Puzzle for January 24, 2019  ( )

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

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

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Hint #1


Subtract F from both sides of eq.4: B - C - F = D + E + F - F which becomes B - C - F = D + E In eq.2, replace D + E with B - C - F: A + C = B - C - F Add (F - C) to both sides of the above equation: A + C + (F - C) = B - C - F + (F - C) which simplifies to eq.2a) A + F = B - 2×C


  

Hint #2


eq.3 may be written as: B = D + A + F In the above equation, replace A + F with B - 2×C (from eq.2a): B = D + B - 2×C Add (2×C - B) to both sides: B + (2×C - B) = D + B - 2×C + (2×C - B) which simplifies to 2×C = D


  

Hint #3


In eq.4, replace E + F with C + D (from eq.5): B - C = D + C + D Add C to each side of the above equation: B - C + C = D + C + D + C which means B = 2×D + 2×C Replace D with (2×C): B = 2×(2×C) + 2×C which becomes B = 4×C + 2×C which makes B = 6×C


  

Hint #4


Add C + F to both sides of eq.6: A - C - F + C + F = F + C + F which becomes eq.6a) A = C + 2×F


  

Hint #5


Substitute C + 2×F for A (from eq.6a), and 6×C for B in eq.2a: C + 2×F + F = 6×C - 2×C which becomes C + 3×F = 4×C Subtract C from both sides of the above equation: C + 3×F - C = 4×C - C which makes 3×F = 3×C Divide both sides by 3: 3×F ÷ 3 = 3×C ÷ 3 which makes F = C


  

Hint #6


In eq.6a, substitute C for F: A = C + 2×C which makes A = 3×C


  

Hint #7


Substitute 3×C for A, and 2×C for D in eq.2: 3×C + C = 2×C + E Subtract 2×C from each side: 3×C + C - 2×C = 2×C + E - 2×C which makes 2×C = E


  

Solution

Substitute 3×C for A, 6×C for B, 2×C for D and E, and C for F in eq.1: 3×C + 6×C + C + 2×C + 2×C + C = 15 which simplifies to 15×C = 15 Divide both sides by 15: 15×C ÷ 15 = 15 ÷ 15 which makes C = 1 making A = 3×C = 3 × 1 = 3 B = 6×C = 6 × 1 = 6 D = E = 2×C = 2 × 1 = 2 F = C = 1 and ABCDEF = 361221