Puzzle for May 19, 2022  ( )

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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) B + F = A eq.3) E + F = B eq.4) A – B + E = B + C eq.5) B – C + E = A + D eq.6) C + D + E = 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.4, replace A with B + F (from eq.2): B + F – B + E = B + C which becomes F + E = B + C which may be written as eq.4a) E + F = B + C


  

Hint #2


In eq.4a, replace E + F with B (from eq.3): B = B + C Subtract B from both sides of the above equation: B – B = B + C – B which makes 0 = C


  

Hint #3


In eq.5, replace C with 0, and A with B + F (from eq.2): B – 0 + E = B + F + D Subtract B from both sides of the above equation: B – 0 + E – B = B + F + D – B which becomes eq.5a) E = F + D


  

Hint #4


In eq.3, substitute F + D for E (from eq.5a): F + D + F = B which becomes eq.3a) 2×F + D = B


  

Hint #5


Substitute 2×F + D for B (from eq.3a) in eq.2: 2×F + D + F = A which becomes eq.2a) 3×F + D = A


  

Hint #6


In eq.6, substitute 0 for C, F + D for E (from eq.5a), 3×F + D for A (from eq.2a), and 2×F + D for B (from eq.3a): 0 + D + F + D = 3×F + D + 2×F + D – D which becomes 2×D + F = 5×F + D Subtract F and D from both sides of the above equation: 2×D + F – F – D = 5×F + D – F – D which makes D = 4×F


  

Hint #7


Substitute 4×F for D in eq.5a: E = F + 4×F which makes E = 5×F


  

Hint #8


Substitute 4×F for D in eq.3a: 2×F + 4×F = B which makes 6×F = B


  

Hint #9


Substitute 4×F for D in eq.2a: 3×F + 4×F = A which makes 7×F = A


  

Solution

Substitute 7×F for A, 6×F for B, 0 for C, 4×F for D, and 5×F for E in eq.1: 7×F + 6×F + 0 + 4×F + 5×F + F = 23 which simplifies to 23×F = 23 Divide both sides of the above equation by 23: 23×F ÷ 23 = 23 ÷ 23 which means F = 1 making A = 7×F = 7 × 1 = 7 B = 6×F = 6 × 1 = 6 D = 4×F = 4 × 1 = 4 E = 5×F = 5 × 1 = 5 and ABCDEF = 760451