Puzzle for September 28, 2021  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) B = C + F eq.3) C + E = A + B eq.4) D + F = B + E eq.5) E – B = B – A eq.6) F – A = C – B

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

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Hint #1


In eq.6, substitute (C + F) for B (from eq.2): F – A = C – (C + F) which becomes F – A = C – C – F which becomes F – A = –F Add A and F to both sides of the above equation: F – A + A + F = –F + A + F which makes 2×F = A


  

Hint #2


In eq.3, replace A with 2×F, and B with C + F (from eq.2): C + E = 2×F + C + F which becomes C + E = 3×F + C Subtract C from each side of the equation above: C + E – C = 3×F + C – C which makes E = 3×F


  

Hint #3


In eq.5, replace E with 3×F, and A with 2×F: 3×F – B = B – 2×F Add B and 2×F to both sides of the equation above: 3×F – B + B + 2×F = B – 2×F + B + 2×F which becomes 5×F = 2×B Divide both sides by 2: 5×F ÷ 2 = 2×B ÷ 2 which makes 2½×F = B


  

Hint #4


In eq.3, substitute 3×F for E, 2×F for A, and 2½×F for B: C + 3×F = 2×F + 2½×F which becomes C + 3×F = 4½×F Subtract 3×F from each side of the above equation: C + 3×F – 3×F = 4½×F – 3×F which makes C = 1½×F


  

Hint #5


Substitute 2½×F for B, and 3×F for E in eq.4: D + F = 2½×F + 3×F which becomes D + F = 5½×F Subtract F from each side of the equation above: D + F – F = 5½×F – F which makes D = 4½×F


  

Solution

Substitute 2×F for A, 2½×F for B, 1½×F for C, 4½×F for D, and 3×F for E in eq.1: 2×F + 2½×F + 1½×F + 4½×F + 3×F + F = 29 which simplifies to 14½×F = 29 Divide both sides of the above equation by 14½: 14½×F ÷ 14½ = 29 ÷ 14½ which means F = 2 making A = 2×F = 2 × 2 = 4 B = 2½×F = 2½ × 2 = 5 C = 1½×F = 1½ × 2 = 3 D = 4½×F = 4½ × 2 = 9 E = 3×F = 3 × 2 = 6 and ABCDEF = 453962