Puzzle for August 6, 2023  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 18 eq.2) D + F = A + C eq.3) B + D = E - D + F eq.4) C - A = A + B - F eq.5) E + F - A = A + C + D - F eq.6) D = average (C, E, F)

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

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Hint #1


eq.6 may be written as: D = (C + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (C + E + F) ÷ 3 which becomes eq.6a) 3×D = C + E + F


  

Hint #2


Add D to both sides of eq.3: B + D + D = E - D + F + D which becomes eq.3a) B + 2×D = E + F


  

Hint #3


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


  

Hint #4


In eq.2, replace D with C + B (from eq.6b): C + B + F = A + C Subtract C from both sides of the above equation: C + B + F - C = A + C - C which becomes eq.2a) B + F = A


  

Hint #5


In eq.4, substitute (B + F) for A (from eq.2a): C - (B + F) = (B + F) + B - F which becomes C - B - F = 2×B Add B and F to both sides of the above equation: C - B - F + B + F = 2×B + B + F which becomes eq.4a) C = 3×B + F


  

Hint #6


Substitute 3×B + F for C (from eq.4a) into eq.6b: D = 3×B + F + B which becomes eq.6c) D = 4×B + F


  

Hint #7


Substitute (4×B + F) for D (from eq.6c) in eq.3a: B + 2×(4×B + F) = E + F which becomes B + 8×B + 2×F = E + F which becomes 9×B + 2×F = E + F Subtract F from each side of the equation above: 9×B + 2×F - F = E + F - F which becomes eq.3b) 9×B + F = E


  

Hint #8


Substitute 9×B + F for E (from eq.3b), (B + F) for A (from eq.2a), 3×B + F for C (from eq.4a), and 4×B + F for D (from eq.6c) in eq.5: 9×B + F + F - (B + F) = (B + F) + 3×B + F + 4×B + F - F which becomes 9×B + 2×F - B - F = 8×B + 2×F which becomes 8×B + F = 8×B + 2×F Subtract 8×B and F from each side of the above equation: 8×B + F - 8×B - F = 8×B + 2×F - 8×B - F which simplifies to 0 = F


  

Hint #9


Substitute 0 for F in eq.2a: B + 0 = A which makes B = A


  

Hint #10


Substitute 0 for F in eq.4a: C = 3×B + 0 which makes C = 3×B


  

Hint #11


Substitute 0 for F in eq.6c: D = 4×B + 0 which makes D = 4×B


  

Hint #12


Substitute 0 for F in eq.3b: 9×B + 0 = E which makes 9×B = E


  

Solution

Substitute B for A, 3×B for C, 4×B for D, 9×B for E, 0 for F in eq.1: B + B + 3×B + 4×B + 9×B + 0 = 18 which simplifies to 18×B = 18 Divide both sides of the above equation by 18: 18×B ÷ 18 = 18 ÷ 18 which means B = 1 making A = B = 1 C = 3×B = 3 × 1 = 3 D = 4×B = 4 × 1 = 4 E = 9×B = 9 × 1 = 9 and ABCDEF = 113490