You are given:
(1) a + b + c + d = e + f + g + h
(2) a + 2b + 3c + 4d = e + 2f + 3g + 4h
Now give me back:
(3) 4a + 3b + 2c + d = 4e + 3f + 2g + h
(show all work in your proof, please).
Hello, wonderboy1953!
You are given: .$\displaystyle \begin{array}{ccccc}(1) & a + b + c + d &=& e + f + g + h\\
(2) &a + 2b + 3c + 4d &=& e + 2f + 3g + 4h \end{array}$
Now give me back: .$\displaystyle (3)\;\; 4a + 3b + 2c + d \;=\; 4e + 3f + 2g + h$
Multiply [1] by 5: .$\displaystyle 5a + 5b + 5c + 5d \;=\;5e + 5f + 5g + 5h$
. . . . . .$\displaystyle (a + 4a) + (2b + 3b) + (4d + d) \;=\;(e+4e) + (2f + 3f) + (3g+2g) + (4h+h)$
. . .$\displaystyle \underbrace{(a+2b + 3c + 4d)} + (4a \;+\; 3b \;+\; 2c \;+\; d) \;=\;\underbrace{(e+2f + 3g + 4h)} + (4e \;+\; 3f \;+\; 2g \;+\; h)$
. . . . . . . . . . . . . $\displaystyle \nwarrow\quad\text{These are equal}\quad\nearrow$
Therefore: .$\displaystyle 4a + 3b + 2c + d \;=\;4e + 3f + 2g + h$
Soroban did a wonderful job with my puzzle (and in record time). Now for my next related puzzle.
I give you a trigrade:
$\displaystyle a^n + b^n + c^n + d^n = e^n + f^n + g^n + h^n; n = 1,2,3$ (although not required, for simplicity's sake, let a through h be positive integers).
Now I also give you $\displaystyle a + 2^2b + 3^2c +4^2d = e + 2^2f + 3^2g + 4^2h$
Can you give me $\displaystyle 4^2a + 3^2b + 2^2c + d = 4^2e + 3^2f + 2^2g + h?$
(I see no one as yet assigned values to the letters so check out 2,8,9,15 = 3,5,12,14; also check out 1,8,10,17 = 2,5,13,16 when you substitute for the letters in my first and second puzzles).
I'll be back next Sunday to check on your progress with my puzzles.