Puzzle for August 12, 2023  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 28 eq.2) D = C + E eq.3) B + D = A + F eq.4) A + D = B + E eq.5) B + F = A + D + E eq.6) C + E + F = A + B + D

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

Scratchpad

 

Help Area

Hint #1


In eq.5, replace A + D with B + E (from eq.4): B + F = B + E + E which becomes B + F = B + 2×E Subtract B from each side of the equation above: B + F - B = B + 2×E - B which makes F = 2×E


  

Hint #2


In eq.6, replace B + D with A + F (from eq.3): C + E + F = A + A + F which becomes C + E + F = 2×A + F Subtract F from each side of the equation above: C + E + F - F = 2×A + F - F which becomes eq.6a) C + E = 2×A


  

Hint #3


In eq.2, replace C + E with 2×A (from eq.6a): D = 2×A


  

Hint #4


eq.1 may be written as: eq.1a) A + B + D + C + E + F = 28   In eq.1a, substitute C + E + F for A + B + D (from eq.6): C + E + F + C + E + F = 28 which becomes 2×(C + E + F) = 28 Divide both sides of the above equation by 2: 2×(C + E + F) ÷ 2 = 28 ÷ 2 which becomes eq.1b) C + E + F = 14


  

Hint #5


In eq.1b, substitute A + B + D for C + E + F (from eq.6): eq.1c) A + B + D = 14


  

Hint #6


Substitute 2×A for D in eq.1c: A + B + 2×A = 14 which becomes 3×A + B = 14 Subtract 3×A from each side of the above equation: 3×A + B - 3×A = 14 - 3×A which becomes eq.1d) B = 14 - 3×A


  

Hint #7


Substitute 2×E for F in eq.1b: C + E + 2×E = 14 which becomes C + 3×E = 14 Subtract 3×E from each side of the equation above: C + 3×E - 3×E = 14 - 3×E which becomes eq.1e) C = 14 - 3×E


  

Hint #8


Substitute 2×A for D, and 14 - 3×A for B (from eq.1d) in eq.4: A + 2×A = 14 - 3×A + E which becomes 3×A = 14 - 3×A + E In the above equation, subtract 14 from both sides, and add 3×A to both sides: 3×A - 14 + 3×A = 14 - 3×A + E - 14 + 3×A which becomes eq.4a) 6×A - 14 = E


  

Hint #9


Substitute 2×A for D, and 14 - 3×E for C (from eq.1e) in eq.2: 2×A = 14 - 3×E + E which becomes eq.2a) 2×A = 14 - 2×E


  

Hint #10


Substitute (6×A - 14) for E (from eq.4a) into eq.2a: 2×A = 14 - 2×(6×A - 14) which becomes 2×A = 14 - 12×A + 28 which becomes 2×A = 42 - 12×A Add 12×A to both sides of the above equation: 2×A + 12×A = 42 - 12×A + 12×A which makes 14×A = 42 Divide both sides by 14: 14×A ÷ 14 = 42 ÷ 14 which makes A = 3 and also makes D = 2×A = 2×3 = 6


  

Hint #11


Substitute 3 for A in eq.4a: 6×3 - 14 = E which becomes 18 - 14 = E which makes 4 = E and also makes F = 2×E = 2×4 = 8


  

Hint #12


Substitute 3 for A, and 6 for D in eq.1c: 3 + B + 6 = 14 which becomes 9 + B = 14 Subtract 9 from both sides of the equation above: 9 + B - 9 = 14 - 9 which makes B = 5


  

Solution

Substitute 4 for E, and 8 for F in eq.1b: C + 4 + 8 = 14 which becomes C + 12 = 14 Subtract 12 from each side of the equation above: C + 12 - 12 = 14 - 12 which makes C = 2 and makes ABCDEF = 352648