際際滷

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C存c b袖i To存n d誰ng ch歎 thay s竪
BTp 1: Cho s竪 c達 2 ch歎 s竪, nu ly t脱ng c存c ch歎 s竪 c辿ng v鱈i tch c存c ch歎
s竪 c単a s竪 速揃 cho th b損ng chnh s竪 速達. Tm ch歎 s竪 h袖ng 速測n v c単a s竪
速揃 cho.
B袖i gi其i
G辰i s竪 c達 2 ch歎 s竪 ph其i tm l袖 ab (a > 0, a, b < 10)
Theo b袖i ra ta c達 ab = a + b + a x b
a x 10 + b = a + b + a x b
a x 10 = a + a x b (c誰ng b鱈t b)
a x 10 = a x (1 + b) (M辿t s竪 nh息n v鱈i m辿t t脱ng)
10 = 1 + b (c誰ng chia cho a)
b = 10 - 1
b = 9
BTp2: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m s竪 21 v袖o b捉n tr存i s竪
速達 th ta 速樽c m辿t s竪 l鱈n gp 31 ln s竪 cn tm.
B袖i gi其i
G辰i s竪 ph其i tm l袖 ab (a > 0, a, b < 0)
Khi vit th捉m s竪 21 v袖o b捉n tr存i s竪 ab ta 速樽c s竪 m鱈i l袖 ab21 .
Theo b袖i ra ta c達:
ab21 = 31 x ab
2100 + ab = 31 x ab (ph息n tch s竪 ab21 = 2100 + ab )
2100 + ab = (30 + 1) x ab
2100 + ab = 30 x ab + ab (m辿t s竪 nh息n m辿t t脱ng)
2100 = ab x 30 (c誰ng b鱈t ab )
ab = 2100 : 30
ab = 70.
Th旦 l孫i
2170 : 70 = 31 (速坦ng)
Vy s竪 ph其i tm l袖: 70
則存p s竪: 70.
B袖i 3 Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng s竪 速達 gp 6 ln ch歎 s竪 h袖ng 速測n v
c単a n達.
B袖i gi其i
C存ch 1:
B鱈c 1: G辰i s竪 ph其i tm l袖 ab (0 < a < 10, b < 10).
Theo 速 b袖i ta c達: ab = 6 x b
B鱈c 2: S旦 d担ng tnh cht ch遜n l hoc ch歎 s竪 tn c誰ng.
V 6 x b l袖 m辿t s竪 ch遜n n捉n ab l袖 m辿t s竪 ch遜n.
b > 0 n捉n b = 2, 4, 6 hoc 8.
B鱈c 3: Tm gi存 tr b損ng ph測ng ph存p th旦 ch辰n
Nu b = 2 th ab = 6 x 2 = 12. (ch辰n)
1
Nu b = 4 th ab = 6 x 4 = 24. (ch辰n)
Nu b = 6 th ab = 6 x 6 = 36. (ch辰n)
Nu b = 8 th ab = 6 x 8 = 48. (ch辰n)
B鱈c 4: Vy ta 速樽c 4 s竪 tho其 m揃n 速 b袖i l袖: 12, 24, 36, 48.
則存p s竪: 12, 24, 36, 48.
C存ch 2:
B鱈c 1: G辰i s竪 ph其i tm l袖 ab (0 < a < 10, b < 10)
Theo 速 b袖i ta c達: ab = 6 x b
B鱈c 2: Xt ch歎 s竪 tn c誰ng
V 6 x b c達 tn c誰ng l袖 b n捉n b ch c達 th l袖: 2, 4, 6 hoc 8.
B鱈c 3: Tm gi存 tr b損ng ph測ng ph存p th旦 ch辰n
Nu b = 2 th ab = 6 x 2 = 12 (ch辰n)
Nu b = 4 th ab = 6 x 4 = 24 (ch辰n)
Nu b = 6 th ab = 6 x 6 = 36 (ch辰n)
Nu b = 8 th ab = 6 x 8 = 48 (ch辰n)
B鱈c 4: Vy ta 速樽c 4 s竪 tho其 m揃n 速 b袖i l袖: 12, 24, 36, 48.
則存p s竪: 12, 24, 36, 48.
B袖i 4: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 4 v袖o b捉n tr存i s竪 速達,
ta 速樽c m辿t s竪 gp 9 ln s竪 ph其i tm.
B袖i gi其i
Nu vit th捉m ch歎 s竪 4 v袖o b捉n ph其i s竪 c達 2 ch歎 s竪 th s竪 速達 t即ng th捉m 400 速測n
v
Theo b袖i ra ta c達:
4ab = ab x 9
400 + ab = ab x 9
400 + ab = ab x (8+ 1)
400 + ab = 8ab + ab
400 = 8ab ( C誰ng b鱈t 2 v 速i ab)
Ab = 400 : 8
Ab = 50
則存p s竪 50
B袖i 5: Tm m辿t s竪 c達 2 ch歎 s竪, khi vit th捉m ch歎 s竪 9 v袖o b捉n tr存i s竪 速達 ta 速樽c
m辿t s竪 gp 13 ln s竪 ph其i tm.
B袖i gi其i
Nu vit th捉m ch歎 s竪 9 v袖o b捉n ph其I s竪 c達 2 ch歎 s竪 th s竪 速達 t即ng th捉m 900 速測n
v
Theo b袖i ra ta c達:
9ab = ab x 13
900 + ab = ab x 13
900 + ab = ab x (12+ 1)
2
900 + ab = 12ab + ab
900 = 12ab ( C誰ng b鱈t 2 v 速i ab)
Ab = 900 : 12
Ab = 75
則存p s竪 75
B袖i 6: Tm m辿t s竪 c達 3 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪
速達 ta 速樽c m辿t s竪 h測n s竪 ph其i tm 1112 速測n v.
B袖i gi其i
Khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其I s竪 速達 th s竪 速達 t即ng th捉m 10 ln v袖 5 速測n v
則 s竪 速達 t即ng th捉m 10 ln th hiu m鱈i l袖:
1112 - 5 = 1107
S竪 cn tm l袖 :
1107 : (10-1) = 123
則存p s竪 123
B袖i 7: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪
速達 ta 速樽c m辿t s竪 h測n s竪 ph其i tm 230 速測n v.
B袖i gi其i
S竪 cn tm l袖:
(230  5) : ( 10  1) = 25
則/ s : 25
B袖i 8: Cho m辿t s竪 c達 2 ch歎 s竪. Nu vit th捉m ch歎 s竪 1 v袖o 速損ng tr鱈c v袖 速損ng
sau s竪 速達 th s竪 速達 t即ng l捉n 21 ln. Tm s竪 速揃 cho.
B袖i gi其i
Theo 速u b袖i ta c達
ab x 21 = 1ab1
ab x 21 = 1000 + ab0 + 1
ab x 21 = 1001 + ab0
ab x 21 = 1001 + ab x 10
ab x11 = 1001 ( C誰ng b鱈t di ab x 10)
ab = 1001 : 11
ab= 91
B袖i 9 : Tm s竪 c達 4 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪 速達
ta 速樽c s竪 l鱈n gp 5 ln s竪 nhn 速樽c khi ta vit th捉m ch歎 s竪 1 v袖o b捉n tr存i s竪
速達.
B袖i 10: Cho s竪 c達 3 ch歎 s竪, nu vit th捉m ch歎 s竪 1 v袖o b捉n ph其i s竪 速達, vit th捉m
ch歎 s竪 2 v袖o b捉n tr存i s竪 速達 ta 速u 速樽c s竪 c達 4 ch歎 s竪 m袖 s竪 n袖y gp 3 ln
s竪 kia
B袖i gi其i
G辰i s竪 ph其i tm l袖 abc ( 0 < a; b,c< 10
Theo 速u b袖i ta c達 hai kh其 n即ng: abc 1 > 2abc hoc abc1 < 2abc
a) Xt kh其 n即ng abc1 > 2abc
3
- Theo 速u b袖i ta c達 :
abc1 = 3 x 2abc
abc x10 + 1 = (2000 +abc) x3 ( Ct孫o s竪 )
abc x 3 + abcx7 + 1 = 6000 + abc x 3 ( M辿t s竪 nh息n v鱈i 1 t脱ng)
abc x 7 + 1 = 6000 ( B鱈t c其 hai v cho abc x 3)
abc x7 = 6000 -1( Tm s竪 h孫ng c単a t脱ng)
abc = 599 : 7857 ( Tm m辿t th探a s竪 )
B袖i 11: Cho m辿t s竪 c達 3 ch歎 s竪, nu xo存 速i ch歎 s竪 h袖ng tr即m th s竪 速達 gi其m 速i 3
ln. Tm s竪 速達.
B袖i gi其i
G辰i s竪 ph其i tm l袖 abc ( 0 < a ; bc< 10)
Theo 速u b袖i ta c達 :
abc = 3 x bc
( 則t tnh theo c辿t d辰c) S竪 50)
B袖i 12: Tm m辿t s竪 c達 4 ch歎 s竪, nu xo存 速i ch歎 s竪 h袖ng nghn th s竪 速達 gi其m
速i 9 ln.
( 則t tnh theo c辿t d辰c) : S竪 500
B袖i 13: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng s竪 速達 gp 5 ln t脱ng c存c ch歎 s竪 c単a
n達.
B袖i gi其i
G辰i s竪 ph其i tm l袖 ab : 則Kin
Theo 速u b袖i ta c達 :
Ab = ( a+b) x5
Ab = a x 5 + b x 5 ( nh息n m辿t s竪 v鱈i m辿t t脱ng)
A x 10 + b = a x 5 + bx5
A x( 5 + 5 ) + b = a x 5 + b x (1 + 4)
A x 5 + a x 5 + b = a x 5 + b + b x 4
A x 5 = b x 4 ( C誰ng b鱈t 速i a x5 + b)
N探u a = 1 th a x 5 = 1 x 5 = 5 Suy ra b kh束ng c達 gi存 tr thch h樽p
Th旦 ch辰n ta c達 a = 5 n捉n b = 4
Vy s竪 cn tm l袖 45.
B袖i 14: Tm m辿t s竪 c達 3 ch歎 s竪, bit r損ng s竪 速達 gp 11 ln t脱ng c存c ch歎 s竪 c単a
n達.
B袖i gi其i
G辰i s竪 ph其i tm l袖 abc : 則Kin
Theo 速u b袖i ta c達 :
Abc = ( a + b+ c) x 11
Abc = 11 x a + 11 x b + 11 x c
A x 100 + bx 10 + c = a x11 + b x 11 + c x11
A x ( 11 + 89 ) + bx 10 + c = a x 11 + b x ( 10 + 1) + c x ( 1 + 10)
A x11 + a x 89 + b x 10 + c = a x11 + b x10 + b + c + c x 10
A x 89 = b+ c x10 ( C誰ng tr探 速i n歎ng s竪 h孫ng gi竪ng nhau)
S竪 cntm l袖 198
4
5
5

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Toan cau tao so lop 4 5 dap an

  • 1. C存c b袖i To存n d誰ng ch歎 thay s竪 BTp 1: Cho s竪 c達 2 ch歎 s竪, nu ly t脱ng c存c ch歎 s竪 c辿ng v鱈i tch c存c ch歎 s竪 c単a s竪 速揃 cho th b損ng chnh s竪 速達. Tm ch歎 s竪 h袖ng 速測n v c単a s竪 速揃 cho. B袖i gi其i G辰i s竪 c達 2 ch歎 s竪 ph其i tm l袖 ab (a > 0, a, b < 10) Theo b袖i ra ta c達 ab = a + b + a x b a x 10 + b = a + b + a x b a x 10 = a + a x b (c誰ng b鱈t b) a x 10 = a x (1 + b) (M辿t s竪 nh息n v鱈i m辿t t脱ng) 10 = 1 + b (c誰ng chia cho a) b = 10 - 1 b = 9 BTp2: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m s竪 21 v袖o b捉n tr存i s竪 速達 th ta 速樽c m辿t s竪 l鱈n gp 31 ln s竪 cn tm. B袖i gi其i G辰i s竪 ph其i tm l袖 ab (a > 0, a, b < 0) Khi vit th捉m s竪 21 v袖o b捉n tr存i s竪 ab ta 速樽c s竪 m鱈i l袖 ab21 . Theo b袖i ra ta c達: ab21 = 31 x ab 2100 + ab = 31 x ab (ph息n tch s竪 ab21 = 2100 + ab ) 2100 + ab = (30 + 1) x ab 2100 + ab = 30 x ab + ab (m辿t s竪 nh息n m辿t t脱ng) 2100 = ab x 30 (c誰ng b鱈t ab ) ab = 2100 : 30 ab = 70. Th旦 l孫i 2170 : 70 = 31 (速坦ng) Vy s竪 ph其i tm l袖: 70 則存p s竪: 70. B袖i 3 Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng s竪 速達 gp 6 ln ch歎 s竪 h袖ng 速測n v c単a n達. B袖i gi其i C存ch 1: B鱈c 1: G辰i s竪 ph其i tm l袖 ab (0 < a < 10, b < 10). Theo 速 b袖i ta c達: ab = 6 x b B鱈c 2: S旦 d担ng tnh cht ch遜n l hoc ch歎 s竪 tn c誰ng. V 6 x b l袖 m辿t s竪 ch遜n n捉n ab l袖 m辿t s竪 ch遜n. b > 0 n捉n b = 2, 4, 6 hoc 8. B鱈c 3: Tm gi存 tr b損ng ph測ng ph存p th旦 ch辰n Nu b = 2 th ab = 6 x 2 = 12. (ch辰n) 1
  • 2. Nu b = 4 th ab = 6 x 4 = 24. (ch辰n) Nu b = 6 th ab = 6 x 6 = 36. (ch辰n) Nu b = 8 th ab = 6 x 8 = 48. (ch辰n) B鱈c 4: Vy ta 速樽c 4 s竪 tho其 m揃n 速 b袖i l袖: 12, 24, 36, 48. 則存p s竪: 12, 24, 36, 48. C存ch 2: B鱈c 1: G辰i s竪 ph其i tm l袖 ab (0 < a < 10, b < 10) Theo 速 b袖i ta c達: ab = 6 x b B鱈c 2: Xt ch歎 s竪 tn c誰ng V 6 x b c達 tn c誰ng l袖 b n捉n b ch c達 th l袖: 2, 4, 6 hoc 8. B鱈c 3: Tm gi存 tr b損ng ph測ng ph存p th旦 ch辰n Nu b = 2 th ab = 6 x 2 = 12 (ch辰n) Nu b = 4 th ab = 6 x 4 = 24 (ch辰n) Nu b = 6 th ab = 6 x 6 = 36 (ch辰n) Nu b = 8 th ab = 6 x 8 = 48 (ch辰n) B鱈c 4: Vy ta 速樽c 4 s竪 tho其 m揃n 速 b袖i l袖: 12, 24, 36, 48. 則存p s竪: 12, 24, 36, 48. B袖i 4: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 4 v袖o b捉n tr存i s竪 速達, ta 速樽c m辿t s竪 gp 9 ln s竪 ph其i tm. B袖i gi其i Nu vit th捉m ch歎 s竪 4 v袖o b捉n ph其i s竪 c達 2 ch歎 s竪 th s竪 速達 t即ng th捉m 400 速測n v Theo b袖i ra ta c達: 4ab = ab x 9 400 + ab = ab x 9 400 + ab = ab x (8+ 1) 400 + ab = 8ab + ab 400 = 8ab ( C誰ng b鱈t 2 v 速i ab) Ab = 400 : 8 Ab = 50 則存p s竪 50 B袖i 5: Tm m辿t s竪 c達 2 ch歎 s竪, khi vit th捉m ch歎 s竪 9 v袖o b捉n tr存i s竪 速達 ta 速樽c m辿t s竪 gp 13 ln s竪 ph其i tm. B袖i gi其i Nu vit th捉m ch歎 s竪 9 v袖o b捉n ph其I s竪 c達 2 ch歎 s竪 th s竪 速達 t即ng th捉m 900 速測n v Theo b袖i ra ta c達: 9ab = ab x 13 900 + ab = ab x 13 900 + ab = ab x (12+ 1) 2
  • 3. 900 + ab = 12ab + ab 900 = 12ab ( C誰ng b鱈t 2 v 速i ab) Ab = 900 : 12 Ab = 75 則存p s竪 75 B袖i 6: Tm m辿t s竪 c達 3 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪 速達 ta 速樽c m辿t s竪 h測n s竪 ph其i tm 1112 速測n v. B袖i gi其i Khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其I s竪 速達 th s竪 速達 t即ng th捉m 10 ln v袖 5 速測n v 則 s竪 速達 t即ng th捉m 10 ln th hiu m鱈i l袖: 1112 - 5 = 1107 S竪 cn tm l袖 : 1107 : (10-1) = 123 則存p s竪 123 B袖i 7: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪 速達 ta 速樽c m辿t s竪 h測n s竪 ph其i tm 230 速測n v. B袖i gi其i S竪 cn tm l袖: (230 5) : ( 10 1) = 25 則/ s : 25 B袖i 8: Cho m辿t s竪 c達 2 ch歎 s竪. Nu vit th捉m ch歎 s竪 1 v袖o 速損ng tr鱈c v袖 速損ng sau s竪 速達 th s竪 速達 t即ng l捉n 21 ln. Tm s竪 速揃 cho. B袖i gi其i Theo 速u b袖i ta c達 ab x 21 = 1ab1 ab x 21 = 1000 + ab0 + 1 ab x 21 = 1001 + ab0 ab x 21 = 1001 + ab x 10 ab x11 = 1001 ( C誰ng b鱈t di ab x 10) ab = 1001 : 11 ab= 91 B袖i 9 : Tm s竪 c達 4 ch歎 s竪, bit r損ng khi vit th捉m ch歎 s竪 5 v袖o b捉n ph其i s竪 速達 ta 速樽c s竪 l鱈n gp 5 ln s竪 nhn 速樽c khi ta vit th捉m ch歎 s竪 1 v袖o b捉n tr存i s竪 速達. B袖i 10: Cho s竪 c達 3 ch歎 s竪, nu vit th捉m ch歎 s竪 1 v袖o b捉n ph其i s竪 速達, vit th捉m ch歎 s竪 2 v袖o b捉n tr存i s竪 速達 ta 速u 速樽c s竪 c達 4 ch歎 s竪 m袖 s竪 n袖y gp 3 ln s竪 kia B袖i gi其i G辰i s竪 ph其i tm l袖 abc ( 0 < a; b,c< 10 Theo 速u b袖i ta c達 hai kh其 n即ng: abc 1 > 2abc hoc abc1 < 2abc a) Xt kh其 n即ng abc1 > 2abc 3
  • 4. - Theo 速u b袖i ta c達 : abc1 = 3 x 2abc abc x10 + 1 = (2000 +abc) x3 ( Ct孫o s竪 ) abc x 3 + abcx7 + 1 = 6000 + abc x 3 ( M辿t s竪 nh息n v鱈i 1 t脱ng) abc x 7 + 1 = 6000 ( B鱈t c其 hai v cho abc x 3) abc x7 = 6000 -1( Tm s竪 h孫ng c単a t脱ng) abc = 599 : 7857 ( Tm m辿t th探a s竪 ) B袖i 11: Cho m辿t s竪 c達 3 ch歎 s竪, nu xo存 速i ch歎 s竪 h袖ng tr即m th s竪 速達 gi其m 速i 3 ln. Tm s竪 速達. B袖i gi其i G辰i s竪 ph其i tm l袖 abc ( 0 < a ; bc< 10) Theo 速u b袖i ta c達 : abc = 3 x bc ( 則t tnh theo c辿t d辰c) S竪 50) B袖i 12: Tm m辿t s竪 c達 4 ch歎 s竪, nu xo存 速i ch歎 s竪 h袖ng nghn th s竪 速達 gi其m 速i 9 ln. ( 則t tnh theo c辿t d辰c) : S竪 500 B袖i 13: Tm m辿t s竪 c達 2 ch歎 s竪, bit r損ng s竪 速達 gp 5 ln t脱ng c存c ch歎 s竪 c単a n達. B袖i gi其i G辰i s竪 ph其i tm l袖 ab : 則Kin Theo 速u b袖i ta c達 : Ab = ( a+b) x5 Ab = a x 5 + b x 5 ( nh息n m辿t s竪 v鱈i m辿t t脱ng) A x 10 + b = a x 5 + bx5 A x( 5 + 5 ) + b = a x 5 + b x (1 + 4) A x 5 + a x 5 + b = a x 5 + b + b x 4 A x 5 = b x 4 ( C誰ng b鱈t 速i a x5 + b) N探u a = 1 th a x 5 = 1 x 5 = 5 Suy ra b kh束ng c達 gi存 tr thch h樽p Th旦 ch辰n ta c達 a = 5 n捉n b = 4 Vy s竪 cn tm l袖 45. B袖i 14: Tm m辿t s竪 c達 3 ch歎 s竪, bit r損ng s竪 速達 gp 11 ln t脱ng c存c ch歎 s竪 c単a n達. B袖i gi其i G辰i s竪 ph其i tm l袖 abc : 則Kin Theo 速u b袖i ta c達 : Abc = ( a + b+ c) x 11 Abc = 11 x a + 11 x b + 11 x c A x 100 + bx 10 + c = a x11 + b x 11 + c x11 A x ( 11 + 89 ) + bx 10 + c = a x 11 + b x ( 10 + 1) + c x ( 1 + 10) A x11 + a x 89 + b x 10 + c = a x11 + b x10 + b + c + c x 10 A x 89 = b+ c x10 ( C誰ng tr探 速i n歎ng s竪 h孫ng gi竪ng nhau) S竪 cntm l袖 198 4
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