Puzzle for December 10, 2019  ( )

Scratchpad

Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 27 eq.2) C + D = A + B eq.3) B + C = A + F eq.4) A + D = B + E eq.5) C = B + D eq.6) F = B + D + E

A, B, C, D, E, and F each represent a one-digit non-negative integer.

Scratchpad

 

Help Area

Hint #1


In eq.2, replace C with B + D (from eq.5): B + D + D = A + B which becomes B + 2×D = A + B Subtract B from both sides of the equation above: B + 2×D – B = A + B – B which makes 2×D = A


  

Hint #2


In eq.4, replace A with 2×D: 2×D + D = B + E which becomes eq.4a) 3×D = B + E


  

Hint #3


eq.6 may be written as: F = B + E + D In the above equation, substitute 3×D for B + E (from eq.4a): F = 3×D + D which makes F = 4×D


  

Hint #4


Substitute 2×D for A in eq.2: C + D = 2×D + B Subtract D from both sides of the above equation: C + D – D = 2×D + B – D which becomes eq.2a) C = D + B


  

Hint #5


Substitute D + B for C, 2×D for A, and 4×D for F in eq.3: B + D + B = 2×D + 4×D which becomes 2×B + D = 6×D Subtract D from each side of the equation above: 2×B + D – D = 6×D – D which means 2×B = 5×D Divide both sides by 2: 2×B ÷ 2 = 5×D ÷ 2 which makes B = 2½×D


  

Hint #6


Substitute 2½×D for B in eq.2a: C = D + 2½×D which makes C = 3½×D


  

Hint #7


Substitute 2½×D for B in eq.4a: 3×D = 2½×D + E Subtract 2½×D from each side of the above equation: 3×D – 2½×D = 2½×D + E – 2½×D which makes ½×D = E


  

Solution

Substitute 2×D for A, 2½×D for B, 3½×D for C, ½×D for E, and 4×D for F in eq.1: 2×D + 2½×D + 3½×D + D + ½×D + 4×D = 27 which simplifies to 13½×D = 27 Divide both sides of the equation above by 13½: 13½×D ÷ 13½ = 27 ÷ 13½ which means D = 2 making A = 2×D = 2 × 2 = 4 B = 2½×D = 2½ × 2 = 5 C = 3½×D = 3½ × 2 = 7 E = ½×D = ½ × 2 = 1 F = 4×D = 4 × 2 = 8 and ABCDEF = 457218