Puzzle for February 5, 2019  ( )

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

eq.1) A + B + C + D + E + F = 24 eq.2) A = C + D eq.3) A + C = B + E + F eq.4) C + E + F = A + B + D eq.5) B + C = D + F eq.6) A - B = E - C

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

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Hint #1


Add B + C to both sides of eq.6: A - B + B + C = E - C + B + C which becomes A + C = E + B which may be written as A + C = B + E In eq.3, replace B + E with A + C: A + C = A + C + F Subtract both A and C from each side of the equation above: A + C - A - C = A + C + F - A - C which simplifies to 0 = F


  

Hint #2


In eq.5, replace F with 0: B + C = D + 0 which becomes B + C = D Substitute B + C for D, and 0 for F in eq.4: C + E + 0 = A + B + B + C which becomes C + E = A + 2×B + C Subtract C from both sides of the above equation: C + E - C = A + 2×B + C - C which becomes eq.4a) E = A + 2×B


  

Hint #3


In eq.6, substitute A + 2×B for E (from eq.4a): A - B = A + 2×B - C Subtract A from both sides of the above equation: A - B - A = A + 2×B - C - A which becomes -B = 2×B - C Add both B and C to each side: -B + B + C = 2×B - C + B + C which makes C = 3×B


  

Hint #4


Substitute 3×B for C, and 0 for F in eq.5: B + 3×B = D + 0 which makes 4×B = D


  

Hint #5


Substitute 3×B for C, and 4×B for D in eq.2: A = 3×B + 4×B which makes A = 7×B


  

Hint #6


Substitute 7×B for A, 3×B for C, and 0 for F in eq.3: 7×B + 3×B = B + E + 0 which becomes 10×B = B + E Subtract B from each side of the above equation: 10×B - B = B + E - B which means 9×B = E


  

Solution

Substitute 7×B for A, 3×B for C, 4×B for D, 9×B for E, and 0 for F in eq.1: 7×B + B + 3×B + 4×B + 9×B + 0 = 24 which becomes 24×B = 24 Divide both sides by 24: 24×B ÷ 24 = 24 ÷ 24 which means B = 1 making A = 7×B = 7 × 1 = 7 C = 3×B = 3 × 1 = 3 D = 4×B = 4 × 1 = 4 E = 9×B = 9 × 1 = 9 and ABCDEF = 713490