Puzzle for April 29, 2021  ( )

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

eq.1) A + B + C + D + E + F = 34 eq.2) D + E = A eq.3) A + E = B eq.4) B + D + E = C + F eq.5) C – D + F = A + D eq.6) B + F = A + C + E

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

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Hint #1


Add D to both sides of eq.5: C – D + F + D = A + D + D which becomes eq.5a) C + F = A + 2×D   In eq.4, replace D + E with A (from eq.2): eq.4a) B + A = C + F


  

Hint #2


In eq.4a, replace C + F with A + 2×D (from eq.5a): B + A = A + 2×D Subtract A from both sides of the above equation: B + A – A = A + 2×D – A which makes B = 2×D


  

Hint #3


Subtract E from both sides of eq.3: A + E – E = B – E which becomes A = B – E In eq.2, substitute B – E for A: D + E = B – E Add E to both sides of the equation above: D + E + E = B – E + E which becomes eq.2a) D + 2×E = B


  

Hint #4


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


  

Hint #5


Substitute 2×E for D in eq.2a: 2×E + 2×E = B which makes 4×E = B


  

Hint #6


Substitute 2×E for D in eq.2: 2×E + E = A which makes 3×E = A


  

Hint #7


Substitute 4×E for B, and 3×E for A in eq.6: 4×E + F = 3×E + C + E which becomes 4×E + F = 4×E + C Subtract 4×E from both sides of the equation above: 4×E + F – 4×E = 4×E + C – 4×E which becomes F = C


  

Hint #8


Substitute 4×E for B, 3×E for A, and C for F in eq.4a: 4×E + 3×E = C + C which makes 7×E = 2×C Divide both sides of the above equation by 2: 7×E ÷ 2 = 2×C ÷ 2 which makes 3½×E = C and also makes F = C = 3½×E


  

Solution

Substitute 3×E for A, 4×E for B, 3½×E for C and F, and 2×E for D in eq.1: 3×E + 4×E + 3½×E + 2×E + E + 3½×E = 34 which simplifies to 17×E = 34 Divide both sides of the above equation by 17: 17×E ÷ 17 = 34 ÷ 17 which means E = 2 making A = 3×E = 3 × 2 = 6 B = 4×E = 4 × 2 = 8 C = F = 3½×E = 3½ × 2 = 7 D = 2×E = 2 × 2 = 4 and ABCDEF = 687427