Puzzle for September 3, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 20 eq.2) C + D = B + F eq.3) B + D + E = A + C + F eq.4) A – C + E = C – E + F eq.5) B + E + F – A = A + C – D – F eq.6) C + F = average (A, B, D, E)

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

Scratchpad

 

Help Area

Hint #1


eq.6 may be written as: C + F = (A + B + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (C + F) = 4 × (A + B + D + E) ÷ 4 which becomes eq.6a) 4×(C + F) = A + B + D + E


  

Hint #2


eq.1 may be written as: A + B + D + E + C + F = 20 In the above equation, substitute 4×(C + F) for A + B + D + E (from eq.6a): 4×(C + F) + C + F = 20 which becomes 5×(C + F) = 20 Divide both sides of the above equation by 5: 5×(C + F) ÷ 5 = 20 ÷ 5 which becomes eq.1a) C + F = 4


  

Hint #3


In eq.6a, replace (C + F) with 4 (from eq.1a): 4×4 = A + B + D + E which becomes 16 = A + B + D + E Subtract A from each side of the equation above: 16 – A = A + B + D + E – A which becomes eq.6b) 16 – A = B + D + E


  

Hint #4


In eq.3, substitute 16 – A for B + D + E (from eq.6b), and 4 for C + F (from eq.1a): 16 – A = A + 4 In the equation above, add A to both sides, and subtract 4 from both sides: 16 – A – 4 + A = A + 4 – 4 + A which becomes 12 = 2×A Divide both sides by 2: 12 ÷ 2 = 2×A ÷ 2 which makes 6 = A


  

Hint #5


Substitute 6 for A in eq.6b: 16 – 6 = B + D + E which becomes eq.6c) 10 = B + D + E


  

Hint #6


Add A, D, and F to both sides of eq.5: B + E + F – A + A + D + F = A + C – D – F + A + D + F which becomes B + E + 2×F + D = 2×A + C which may be written as eq.5a) B + D + E + 2×F = 2×A + C


  

Hint #7


In eq.5a, substitute 10 for B + D + E (from eq.6c), and 6 for A: 10 + 2×F = 2×6 + C which becomes 10 + 2×F = 12 + C Subtract 12 from each side of the above equation: 10 + 2×F – 12 = 12 + C – 12 which becomes eq.5b) 2×F – 2 = C


  

Hint #8


Substitute 2×F – 2 for C (from eq.5b) into eq.1a: 2×F – 2 + F = 4 which becomes 3×F – 2 = 4 Add 2 to both sides of the above equation: 3×F – 2 + 2 = 4 + 2 which becomes 3×F = 6 Divide both sides of the above equation by 3: 3×F ÷ 3 = 6 ÷ 3 which makes F = 2


  

Hint #9


Substitute 2 for F in eq.5b: 2×2 – 2 = C which becomes 4 – 2 = C which makes 2 = C


  

Hint #10


Substitute 2 for both C and F in eq.2: 2 + D = B + 2 Subtract 2 from each side of the equation above: 2 + D – 2 = B + 2 – 2 which makes D = B


  

Hint #11


Substitute 6 for A, and 2 for C and F in eq.4: 6 – 2 + E = 2 – E + 2 which becomes 4 + E = 4 – E In the equation above, subtract 4 from both sides, and add E to both sides: 4 + E – 4 + E = 4 – E – 4 + E which makes 2×E = 0 which means E = 0


  

Solution

Substitute B for D, and 0 for E in eq.6c: 10 = B + B + 0 which makes 10 = 2×B Divide both sides of the above equation by 2: 10 ÷ 2 = 2×B ÷ 2 which makes 5 = B and makes D = B = 5 and makes ABCDEF = 652502