Puzzle for July 4, 2020  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) E = B – C eq.3) B + C = A + F eq.4)* CD = A + B eq.5)* AB = BC + CD

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB, BC, and CD are 2-digit numbers (not A×B, B×C, or C×D).

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Hint #1


eq.5 may be written as: 10×A + B = 10×B + C + 10×C + D which becomes 10×A + B = 10×B + 11×C + D Subtract B from both sides of the above equation: 10×A + B – B = 10×B + 11×C + D – B which becomes eq.5a) 10×A = 9×B + 11×C + D


  

Hint #2


Multiply both sides of eq.3 by 10: 10×(B + C) = 10×(A + F) which is equivalent to 10×B + 10×C = 10×A + 10×F In the equation above, replace 10×A with 9×B + 11×C + D (from eq.5a): 10×B + 10×C = 9×B + 11×C + D + 10×F Subtract 9×B and 11×C from both sides: 10×B + 10×C – 9×B – 11×C = 9×B + 11×C + D + 10×F – 9×B – 11×C which becomes eq.3a) B – C = D + 10×F


  

Hint #3


In eq.3a, replace B – C with E (from eq.2): eq.3b) E = D + 10×F   To make eq.3b true, check several possible values for F and E:   If F = 0, then E = D + 10×0 = D + 0 which means E = D If F = 1, then E = D + 10×1 = D + 10 which makes E ≥ 10 (since D ≥ 0) If F > 1, then E > 10   Since E must be a one-digit integer, then E < 10 and therefore makes E = D and F = 0


  

Hint #4


In eq.2, substitute D for E: D = B – C Add C to both sides of the equation above: D + C = B – C + C which becomes D + C = B which may be written as eq.2a) C + D = B


  

Hint #5


eq.4 may be written as: 10×C + D = A + B which is equivalent to 9×C + C + D = A + B Substitute B for C + D (from eq.2a) in the above equation: 9×C + B = A + B Subtract B from each side of the equation above: 9×C + B – B = A + B – B which makes 9×C = A


  

Hint #6


Substitute 9×C for A, and 0 for F in eq.3: B + C = 9×C + 0 Subtract C from both sides of the equation above: B + C – C = 9×C + 0 – C which makes B = 8×C


  

Hint #7


Substitute 8×C for B in eq.2a: C + D = 8×C Subtract C from each side of the equation above: C + D – C = 8×C – C which makes D = 7×C and also makes E = D = 7×C


  

Solution

Substitute 9×C for A, 8×C for B, 7×C for D and E, and 0 for F in eq.1: 9×C + 8×C + C + 7×C + 7×C + 0 = 32 which simplifies to 32×C = 32 Divide both sides of the equation above by 32: 32×C ÷ 32 = 32 ÷ 32 which means C = 1 making A = 9×C = 9 × 1 = 9 B = 8×C = 8 × 1 = 8 D = E = 7×C = 7 × 1 = 7 and ABCDEF = 981770