Puzzle for December 12, 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 EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
eq.6 may be written as: 10×B + C + D = A - B + 10×E + F Substitute B + C for D (from eq.4), and add B to both sides of the equation above: 10×B + C + B + C + B = A - B + 10×E + F + B which becomes 12×B + 2×C = A + 10×E + F From eq.5, substitute (B + F) for C: eq.6a) 12×B + 2×(B + F) = A + 10×E + F
Hint #2
In eq.6a, replace F with A + B (from eq.2): 12×B + 2×(B + A + B) = A + 10×E + A + B which is equivalent to 12×B + 2×B + 2×A + 2×B = 2×A + 10×E + B which becomes 16×B + 2×A = 2×A + 10×E + B Subtract both 2×A and B from each side: 16×B + 2×A - 2×A - B = 2×A + 10×E + B - 2×A - B which simplifies to 15×B = 10×E Divide both sides by 10: 15×B ÷ 10 = 10×E ÷ 10 which makes 1½×B = E
Hint #3
Substitute B + C for D (from eq.4) in eq.3: C - (B + C) = E - F which is equivalent to C - B - C = E - F which becomes -B = E - F Substitute 1½×B for E in the equation above: -B = 1½×B - F Add B + F to each side: -B + B + F = 1½×B - F + B + F which makes F = 2½×B
Hint #4
Substitute 2½×B for F in eq.5: C = B + 2½×B which makes C = 3½×B
Hint #5
Replace C with 3½×B in eq.4: B + 3½×B = D which makes 4½×B = D
Hint #6
Substitute 2½×B for F in eq.2: A + B = 2½×B Subtract B from both sides of the above equation: A + B - B = 2½×B - B which makes A = 1½×B
Solution
Substitute 1½×B for A and E, 3½×B for C, 4½×B for D, and 2½×B for F in eq.1: 1½×B + B + 3½×B + 4½×B + 1½×B + 2½×B = 29 which simplifies to 14½×B = 29 Divide each side of the above equation by 14½: 14½×B ÷ 14½ = 29 ÷ 14½ which makes B = 2 making A = E = 1½×B = 1½ × 2 = 3 C = 3½×B = 3½ × 2 = 7 D = 4½×B = 4½ × 2 = 9 F = 2½×B = 2½ × 2 = 5 and ABCDEF = 327935