Puzzle for April 21, 2019 ( )
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.
* BC and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
Add C + F to both sides of eq.3: C - F + C + F = A - C + C + F which becomes eq.3a) 2×C = A + F
Hint #2
Subtract the left and right sides of eq.5 from the left and right sides of eq.2, respectively: B + C - (B - C) = D + E - (E - A) which is equivalent to B + C - B + C = D + E - E + A which becomes eq.2a) 2×C = D + A
Hint #3
In eq.3a, replace 2×C with D + A (from eq.2a): D + A = A + F Subtract A from each side of the above equation: D + A - A = A + F - A which makes D = F
Hint #4
In eq.4, replace F with D: A - D = D - B + C - D which becomes A - D = -B + C Multiply both sides of the above equation by (-1): (-1) × (A - D) = (-1) × (-B + C) which becomes eq.4a) -A + D = B - C
Hint #5
In eq.4a, replace B - C with E - A (from eq.5): -A + D = E - A Add A to both sides: -A + D + A = E - A + A which makes D = E and also makes F = D = E
Hint #6
In eq.2, substitute E for D: B + C = E + E which is the same as B + C = 2×E Add the left and right sides of the above equation to the left and right sides of eq.5, respectively: B - C + B + C = E - A + 2×E which becomes eq.5a) 2×B = 3×E - A
Hint #7
eq.6 may be written as: 10×D + E = A + 10×B + C which may be written as 10×D + E = A + 5×(2×B) + C Substitute E for D, and 3×E - A for 2×B (from eq.5a) in the above equation: 10×E + E = A + 5×(3×E - A) + C which becomes 11×E = A + 15×E - 5×A + C which becomes 11×E = 15×E - 4×A + C Add (4×A - 11×E) to both sides: 11×E + (4×A - 11×E) = A + 15×E - 5×A + C + (4×A - 11×E) which simplifies to eq.6a) 4×A = 4×E + C
Hint #8
Multiply both sides of eq.6a by 2: 2×4×A = 2×(4×E + C) which becomes 8×A = 8×E + 2×C Substitute D for E, and D + A for 2×C (from eq.2a): 8×A = 8×D + D + A Subtract A from each side: 8×A - A = 8×D + D + A - A which becomes eq.6b) 7×A = 9×D
Hint #9
Multiply both sides of eq.2a by 9: 9×2×C = 9×(D + A) which is equivalent to 18×C = 9×D + 9×A Substitute 7×A for 9×D (from eq.6a) in the above equation: 18×C = 7×A + 9×A which makes 18×C = 16×A Divide both sides by 16: 18×C ÷ 16 = 16×A ÷ 16 which makes 1⅛×C = A
Hint #10
Substitute 1⅛×C for A in eq.6b: 7×1⅛×C = 9×D which means 7⅞×C = 9×D Divide both sides by 9: 7⅞×C ÷ 9 = 9×D ÷ 9 which makes ⅞×C = D and which also makes F = E = D = ⅞×C
Hint #11
Substitute ⅞×C for D and E in eq.2: B + C = ⅞×C + ⅞×C which becomes B + C = 1¾×C Subtract C from each side of the above equation: B + C - C = 1¾×C - C which makes B = ¾×C
Solution
Substitute 1⅛×C for A, ¾×C for B, and ⅞×C for D and E and F in eq.1: 1⅛×C + ¾×C + C + ⅞×C + ⅞×C + ⅞×C = 44 which simplifies to 5½×C = 44 Divide each side by 5½: 5½×C ÷ 5½ = 44 ÷ 5½ which means C = 8 making A = 1⅛×C = 1⅛ × 8 = 9 B = ¾×C = ¾ × 8 = 6 D = E = F = ⅞×C = ⅞ × 8 = 7 and ABCDEF = 968777