Metoda mersului invers. Rezolvăm problema în ordine inversă:
Bidon 1
Bidon 2
Bidon 3
Final
15 15 15 9 9 9 10 10 10
( B 1 , B 2 ) → B 3 (B_1,B_2) \to B_3 \quad ( B 1 , B 2 ) → B 3 10 : 2 = 5 10:2=5 10 : 2 = 5 15 − 5 = 10 15-5=10 15 − 5 = 10 9 − 5 = 4 9-5=4 9 − 5 = 4 10 + 10 = 20 10+10=20 10 + 10 = 20
( B 1 , B 3 ) → B 2 (B_1,B_3) \to B_2 \quad ( B 1 , B 3 ) → B 2 4 : 2 = 2 4:2=2 4 : 2 = 2 10 − 2 = 8 10-2=8 10 − 2 = 8 4 + 4 = 8 4+4=8 4 + 4 = 8 20 − 2 = 18 20-2=18 20 − 2 = 18
( B 2 , B 3 ) → B 1 (B_2,B_3) \to B_1 \quad ( B 2 , B 3 ) → B 1 8 : 2 = 4 8:2=4 8 : 2 = 4 8 + 8 = 16 8+8=16 8 + 8 = 16 8 − 4 = 4 8-4=4 8 − 4 = 4 18 − 4 = 14 18-4=14 18 − 4 = 14
Așadar, în cele 3 3 3 16 , 4 , 16, 4, 16 , 4 , 14 14 14
Metoda algebrică. Notăm cu x , y x,y x , y z z z 1 , 2 1,2 1 , 2 3. 3. 3.
Bidon 1
Bidon 2
Bidon 3
Inițial
x x x y y y z z z
B 1 → ( B 2 , B 3 ) B_1 \to (B_2, B_3) B 1 → ( B 2 , B 3 ) x 2 \dfrac{x}{2} 2 x x 4 + y \dfrac{x}{4}+y 4 x + y x 4 + z \dfrac{x}{4}+z 4 x + z
B 2 → ( B 1 , B 3 ) B_2 \to (B_1, B_3) B 2 → ( B 1 , B 3 ) x 2 + 1 4 ( x 4 + y ) = 9 x 16 + y 4 \dfrac{x}{2} + \dfrac{1}{4} \bigg(\dfrac{x}{4} + y \bigg) = \dfrac{9x}{16} + \dfrac{y}{4} 2 x + 4 1 ( 4 x + y ) = 16 9 x + 4 y 1 2 ( x 4 + y ) = x 8 + y 2 \dfrac{1}{2} \bigg(\dfrac{x}{4} +y\bigg) = \dfrac{x}{8} + \dfrac{y}{2} 2 1 ( 4 x + y ) = 8 x + 2 y x 4 + z + 1 4 ( x 4 + y ) = 5 x 16 + y 4 + z \dfrac{x}{4}+z + \dfrac{1}{4} \bigg(\dfrac{x}{4} +y\bigg)=\dfrac{5x}{16}+\dfrac{y}{4}+z 4 x + z + 4 1 ( 4 x + y ) = 16 5 x + 4 y + z
B 3 → ( B 1 , B 2 ) B_3 \to (B_1, B_2) B 3 → ( B 1 , B 2 ) 9 x 16 + y 4 + 1 4 ( 5 x 16 + y 4 + z ) = 41 x 64 + 5 y 16 + z 4 \dfrac{9x}{16} + \dfrac{y}{4} + \dfrac{1}{4} \bigg(\dfrac{5x}{16} + \dfrac{y}{4} + z \bigg) =\dfrac{41x}{64}+\dfrac{5y}{16} + \dfrac{z}{4} 16 9 x + 4 y + 4 1 ( 16 5 x + 4 y + z ) = 64 41 x + 16 5 y + 4 z x 8 + y 2 + 1 4 ( 5 x 16 + y 4 + z ) = 13 x 64 + 9 y 16 + z 4 \dfrac{x}{8}+\dfrac{y}{2}+\dfrac{1}{4} \bigg(\dfrac{5x}{16} + \dfrac{y}{4} + z \bigg)=\dfrac{13x}{64}+\dfrac{9y}{16} + \dfrac{z}{4} 8 x + 2 y + 4 1 ( 16 5 x + 4 y + z ) = 64 13 x + 16 9 y + 4 z 1 2 ( 5 x 16 + y 4 + z ) = 5 x 32 + y 8 + z 2 \dfrac{1}{2} \bigg(\dfrac{5x}{16} + \dfrac{y}{4} + z \bigg)=\dfrac{5x}{32} + \dfrac{y}{8} + \dfrac{z}{2} 2 1 ( 16 5 x + 4 y + z ) = 32 5 x + 8 y + 2 z
Egalând valorile obținute după ultimul transfer cu 15 , 9 , 15,9, 15 , 9 , 10 , 10, 10 ,
{ 41 x + 20 y + 16 z = 960 ( 1 ) 13 x + 36 y + 16 z = 576 ( 2 ) 5 x + 4 y + 16 z = 320 ( 3 ) ⇒ { ( 2 ) − ( 3 ) : 8 x + 32 y = 256 ∣ : 2 ( 1 ) − ( 3 ) : 36 x + 16 y = 640 ⇔ { 4 x + 16 y = 128 36 x + 16 y = 640 ⇒ ( − ) 32 x = 512 ⇒ x = 16 .
\begin{cases}
41x + 20y + 16z=960 \quad (1) \\
13x+36y+16z=576 \quad (2) \\
5x+4y+16z=320 \quad (3)
\end{cases} \Rightarrow
\begin{cases}
(2)-(3): 8x+32y=256 \quad |:2 \\
(1)-(3): 36x+16y=640
\end{cases} \Leftrightarrow
\begin{cases}
4x+16y=128 \\
36x+16y=640
\end{cases} \overset{(-)}{\Rightarrow} 32x=512 \Rightarrow \boxed{x=16}.
⎩ ⎨ ⎧ 41 x + 20 y + 16 z = 960 ( 1 ) 13 x + 36 y + 16 z = 576 ( 2 ) 5 x + 4 y + 16 z = 320 ( 3 ) ⇒ { ( 2 ) − ( 3 ) : 8 x + 32 y = 256 ∣ : 2 ( 1 ) − ( 3 ) : 36 x + 16 y = 640 ⇔ { 4 x + 16 y = 128 36 x + 16 y = 640 ⇒ ( − ) 32 x = 512 ⇒ x = 16 . 4 ⋅ 16 + 16 y = 128 ⇒ y = 4 . 4 \cdot 16+16y=128 \Rightarrow \boxed{y=4}. 4 ⋅ 16 + 16 y = 128 ⇒ y = 4 . 5 ⋅ 16 + 4 ⋅ 4 + 16 z = 320 ⇒ z = 14 . 5 \cdot 16+4 \cdot 4+16z=320 \Rightarrow \boxed{z=14}. 5 ⋅ 16 + 4 ⋅ 4 + 16 z = 320 ⇒ z = 14 .