Puzzle for September 9, 2020  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) B = D + E eq.3) A + E + F = B + D eq.4) C + D = A + F – (B – C) eq.5) D + F = A + B eq.6) C – D + E = A – C + F

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

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Hint #1


In eq.3, replace B with D + E (from eq.2): A + E + F = D + E + D which becomes A + E + F = 2×D + E Subtract E from both sides of the equation above: A + E + F – E = 2×D + E – E which becomes eq.3a) A + F = 2×D


  

Hint #2


In eq.4, replace A + F with 2×D (from eq.3a): C + D = 2×D – (B – C) which is equivalent to C + D = 2×D – B + C Subtract C and D from both sides of the above equation: C + D – C – D = 2×D – B + C – C – D which becomes 0 = D – B Add B to both sides: 0 + B = D – B + B which makes B = D


  

Hint #3


In eq.5, substitute B for D: B + F = A + B Subtract B from each side of the equation above: B + F – B = A + B – B which makes F = A


  

Hint #4


Substitute A for F in eq.3a: A + A = 2×D which makes 2×A = 2×D Divide both sides of the above equation by 2: 2×A ÷ 2 = 2×D ÷ 2 which makes A = D and makes F = A = D = B


  

Hint #5


Substitute B for D in eq.2: B = B + E Subtract B from each side of the above equation: B – B = B + E – B which means 0 = E


  

Hint #6


Substitute A for D and F, and 0 for E in eq.6: C – A + 0 = A – C + A which becomes C – A + 0 = 2×A – C Add A and C to both sides of the equation above: C – A + 0 + A + C = 2×A – C + A + C which makes 2×C = 3×A Divide both sides by 2: 2×C ÷ 2 = 3×A ÷ 2 which makes C = 1½×A


  

Solution

Substitute A for B and D and F, 1½×A for C, and 0 for E in eq.1: A + A + 1½×A + A + 0 + A = 22 which simplifies to 5½×A = 22 Divide both sides of the equation above by 5½: 5½×A ÷ 5½ = 22 ÷ 5½ which means A = 4 making B = D = F = A = 4 C = 1½×A = 1½ × 4 = 6 and ABCDEF = 446404