Puzzle for December 28, 2018 ( )
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
In eq.2, replace D + F with E (from eq.4): A + B = C + E In eq.3, replace C + E with A + B: B + D = A + B Subtract B from each side: B + D - B = A + B - B which makes D = A
Hint #2
Replace D with A in eq.2: A + B = C + A + F Subtract A from both sides: A + B - A = C + A + F - A which becomes eq.2a) B = C + F
Hint #3
eq.6 may be written as: 10×B + C + F = 10×D + E - F Add F to both sides of the equation above: 10×B + C + F + F = 10×D + E - F + F which becomes eq.6a) 10×B + C + F + F = 10×D + E
Hint #4
In eq.6a, substitute D + F for E (from eq.4), and B for C + F (from eq.2a): 10×B + B + F = 10×D + D + F which becomes 11×B + F = 11×D + F Subtract F from each side: 11×B + F - F = 11×D + F - F which simplifies to 11×B = 11×D Divide both sides by 11: 11×B ÷ 11 = 11×D ÷ 11 which means B = D
Hint #5
Substitute D for B, and D + F for E (from eq.4) in eq.5: D + C - F = D + F + F - C which becomes D + C - F = D + 2×F - C Add (C - D + F) to both sides of the above equation: D + C - F + (C - D + F) = D + 2×F - C + (C - D + F) which simpifies to 2×C = 3×F Divide both sides by 2: 2×C ÷ 2 = 3×F ÷ 2 which makes C = 1½×F
Hint #6
Substitute 1½×F for C in eq.2a: B = 1½×F + F = 2½×F which means A = D = B = 2½×F
Hint #7
In eq.4, replace D with 2½×F: F + 2½×F = E which makes 3½×F = E
Solution
Substitute 2½×F for A and B and D, 1½×F for C, and 3½×F for E in eq.1: 2½×F + 2½×F + 1½×F + 2½×F + 3½×F + F = 27 which simplifies to 13½×F = 27 Divide both sides by 13½: 13½×F ÷ 13½ = 27 ÷ 13½ which makes F = 2 making A = B = D = 2½×F = 2½ × 2 = 5 C = 1½×F = 1½ × 2 = 3 E = 3½×F = 3½ × 2 = 7 and ABCDEF = 553572