Puzzle for April 1, 2020  ( )

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

eq.1) A + B + C + D + E + F = 21 eq.2) C + D = E eq.3) D + F = A eq.4) F – E = B + C + D + E eq.5) E + F – B = B eq.6) A – D – E = B + D + E

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 C + D with E (from eq.2): F – E = B + E + E which becomes F – E = B + 2×E Add E to both sides of the above equation: F – E + E = B + 2×E + E which becomes eq.4a) F = B + 3×E


  

Hint #2


In eq.6, replace A with D + F (from eq.3): D + F – D – E = B + D + E which becomes F – E = B + D + E Add E to both sides of the equation above: F – E + E = B + D + E + E which becomes eq.6a) F = B + D + 2×E


  

Hint #3


In eq.6a, substitute B + 3×E for F (from eq.4a): B + 3×E = B + D + 2×E Subtract B and 2×E from each side of the above equation: B + 3×E – B – 2×E = B + D + 2×E – B – 2×E which simplifies to E = D


  

Hint #4


Substitute E for D in eq.2: C + E = E Subtract E from both sides of the equation above: C + E – E = E – E which makes C = 0


  

Hint #5


In eq.5, add B to both sides, and subtract E from each side: E + F – B + B – E = B + B – E which becomes F = 2×B – E Substitute 2×B – E for F in eq.4a: 2×B – E = B + 3×E In the above equation, add E to both sides, and subtract B from both sides: 2×B – E + E – B = B + 3×E + E – B which makes B = 4×E


  

Hint #6


Substitute 4×E for B in eq.4a: F = 4×E + 3×E which makes F = 7×E


  

Hint #7


Substitute E for D, and 7×E for F in eq.3: E + 7×E = A which makes 8×E = A


  

Solution

Substitute 8×E for A, 4×E for B, 0 for C, E for D, and 7×E for F in eq.1: 8×E + 4×E + 0 + E + E + 7×E = 21 which simplifies to 21×E = 21 Divide both sides of the equation above by 21: 21×E ÷ 21 = 21 ÷ 21 which means E = 1 making A = 8×E = 8 × 1 = 8 B = 4×E = 4 × 1 = 4 D = E = 1 F = 7×E = 7 × 1 = 7 and ABCDEF = 840117