Puzzle for May 5, 2020  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) A + B = D eq.3) B + E = F eq.4) D – A = E eq.5) C – F = A eq.6) E + F = A + B + D – F

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

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Hint #1


In eq.4, replace D with A + B (from eq.2): A + B – A = E which becomes B = E


  

Hint #2


In eq.3, replace E with B: B + B = F which becomes 2×B = F


  

Hint #3


In eq.4, add A to both sides, and subtract E from both sides: D – A + A – E = E + A – E which becomes D – E = A In the above equation, substitute B for E: eq.4a) D – B = A


  

Hint #4


Substitute B for E, 2×B for F, and D – B for A (from eq.4a) in eq.6: B + 2×B = D – B + B + D – 2×B which becomes 3×B = 2×D – 2×B Add 2×B to both sides of the equation above: 3×B + 2×B = 2×D – 2×B + 2×B which makes 5×B = 2×D Divide both sides by 2: 5×B ÷ 2 = 2×D ÷ 2 which makes 2½×B = D


  

Hint #5


Substitute 2½×B for D in eq.4a: 2½×B – B = A which makes 1½×B = A


  

Hint #6


Substitute 1½×B for A, and 2×B for F in eq.5: C – 2×B = 1½×B Add 2×B to each side of the above equation: C – 2×B + 2×B = 1½×B + 2×B which makes C = 3½×B


  

Solution

Substitute 1½×B for A, 3½×B for C, 2½×B for D, B for E, and 2×B for F in eq.1: 1½×B + B + 3½×B + 2½×B + B + 2×B = 23 which simplifies to 11½×B = 23 Divide both sides of the above equation by 11½: 11½×B ÷ 11½ = 23 ÷ 11½ which means B = 2 making A = 1½×B = 1½ × B = 3 C = 3½×B = 3½ × B = 7 D = 2½×B = 2½ × B = 5 E = B = 2 F = 2×B = 2 × B = 4 and ABCDEF = 327524