Puzzle for July 31, 2019  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) D – F = A + B eq.3) C + F = D + E eq.4) B + E = D eq.5) E – F = B – A eq.6) A + B + D = C + E + F

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

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Hint #1


In eq.2, replace D with B + E (from eq.4): B + E – F = A + B Subtract B from both sides of the equation above: B + E – F – B = A + B – B which becomes eq.2a) E – F = A


  

Hint #2


In eq.5, replace E – F with A (from eq.2a): A = B – A Add A to both sides of the above equation: A + A = B – A + A which makes 2×A = B


  

Hint #3


eq.6 may be written as: A + B + D = E + C + F In the above equation, replace B with 2×A, and replace C + F with D + E (from eq.3): A + 2×A + D = E + D + E which becomes 3×A + D = D + 2×E Subtract D from each side: 3×A + D – D = D + 2×E – D which becomes 3×A = 2×E Divide both sides by 2: 3×A ÷ 2 = 2×E ÷ 2 which makes 1½×A = E


  

Hint #4


In eq.2a, substitute 1½×A for E: 1½×A – F = A In the above equation, add F to both sides, and subtract A from both sides: 1½×A – F + F – A = A + F – A which makes ½×A = F


  

Hint #5


Substitute 2×A for B, and 1½×A for E in eq.4: 2×A + 1½×A = D which makes 3½×A = D


  

Hint #6


Substitute ½×A for F, 3½×A for D, and 1½×A for E in eq.3: C + ½×A = 3½×A + 1½×A which becomes C + ½×A = 5×A Subtract ½×A from both sides: C + ½×A – ½×A = 5×A – ½×A which makes C = 4½×A


  

Solution

Substitute 2×A for B, 4½×A for C, 3½×A for D, 1½×A for E, and ½×A for F in eq.1: A + 2×A + 4½×A + 3½×A + 1½×A + ½×A = 26 which simplifies to 13×A = 26 Divide both sides of the above equation by 13: 13×A ÷ 13 = 26 ÷ 13 which means A = 2 making B = 2×A = 2 × 2 = 4 C = 4½×A = 4½ × 2 = 9 D = 3½×A = 3½ × 2 = 7 E = 1½×A = 1½ × 2 = 3 F = ½×A = ½ × 2 = 1 and ABCDEF = 249731