ed. sm - 3º eso - matemáticas - unidad 1 - números naturales. divisibilidad
DESCRIPTION
Ed.SM - 3ºESO - Matemáticas - Resolución problemasTRANSCRIPT
Pg
3 ESO - MATEMTICAS - UNIDAD 1 (Nmeros naturales. Divisibilidad)
Pg.8 1a) 6 3 2 7 = 6 UM + 3 C + 2 D + 7 U
b) 5 1 0 4 = 5 UM + 1 C + 0 D + 4 U
c) 3 7 9 5 = 3 UM + 7 C + 9 D + 5 U
d) 4 3 8 0 = 4 UM + 3 C + 8 D + 0 U.
Pg.8 2a) 3 7 2
b) 48.516
c) 37.370.
Pg.8 3a) 6 C + 7 D + 8 U = seiscientos setenta y ocho.
b) 4 UM + 8 C + 3 D + 5 U = cuatro mil ochocientos treinta y cinco.
c) 2 DM + 3 UM + 4 C + 0 D + 5 U = veintitrs mil cuatrocientos cincod) 9CM + 8DM + 6UM + 5C + 2D + 3U = novecientos ochenta y seis mil quinientos veintitrs.e) 1CM + 0DM + 0UM + 8C + 0D + 0U = cien mil ochocientos.f) 3MM + 0CM + 0DM + 0UM + 9UM + 0DM + 9U = tres millones novecientos nueve.Pg.8 4a) MMXXXIV
b) MCMLXXXIV
c) CCCXV
d) MMCXVII
e) MMXII
f) MCMXC.
Pg.8 5a) Dos mil unob) Quinientos cuarenta y cinco c) Ciento veintitrs.
Pg.8 6LibrosVivos.NET.
Pg.9 7 S, porque tienen el mismo cdigo de entidad = 4735.
S, porque tienen el mismo cdigo de sucursal = 1003.
Pg.9 8a) Lugo
b) Zaragoza
c) Crdoba.
Pg.9 9En mi gua telefnica, slo figuran prefijos de Asturias. En www.telefonica se ven.
a) Mlaga
b) La Rioja
c) Crdoba.
Pg.10 10a)
EMBED Equation.3
b)
EMBED Equation.3
c)
EMBED Equation.3 .
Pg.10 11a) 13 + 7 = 7 + 13 = 20 ( conmutativa.
b) ( 8 3 ) 5 = 8 (3 5) = 120 ( asociativa.
c) 12(3+5) = 36 + 60 = 96 ( distributiva.
Pg.10 12a) 4 + 4 + 8 = 16
b) 22 ( 11 = 11.
Pg.10 13a) 13 ( 7 = 6 ( 18 12 = 6 ( Propiedad de la resta.
b) 2( 6 + 9 ) = 12 + 18 ( distributiva.
Pg.10 14LibrosVivos.NET.
Pg.11 15Con ( ( ( (
Mltiplos (
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3 etc.
Pg.11 16S, es mltiplo de 4 ( 556 / 4 = 139.
Pg.11 17Con ( ( (
EMBED Equation.3 .
Pg.11 18Con ( ( ( .
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3 etc.
Pg.11 19S, son divisores ( .
Pg.11 20
( Son divisores 7 y 13 ( Opciones b) y d).
Pg.11 21a) ( 1, 2, 3, 4, 6, 8, 12, 24.
b) ( 1, 3, 9, 27.
c) ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
d) ( 1, 5, 25.
e) ( 1, 7.
f) ( 1, 2, 4, 7, 8, 14, 28, 56.
Pg.11 22a) ( ( Tiene 3 divisores 1, 2, 4.
b) ( ( Tiene 3 divisores 1, 5, 25.
c) ( ( Tiene 4 divisores.
d) ( ( Tiene 3 divisores 1, 7, 49.
e) ( ( Tiene 6 divisores.
f) ( ( Tiene 12 divisores.
Pg.11 23
( ( Tiene 8 divisores 1, 2, 3, 6, 9, 18, 27, 54.
Se pueden hacer 8 rectngulos posibles.
1 ( 54, 2 ( 27, 3 ( 18, 6 ( 9, 9 ( 6, 18 + 3, 27 ( 2 y 54 ( 1.
Si slo reconsideran los de distintas dimensiones, slo hay 4 rectngulos.
Pg.11 24LibrosVivos.NET
Pg.13 25
No es divisible entre 2 No acaba en cifra par
S es divisible entre 3 La suma de sus cifras es mltiplo de 3
S es divisible entre 5 Termina en 5
No es divisible entre 6 No lo es entre 2
No es divisible entre 10 No termina en 0
S es divisible entre 25 Sus dos ltimas cifras (75) forman mltiplo de 25
No es divisible entre 100 No termina en 00.
Pg.13 26Divisible entre23459101125100
375
nosno s nononosno
990
ssnossssnono
1.848
sssnononosnono
12.300
ssssnosnoss
14.240
snos snosnonono.
Pg.13 27El nmero N (10 k para ser divisible entre 2 y 5, y k no puede acabar en 0.
10000 < N < 99999 ( 10000 < 10k < 99999 ( 1000 < k < 9999,9.
El nmero k puede tomar valores enteros no terminados en 0. Es decir,
1001, 1002, 1003, 1004, 1005, 1006, 1007, 1008, 1009, 1011, 1012, etc.
Los dos primeros nmeros sern 10010 y 1020.
Pg.13 28a) El nmero ser N ( 33 k y k no puede ser mltiplo de 3.
( (
k = 304, 305, 307, 308, 310, 311, 313, 314, 316, 317, 319, etc.
Los dos primeros nmeros sern
EMBED Equation.3 y
EMBED Equation.3 .
b) El nmero ser N(99k. En cualquier caso, ser divisible entre 3.
( (
k = 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, etc.
Los dos ms bajos sern
EMBED Equation.3 y
EMBED Equation.3 .
Pg.13 29Las dos ltimas cifras deben ser un mltiplo de 4.
Los mltiplos de dos cifras, empezando por 8 son 80, 84 y 88 ( A=0, 4, 8.
Pg.13 30Divisibilidad entre 3 ( Divisibilidad entre 11 (
(
( El nico valor comn es .
Pg.13 31Divisibilidad entre 3 (
No divisibilidad entre 9 (
Las soluciones son los nmeros , y .
Pg.13 32Mltiplo de 3 y de 5 es mltiplo de 15. Luego el nmero ser
( (
Los nmeros solucin son cada uno de estos ltimos, multiplicado por 15:
, , , , .. , y .
Pg.13 33a) 3 1 5 0, pues 50 es el nico divisible entre 25 y no entre 100.
b) 4 5 X 4, con , luego puede ser .
c) ( .
Pg.13 34No se especifica el mximo nmero que puede tener el instituto. Hay infinitas soluciones.
El nmero es mltiplo de 3, de 4 y de 5. Luego lo es de 3(4( 5 = 60.
( ( .
Todo entero igual o superior a 4, multiplicado por 60, da un nmero solucin:
, , , , , , ..
Pg.14 35De entrada se quitan los pares, divisibles por 2, y queda la serie siguiente
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, .
A partir del 3, se van tachando los que estn sucesivamente cada tres lugares.
A partir del 5, anlogamente, cada cinco lugares (contando sobre tachados).
A partir del 7, de igual forma, cada siete lugares.
Los restantes son primos 1, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ..
Pg.14 36La regla es ( estudiar divisibilidad entre 1 < primos ( 5.
Primos a considerar ( 3 y 5.
31 no es divisible entre 3, porque
31 no es divisible entre 5, porque no termina en 0 ni en 5 ( 31 es primo.
Pg.14 37a) No, divisible entre 2, acaba e cifra par.
b) S, no divisible entre 2 ni 5 (acaba en 1), ni 3 (suma cifras = 2), ni 7
ltimo a ensayar, pues , y el siguiente primo es 11.
c) No, es divisible entre 3 (suma cifras = 12).
d) No, es divisible entre 7 (cuadrado perfecto).
e) S, no es divisible entre 2 ni 5 (acaba en 1), ni 3 (suma cifras 7), ni 7.f) No, es divisible por 3 (suma de cifras 9). Pg.14 38 Nmero primo es el que slo tiene por divisores la unidad y l mismo.
El nico par, que es primo, es el nmero 2.
Cualquier nmero par, distinto del 2, no puede ser primo, al ser divisible por 2.
Pg.14 39La serie de nmeros a estudiar, para determinar los primos, se concreta en los 25:
501, 503, 505, 507, 509, 511, 513, 515, 517, 519, 521, 523, 535, 527, 529, 531, 533,
535, 537, 539, 5411, 543, 545, 547 y 549. .
Se suprimen los 5 terminados en 5 (divisibles entre 5) ( Quedan 20
Se suprimen los 8 con suma cifras = (divisibles entre 3) ( Quedan 12
Se suprimen los 2 divisibles entre 7 (511 y 539) ( Quedan 10
Se suprime el divisible entre 11 (517) ( Quedan 9
Se suprime el divisible entre 13 (533) ( Quedan 8
Se suprime el divisible entre 17 (527) ( Quedan 7
Finalmente, se suprime el divisible entre 23 (529) ( Quedan los 6 primos.
Los nmeros primos sern 503, 509, 521, 523, 541 y 547.
Pg.15 40
a) 2 1 0 | 2
b) 3 9 6 | 2
c) 8 0 | 2
1 0 5 | 3
1 9 8 | 2 4 0| 2
3 5 | 5 9 9 | 3 2 0| 2 7 | 7 3 3 | 3
10 | 2 1 | 1 1 | 11
5 | 5
1 | 1 |
Pg.15 41
3 ( 13 ( 23 ( 33 ( 43 ( 53 (
63 ( 73 ( 83 ( 93 ( .
Pg.15 42a) 108| 2 b) 99| 3 c) 42| 2 d) 37| 37e) 100| 2f) 840| 2
54| 2 33| 3 21| 3 1| 50| 2 420| 2
27| 3 11| 11 7| 7
25| 5 210| 2
9| 3 1| 1|
5| 5 105| 3
3| 3
1| 35| 5
1|
7| 7
1|
.
g) 120| 2 h) 2100| 2 i) 2294| 2
60| 2
1050| 2
1147| 31
30| 2
525| 3
37| 37
15| 3
175| 5
1|
5| 5
35| 5
1|
7| 7
1|
.
Pg.15 43 a) 360| 2b) 300| 2
c) 520| 2 d) 750| 2
180| 2 150| 2
260| 2
375| 3
90| 2 75| 3
130| 2
125| 5
45| 3 25| 5
65| 5
25| 5
15| 3 5| 5 13| 13
5| 5
5| 5 1|
1| 1 |
1|
360 = 23325
300 = 22352
520 = 23513750 = 2353.
Pg.15 44LibrosVivos.NET
Pg.16 45
18 =
36 =
54 =
mcd ( factores comunes con los menores exponentes ( 21 ( 32 ( 18.
Pg.16 46a) , (
b) (
c) , (
d) , , (
e) ,, (
f) , , ( .
Pg.16 47a) , , ( =
b) , , ( =
c) , , ( =
d) , , , ( =
e) , , , , ( =
f) , , , , ( = .
Pg.16 48
( = .
a) Debe tener
b) N parcelas = .
Pg.16 49Se busca el mnimo de mesas.
( =
Este es el nmero de comensales por mesa.
Para chicos =
Para chicas = ( = .
Pg.16 50
( =
El nmero de napolitanas, de cada bandeja, es 24.
Se obtienen as
EMBED Equation.3 y =.
Pg.17 51mcm ( factores comunes y no comunes con los mayores exponentes.
( =
Pg.17 52a) ( =
b) ( =
c) ( =
d) ( =
e) ( =
f) ( =
Pg.17 53a) ( =
b) ( =
c) ( =
d) ( = .
Pg.17 54
( ( .
Pg.17 55
EMBED Equation.3 .
Pg.17 56
( =
1 coincidencia = 16:00 ( 2 coincidencia = 16:00 + 12 = 28:00 = 04:00 h.
Pg.17 57
( =
Colocar
EMBED Equation.3 en la direccin de la dimensin 8 cm
en la direccin de la dimensin 15 cm.
Pg.18 58a) ( =
b) ( =
c) ( =
d) ( =
Pg.18 59a) ( =
b) ( =
c) ( =
d) ( = .
Pg.18 60LibrosVivos.NET.
Pg.18 61
( a = 23 3 5 b = 22 3 52 7
Pg.18 62
( =
a) Vuelvan a coincidir al cabo de 7560 (s) = 126 (min) = 2 h 6 min.
b) Vueltas
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3 .
Pg.20 63
Nmero M
C
D
U
7816 7
8
1
6
69513
69
5
1
3
2754027
5
4
0
2318
2
3
1
8
1358
13
5
8
107890107
8
9
0.
Pg.20 64a) En 345, 34D. En 2621, 262D. En 94013, 9401D.
b) En 345, 15U. En 2621, 11U. En 94013, 13U.
Pg.20 65a) 9.502b) 8.000.413
c) 1.566d) 70.070.
Pg.20 66a) Veinte mil doce.
b) Doscientos treinta y cuatro millones doscientos treinta y cuatro mil.
c) Treinta y tres mil ochocientos cuarenta.
d) Trescientos cinco millones trescientos cinco mil trescientos.
Pg.20 67a) Catorce
d) Trescientos tresb) Quinientos cincuenta y seis
e) Quinientos nuevec) Mil novecientos sesenta y tresf) Doscientos cuarenta y siete.
Pg.20 68a) LXXXIII
c) CCCXVI
e) CMXCIX
b) CCLXXVIII
d) DCCXLV
f) MDCXXXVII.
Pg.20 69a) Tres (3) de diciembre (12)
b) En el ao actual, 2013, tiene 2013 ( 2003 = 10 aos.
Pg.20 70a) 14004b) 46810c) 28230d) 40001.
Pg.20 71a) Meres (Siero) = 33199
b)Alimerka (Oviedo) = 33008.
Pg.20 72a) 140 68 = 72 142 70 = 72 ( Propiedad de la resta
b) 431 88 = 343 421 78 = 343 ( Propiedad de la resta
c) 20 (15 + 2) = 20 15 + 20 2 ( Propiedad distributiva
d) 5 (10 + 4) = 5 10 + 5 4 ( Propiedad distributiva.
Pg.20 73a) 10 9 = 90 ( 4 9 + 6 9 = 36 + 54 = 90
b) 35 8 = 280 ( 12 8 + 23 8 = 96 + 184 = 280
c) 130 : 10 = 13 ( 19 8 + 2 = 13
d) 117 : 13 = 9 ( 39:13 + 13:13 + 65:13 = 3 + 1 + 5 = 9
Pg.20 74Si existe tal nmero N se cumple (
Pg.20 75a)
EMBED Equation.3 (
EMBED Equation.3
b)
EMBED Equation.3 (
c)
EMBED Equation.3 (
d)
EMBED Equation.3 ( .
Pg.20 76Tengo en cuenta que Dividendo = Divisor ( Cociente + Resto.
Dividendo
Divisor Cociente Resto
364
148
2
68
91
37
2
17
888
444 2 0
1564
27 57 25
2456 13 188
12.
Pg.21 77a)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
b)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
c)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
d)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
e)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
f)
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Pg.21 78a) (
b) (
c) (
d) (
e) (
f) (
g) (
h) (
i) ( .
Pg.21 79Multiplicacin
Divisiones asociadas Mltiplos
Divisores
5 6 = 30
30:5 = 6 30:6 = 5
30 de 5 y 6
5 y 6 de 30
7 4 = 28
28:7 = 4 28:4 = 7
28 de 4 y 7
4 y 7 de 28
7 8 = 56
56:8 = 7 56:7 = 8
56 de 7 y 8
7 y 8 de 56.
Pg.21 80
( ( (
EMBED Equation.3 Pg.21 81
( (
EMBED Equation.3 Pg.21 82
(
(
(
( .
El producto de los exponentes, aumentados en 1, de la descomposicin en factores primos, es el nmero de divisores que existen.
Pg.21 83El acabado en 0 en cifra par ( S = a), b), d). No, c).
Pg.21 84Los acabados en 0 ( S = b) y d). No = a) y c).
Pg.21 85a) Suma de cifras = ( S = 3033, 18951, 90. No = 21073.
b) Suma de cifras = ( S = 3033, 90. No = 18951, 21073.
c) Lo son 3033 y 90.
Pg.21 . 86a) Los acabados en 4 y en 00 ( S = 144, 900. No = 4255, 1875.
b) Los acabados en y en 00. ( S = 900, 1875. No = 144, 4255.
c) Los acabados en 00 ( S = 900. No = 144, 4255, 1875.
Pg.21 87
( S = b), c), d), e), f). No = a).
Pg.21 88a) b) c) d) e) f)
Son primos los nmeros 1 y 31.
Pg.21 89
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3 .
Pg.21 90a) (
b) (
c) ( Tiene 6 divisores 1, 2, 3, 4, 6, 12.
d) (
e) (
f) ( Tiene 8 divisores 1, 2, 3, 5, 6, 10, 15, 30
g) (
h) ( .
Pg.21 91Ha de ser mltiplo de ( (
( N ( Cualquier nmero que no sea ni .
(
EMBED Equation.3
Pg.21 92a) (
b) (
c) (
d) (
e) (
f) (
g) (
h) (
i) (
j) ( .
Pg.21 93a) (
b) (
c) (
d) (
e) (
f) (
g) (
h) (
i) (
j) ( .
Pg.22 94El 61 es un nmero vlido, porque es mayor que 60.
El otro tendr que ser 61+ 56 = 117. As 117 61 = 56.
Pg.22 95a) AB+BC = 235+134 = 389 km
b) BC+CA = 134+49 = 180 km.
Pg.22 96De tantos como divisores tenga el nmero 120 (
Nmero de divisores =
EMBED Equation.3 .
Pg.22 97Ha de terminar en 0 cifra par. 3 5 7 X ( X = .Pg.22 98La suma de sus cifras debe ser . X 4 5 1 ( .
( ( (
( (
Pg.22 99Nmero ( a b c d. Valor posicional del 5 500 ( . Unidades .
Capica a ( d y b ( c ( a b c d = .
Pg.22 100
( se vuelven a encontrar cada 15 das.
De 20 a 31 de mayo van 11 das.
EMBED Equation.3 .
Se encuentran, de nuevo, el da 4 de junio.
Pg.22 101
Un cuadrado exige potencias de de ndices pares.
Las mayores potencias pares, contenidas, son
El mayor lado posible ser
EMBED Equation.3 .
Pg.22 - 102
(
( (
a) Puede tener
b) Con ( pondr 2 en cada una de las 31 pginas.
Con ( pondr 2 en cada una de las 61 pginas.
Con ( pondr 2 en cada una de las 91 pginas, o bien
7 en cada una de las 26 pginas, o bien
13 en cada una de las 14 pginas, o bien
14 en cada una de las 13 pginas, etc.
Pg.22 103
EMBED Equation.3 ( .Pg.22 104 Dientes rueda mayor = 18
Dientes rueda menor = 12.
a) ( Vueltas rueda menor =
EMBED Equation.3 .
b) Vueltas rueda mayor =
EMBED Equation.3 (
EMBED Equation.3 Pg.22 105
a) ( =
El menor nmero que cumple es 251.
b) Los dos siguientes son 503 y 755.
Pg.22 106 ( (
No hay cuadrado perfecto que, por , d menos de 30.000
El nmero de espectadores tiene que ser . Pg.23 107
Sesenta (60) cifras.
Pg.23 108
( ( .
Pg.23 109
El siglo XXI comprende de 2001 a 2100, ambos inclusive.
Un capica del XXI tiene fijos las dos primeras cifras 2 y 0.
El nico existente es 2002 ( a) En ninguno.
Pg.23 110
El nmero 2009, como nmero impar, es siempre suma de par+impar.
El nico par primo es 2, y entonces la suma sera 2009=2+2007.
Pero 2007 es . Luego d) De ninguna.
Pg.23 111
6 3 + 2 5 = (
( (
Pero y son dgitos, con suma mxima 9+9=18 ( .
Pg.23 112
( N divisores =
Divisores = 1, 2, 3, 6, 7, 11, 14, 21, 22, 33, 42, 66, 77, 154, 231, 462.
( ( ( .
( Edades = 21 y 22 ( Opcin c) 43Pg.23 113 ( ( (
( .
Pg.23 114
( ltimo nmero .
Los nmeros son ( .Pg.23 115 ( (
( No vlidos = 1 y 2.
EMBED Equation.3 ( Opcin d) 57.Pg.23 1a)
mil quinientos siete.
b)
EMBED Equation.3 ochenta y ocho mil seiscientos noventa y nueve.
c)
EMBED Equation.3 seis mil ochocientos veinte.
Pg.23 2Por ejemplo, la propiedad posicional.
Pg.23 3a) 44 + 13 = 13 + 44 ( Propiedad conmutativa.
b) 5(7 + 8) = 35 + 40 ( Propiedad distributiva.
c) 133 ( 47 = 86 100 ( 14 = 86 ( Propiedad de la resta.
d) 12(5 + 17) = (12 5) + (12 17) ( Propiedad distributiva.Pg.23 4
( (
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
.Pg.23 5a) Acabe en 0 par ( 4158, 7058, 1800, 1530.
b) Suma cifras ( 4158, 1800, 14727, 1530.
c) Dos ltimas cifras 00 ( 1800.
d) Acabe en 0 5 ( 1800, 1530.
e) Suma cifras ( 4158, 1800, 1530.
f) Diferencia de suma nones suma pares = ( 4158.
Pg.23 6No existe el concepto de primos entre s.
Hay parejas de nmeros que no tienen divisin exacta entre, s.
Basta comparar las mutuas descomposiciones en factores.
Si uno de ellos contiene algn factor primo distinto, sern primos entre s
Y, anlogamente, si los exponentes respectivos no son idnticos.
Ejemplos:
y
EMBED Equation.3 y
EMBED Equation.3
EMBED Equation.3 y
EMBED Equation.3
y
EMBED Equation.3 .Pg.23 7Tantos como divisores tenga el nmero 40.
( ( De ocho formas.
Divisores = 1, 2, 4, 5, 8, 10, 20, 40.
Grupos(Miembros = 1(40 = 2(20 = 4(10 = 5(8 = 8(5 = 10(4 = 20(2 = 40(1. Pg.23 8729 es mltiplo de 9, pues 7+ 2+ 9 = 18 (
81 es 92 (
EMBED Equation.3 .
Pg.23 9a) (
EMBED Equation.3
EMBED Equation.3 .
b) (
EMBED Equation.3
EMBED Equation.3 .1/14
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