Puzzle for January 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 = 13 eq.2) C = B + E eq.3) D + E = A + B eq.4) E + F = A + C eq.5) B + C = D + E + F eq.6) A + D + F = C + E – A

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

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Hint #1


In eq.5, replace E + F with A + C (from eq.4): B + C = D + A + C Subtract C from each side of the above equation: B + C – C = D + A + C – C which becomes eq.5a) B = D + A


  

Hint #2


In eq.3, replace B with D + A (from eq.5a): D + E = A + D + A which becomes D + E = 2×A + D Subtract D from each side of the equation above: D + E – D = 2×A + D – D which makes E = 2×A


  

Hint #3


In eq.6, substitute 2×A for E: A + D + F = C + 2×A – A which becomes A + D + F = C + A Subtract A from both sides of the equation above: A + D + F – A = C + A – A which becomes eq.6a) D + F = C


  

Hint #4


eq.5 may be written as: B + C = D + F + E Substitute C for D + F (from eq.6a): B + C = C + E Subtract C from both sides of the above equation: B + C – C = C + E – C which makes B = E and also makes B = E = 2×A


  

Hint #5


Substitute 2×A for B in eq.5a: 2×A = D + A Subtract A from each side of the equation above: 2×A – A = D + A – A which makes A = D


  

Hint #6


Substitute 2×A for B and E in eq.2: C = 2×A + 2×A which makes C = 4×A


  

Hint #7


Substitute 2×A for E, and 4×A for C in eq.4: 2×A + F = A + 4×A which becomes 2×A + F = 5×A Subtract 2×A from both sides of the equation above: 2×A + F – 2×A = 5×A – 2×A which makes F = 3×A


  

Solution

Substitute 2×A for B and E, 4×A for C, A for D, and 3×A for F in eq.1: A + 2×A + 4×A + A + 2×A + 3×A = 13 which simplifies to 13×A = 13 Divide both sides of the above equation by 13: 13×A ÷ 13 = 13 ÷ 13 which means A = 1 making B = E = 2×A = 2×1 = 2 C = 4×A = 4×1 = 4 D = A = 1 F = 3×A = 3×1 = 3 and ABCDEF = 124123