Puzzle for February 15, 2020  ( )

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

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

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Hint #1


Add D and F to both sides of eq.2: B – F + D + F = A – D + D + F which becomes eq.2a) B + D = A + F   Add C and F to both sides of eq.4: D – F + C + F = A – C + C + F eq.4a) D + C = A + F


  

Hint #2


In eq.2a, replace A + F with D + C (from eq.4a): B + D = D + C Subtract D from both sides of the equation above: B + D – D = D + C – D which makes B = C


  

Hint #3


Add A and B to both sides of eq.6: F – A – B + A + B = A + B + A + B which becomes eq.6a) F = 2×A + 2×B   In eq.2a, replace F with 2×A + 2×B (from eq.6a): B + D = A + 2×A + 2×B which becomes B + D = 3×A + 2×B Subtract B from each side of the above equation: B + D – B = 3×A + 2×B – B which becomes eq.2b) D = 3×A + B


  

Hint #4


Add D to both sides of eq.5: E – D + D = A + D + D which becomes eq.5a) E = A + 2×D   Add F to each side of eq.3: C + D + F + F = E – F + F which becomes eq.3a) C + D + 2×F = E


  

Hint #5


Substitute B for C, (2×A + 2×B) for F (from eq.6a), and A + 2×D for E (from eq.5a) in eq.3a: B + D + 2×(2×A + 2×B) = A + 2×D which is the same as B + D + 4×A + 4×B = A + 2×D which becomes D + 4×A + 5×B = A + 2×D Subtract A and D from both sides of the above equation: D + 4×A + 5×B – A – D = A + 2×D – A – D which becomes eq.3b) 3×A + 5×B = D


  

Hint #6


Substitute 3×A + 5×B for D (from eq.3b) in eq.2b: 3×A + B = 3×A + 5×B Subtract 3×A and B from each side of the above equation: 3×A + B – 3×A – B = 3×A + 5×B – 3×A – B which makes 0 = 4×B which means 0 = B and also means C = B = 0


  

Hint #7


Substitute 0 for B in eq.6a: F = 2×A + 2×0 which makes F = 2×A


  

Hint #8


Substitute 0 for B in eq.2b: D = 3×A + 0 which makes D = 3×A


  

Hint #9


Substitute (3×A) for D in eq.5a: E = A + 2×(3×A) which is equivalent to E = A + 6×A which makes E = 7×A


  

Solution

Substitute 0 for B and C, 3×A for D, 7×A for E, and 2×A for F in eq.1: A + 0 + 0 + 3×A + 7×A + 2×A = 13 which simplifies to 13×A = 13 Divide both sides of the equation above by 13: 13×A ÷ 13 = 13 ÷ 13 which means A = 1 making D = 3×A = 3 × 1 = 3 E = 7×A = 7 × 1 = 7 F = 2×A = 2 × 1 = 2 and ABCDEF = 100372