Puzzle for August 17, 2021  ( )

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

eq.1) A + B + C + D + E + F = 15 eq.2) F – C = D eq.3) C + E = B – C eq.4) D + F – C = A + C – D eq.5)* EF = BC + D eq.6)* BC – E = AB + B + E

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

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Hint #1


Add C to both sides of eq.2: F – C + C = D + C which becomes F = D + C which may be written as: eq.2a) F = C + D   eq.5 may be written as: eq.5a) 10×E + F = 10×B + C + D


  

Hint #2


In eq.5a, replace C + D with F (from eq.2a): 10×E + F = 10×B + F Subtract F from each side of the equation above: 10×E + F – F = 10×B + F – F which becomes 10×E = 10×B Divide both sides by 10: 10×E ÷ 10 = 10×B ÷ 10 which makes E = B


  

Hint #3


In eq.3, replace E with B: C + B = B – C In the equation above, add C to each side, and subtract B from each side: C + B + C – B = B – C + C – B which simplifies to 2×C = 0 which means C = 0


  

Hint #4


In eq.2, substitute 0 for C: F – 0 = D which makes F = D


  

Hint #5


Substitute D for F, and 0 for C in eq.4: D + D – 0 = A + 0 – D which becomes 2×D = A – D Add D to both sides of the above equation: 2×D + D = A – D + D which makes 3×D = A


  

Hint #6


eq.6 may be written as: 10×B + C – E = 10×A + B + B + E which becomes 10×B + C – E = 10×A + 2×B + E Substitute 0 for C, B for E, and (3×D) for A in the equation above: 10×B + 0 – B = 10×(3×D) + 2×B + B which becomes 9×B = 30×D + 3×B Subtract 3×B from each side: 9×B – 3×B = 30×D + 3×B – 3×B which becomes 6×B = 30×D Divide both sides by 6: 6×B ÷ 6 = 30×D ÷ 6 which makes B = 5×D and also makes E = B = 5×D


  

Solution

Substitute 3×D for A, 5×D for B and E, 0 for C, and D for F in eq.1: 3×D + 5×D + 0 + D + 5×D + D = 15 which simplifies to 15×D = 15 Divide both sides of the above equation by 15: 15×D ÷ 15 = 15 ÷ 15 which means D = 1 making A = 3×D = 3 × 1 = 3 B = E = 5×D = 5 × 1 = 5 F = D = 1 and ABCDEF = 350151