Puzzle for November 13, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
Help Area
Hint #1
Subtract F and C from both sides of eq.2: D + F - F - C = A + C - F - C which becomes eq.2a) D - C = A - F Subtract F from both sides of eq.3: C + F - F = A + D + E - F which becomes C = A + D + E - F which may be written as eq.3a) C = A - F + D + E
Hint #2
In eq.3a, replace A - F with D - C (from eq.2a): C = D - C + D + E which becomes C = 2×D - C + E Add C to both sides of the above equation: C + C = 2×D - C + E + C which becomes eq.3b) 2×C = 2×D + E
Hint #3
eq.6 may be written as: A = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (C + D + F) ÷ 3 which becomes 3×A = C + D + F which may be written as eq.6a) 3×A = C + F + D
Hint #4
In eq.6a, replace C + F with A + D + E (from eq.3): 3×A = A + D + E + D which becomes 3×A = A + 2×D + E Subtract A and E from both sides of the equation above: 3×A - A - E = A + 2×D + E - A - E which becomes eq.6b) 2×A - E = 2×D
Hint #5
In eq.3b, substitute 2×A - E for 2×D (from eq.6b): 2×C = 2×A - E + E which means 2×C = 2×A Divide both sides of the above equation by 2: 2×C ÷ 2 = 2×A ÷ 2 which makes C = A
Hint #6
Substitute A for C in eq.3: A + F = A + D + E Subtract A from each side of the equation above: A + F - A = A + D + E - A which becomes eq.3c) F = D + E
Hint #7
Divide both sides of eq.6b by 2: (2×A - E) ÷ 2 = 2×D ÷ 2 which becomes A - ½×E = D Add ½×E to both sides of the above equation: A - ½×E + ½×E = D + ½×E which becomes eq.6c) A = D + ½×E
Hint #8
eq.5 may be written as: D = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (A + B + C) ÷ 3 which becomes eq.5a) 3×D = A + B + C
Hint #9
Substitute A for C in eq.5a: 3×D = A + B + A which becomes eq.5b) 3×D = 2×A + B
Hint #10
Substitute (D + ½×E) for A (from eq.6c) in eq.5b: 3×D = 2×(D + ½×E) + B which becomes 3×D = 2×D + E + B Subtract 2×D and E both sides of the equation above: 3×D - 2×D - E = 2×D + E + B - 2×D - E which becomes eq.5c) D - E = B
Hint #11
Substitute (D - E) for B (from eq.5c), D + ½×E for A (from eq.6c), and D + E for F (from eq.3c) in eq.4: (D - E) + D = D + ½×E - (D - E) + D + E which becomes 2×D - E = 2×D + 1½×E - D + E which becomes 2×D - E = D + 2½×E In the above equation, add E to both sides, and subtract D from both sides: 2×D - E + E - D = D + 2½×E + E - D which makes D = 3½×E
Hint #12
Substitute 3½×E for D in eq.6c: A = 3½×E + ½×E which makes A = 4×E and also makes C = A = 4×E
Hint #13
Substitute 3½×E for D in eq.5c: 3½×E - E = B which makes 2½×E = B
Hint #14
Substitute 3½×E for D in eq.3c: F = 3½×E + E which makes F = 4½×E
Solution
Substitute 4×E for A and C, 2½×E for B, 3½×E for D, and 4½×E for F in eq.1: 4×E + 2½×E + 4×E + 3½×E + E + 4½×E = 39 which simplifies to 19½×E = 39 Divide both sides of the above equation by 19½: 19½×E ÷ 19½ = 39 ÷ 19½ which means E = 2 making A = C = 4×E = 4 × 2 = 8 B = 2½×E = 2½ × 2 = 5 D = 3½×E = 3½ × 2 = 7 F = 4½×E = 4½ × 2 = 9 and ABCDEF = 858729