Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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SOLUCIONARIO DE ARITMETICA DE BALDOR1º el m.c.d. de 221 y 204:-5. 153, 357 y 1871º el m.c.d. de 187 y 153:-9. 168, 252, 280 y 9171º m.c.d. de 280 y 252:-12. 31 740, 47 610, 95 220 y 126 960-6. 236, 590 y 1 2391º el m.c.d. de 590 y 236:Luego el m.c.d. de 236, 590 y 1 239 es:-10. 770, 990, 1 265 y 3 388-13. 45 150, 51 600, 78 045 y 108 489 es:-7. 465, 651 y 6821º el m.c.d. de 682 y 651:-8. 136, 204, 221 y 272-11. 1 240, 1 736, 2 852 y 3 131-14. 63 860, 66 340, 134 385 y 206 305LEONARDO F. APALA TITO 112
SOLUCIONARIO DE ARITMETICA DE BALDOR1º m.c.d. de 66 340 y 63 860:EJERCICIO 88-1. Citar tres divisores comunes de losnúmeros 12, 24 y 48.-17. 3 240, 5 400, 5 490, 6 300 y 7 110Luego los tres divisores comunes son 2, 4y 3-15. 500, 560, 725, 4 350 y 8 200-2. Decir, por inspección, cual es el m.c.d.de 7 y 11; de 8, 9 y10; de 25, 27 y 36.-18. 486, 729, 891, 1 944 y 4 5271º m.c.d. de 891 y 729:-3. Si 24 es el divisor y 8 el residuo de unadivisión inexacta, ¿Sera 4 factor común deldividendo y el divisor? ¿Por qué?-16. 432, 648, 756, 702 y 621Luego: D = 24 + 8 = 32 y C = 1Donde 24 y D = 32Entonces 4 es un factor común, porque 24y 32 son múltiplos de 4.LEONARDO F. APALA TITO 113
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Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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