encontramos una solución particular: n=8n = 8 y m=7m = 7.

La solución general de la ecuación diofántica es:

n=8+9k n = 8 + 9k m=7+8k m = 7 + 8k

donde kk es un entero.

Para encontrar el mayor entero común a ambas secuencias dentro del límite n500n \leq 500, buscamos el mayor nn tal que:

8+9k500 8 + 9k \leq 500 9k492 9k \leq 492 k492954.666 k \leq \frac{492}{9} \approx 54.666

El mayor valor entero de kk es 54. Sustituimos k=54k = 54 en la solución general:

n=8+9(54)=8+486=494 n = 8 + 9(54) = 8 + 486 = 494

Entonces, el mayor entero común a ambas secuencias es:

A494=8(494)+3=3952+3=3955 A_{494} = 8(494) + 3 = 3952 + 3 = 3955

Por lo tanto, el mayor entero común a ambas secuencias es 39553955.

This problem has been solved

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