Puzzle for February 14, 2023  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) D = A + E eq.3) B + C = E + F eq.4) E - B = B - F eq.5) D - B = A + C eq.6) C - A = (E ÷ B) + F

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

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Hint #1


Subtract B and F from both sides of eq.3: B + C - B - F = E + F - B - F which becomes eq.3a) C - F = E - B


  

Hint #2


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


  

Hint #3


In eq.5, replace D with A + E (from eq.2), and C with B: A + E - B = A + B In the above equation, subtract A from both sides, and add B to both sides: A + E - B - A + B = A + B - A + B which makes E = 2×B


  

Hint #4


In eq.4, substitute 2×B for E: 2×B - B = B - F which becomes B = B - F Subtract B from both sides of the equation above: B - B = B - F - B which makes 0 = -F which means 0 = F


  

Hint #5


Substitute B for C, 2×B for E, and 0 for F in eq.6: B - A = (2×B ÷ B) + 0 which becomes B - A = 2 + 0 which becomes B - A = 2 In the above equation, add A to both sides, and subtract 2 from both sides: B - A + A - 2 = 2 + A - 2 which makes eq.6a) B - 2 = A


  

Hint #6


Substitute B - 2 for A (from eq.6a), and 2×B for E in eq.2: D = B - 2 + 2×B which becomes eq.2a) D = 3×B - 2


  

Hint #7


Substitute B - 2 for A (from eq.6a), B for C, 3×B - 2 for D (from eq.2a), 2×B for E, and 0 for F in eq.1: B - 2 + B + B + 3×B - 2 + 2×B + 0 = 20 which simplifies to 8×B - 4 = 20 Add 4 to both sides of the above equation: 8×B - 4 + 4 = 20 + 4 which makes 8×B = 24 Divide both sides by 8: 8×B ÷ 8 = 24 ÷ 8 which means B = 3


  

Solution

Since B = 3, then: A = B - 2 = 3 - 2 = 1 (from eq.6a) C = B = 3 D = 3×B - 2 = (3 × 3) - 2 = 9 - 2 = 7 (from eq.2a) E = 2×B = 2 × 3 = 6 and ABCDEF = 133760