Puzzle for April 4, 2022  ( )

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

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


  

Hint #2


In eq.2, replace D with 2×F: C = 2×F + F which makes C = 3×F


  

Hint #3


In eq.4, replace D with 2×F: B + F = 2×F – F which becomes B + F = F Subtract F from both sides of the above equation: B + F – F = F – F which makes B = 0


  

Hint #4


In eq.5, substitute 2×F for D, and 0 for B: A + F = 2×F – A + 0 which becomes A + F = 2×F – A In the above equation, add A to both sides, and subtract F from both sides: A + F + A – F = 2×F – A + A – F which makes 2×A = F Divide both sides by 2: 2×A ÷ 2 = F ÷ 2 which makes A = ½×F


  

Hint #5


Substitute 2×F for D, ½×F for A, and 0 for B in eq.6: E – 2×F = ½×F – 0 + 2×F which becomes E – 2×F = 2½×F Add 2×F to both sides of the equation above: E – 2×F + 2×F = 2½×F + 2×F which becomes E = 4½×F


  

Solution

Substitute ½×F for A, 0 for B, 3×F for C, 2×F for D, and 4½×F for E in eq.1: ½×F + 0 + 3×F + 2×F + 4½×F + F = 22 which simplifies to 11×F = 22 Divide both sides of the above equation by 11: 11×F ÷ 11 = 22 ÷ 11 which means F = 2 making A = ½×F = ½ × 2 = 1 C = 3×F = 3 × 2 = 6 D = 2×F = 2 × 2 = 4 E = 4½×F = 4½ × 2 = 9 and ABCDEF = 106492