Olimpiada Caras-Severin 2004 - faza municipala (1)
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Marcelina Popa
- Bernoulli
- Posts: 208
- Joined: Wed Mar 05, 2008 3:25 pm
- Location: Tulcea
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Olimpiada Caras-Severin 2004 - faza municipala (1)
Aflati numerele \( \overline{abc} \) stiind ca \( \overline{bc} \) reprezinta \( \frac{1}{6} \) din \( \overline{abc} \).
-
thekrisser
- Arhimede
- Posts: 9
- Joined: Mon Nov 03, 2008 2:36 pm
Aceasta problema s-a dat si la cls V la Scoala cu Ceas Rm Valcea acum 2 ani.( sau ceva asemanator )
Abc=6*bc
100 a + bc =6 * bc | - cb
=> 100 a = 5 * bc | :5
=> 20 a = bc
Cum b , c cifre => b, c < 10
=> b*c <= 81
=> 20 a <= 81
=> a <=4 a<>0
=> a € { 1 ; 2 ;3 ;4 }
Dc a = 1 => bc =20
B,c cifre =>(a=1,b=5,c=4) , (a=1,b=4,c=5)
Dc a=2 => bc=40
B c cifre => (a=2,b=8,c=5) , (a=2,b=5,c=8)
Dc a =3 bc=60=>
B,c cifre => nu exista sol
Dc a=4 bc=80
B,c cifre=> nu exista sol
Abc=6*bc
100 a + bc =6 * bc | - cb
=> 100 a = 5 * bc | :5
=> 20 a = bc
Cum b , c cifre => b, c < 10
=> b*c <= 81
=> 20 a <= 81
=> a <=4 a<>0
=> a € { 1 ; 2 ;3 ;4 }
Dc a = 1 => bc =20
B,c cifre =>(a=1,b=5,c=4) , (a=1,b=4,c=5)
Dc a=2 => bc=40
B c cifre => (a=2,b=8,c=5) , (a=2,b=5,c=8)
Dc a =3 bc=60=>
B,c cifre => nu exista sol
Dc a=4 bc=80
B,c cifre=> nu exista sol
- miruna.lazar
- Bernoulli
- Posts: 224
- Joined: Wed Oct 08, 2008 8:41 pm
- Location: Tulcea
\( \overline {bc}= 10b +c \)
\( \overline {abc} = 100a + \overline {bc} = 100a + 10b + c \)
\( \overline {bc} = \frac {1}{6} \) din \( \overline {abc} \)
\( \overline {bc} = \frac {1}{6} \overline {abc} / \cdot 6 \)
\( 6\overline {bc} =\overline { abc} \)
\( \overline {abc} = 6\overline{bc} \)
\( 100a+\overline{bc} = 6\overline{bc} /- bc \)
\( 100a = 5\overline {bc} /:5 => 20a = \overline {bc} \)
SOLUTII
Daca a = 1 =>\( \overline{bc}= 20 = > \overline {abc} = 120 \)
Daca a = 2 , => \( \overline{bc} = 40 => \overline {abc} = 240 \)
Daca a = 3 , => \( \overline{bc} = 60 => \overline {abc} = 360 \)
Daca a = 4 => \( \overline{bc} =80 => \overline {abc} = 480 \)
\( \overline {abc} = 100a + \overline {bc} = 100a + 10b + c \)
\( \overline {bc} = \frac {1}{6} \) din \( \overline {abc} \)
\( \overline {bc} = \frac {1}{6} \overline {abc} / \cdot 6 \)
\( 6\overline {bc} =\overline { abc} \)
\( \overline {abc} = 6\overline{bc} \)
\( 100a+\overline{bc} = 6\overline{bc} /- bc \)
\( 100a = 5\overline {bc} /:5 => 20a = \overline {bc} \)
SOLUTII
Daca a = 1 =>\( \overline{bc}= 20 = > \overline {abc} = 120 \)
Daca a = 2 , => \( \overline{bc} = 40 => \overline {abc} = 240 \)
Daca a = 3 , => \( \overline{bc} = 60 => \overline {abc} = 360 \)
Daca a = 4 => \( \overline{bc} =80 => \overline {abc} = 480 \)