Puzzle for February 10, 2023 ( )
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
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Hint #1
Subtract F and B from both sides of eq.5: E + F - B - F - B = A + B + D - F - B which becomes eq.5a) E - 2×B = A + D - F In eq.6, add C and D to both sides, and subtract F from both sides: A - C + E + C + D - F = C - D + F + C + D - F which becomes A + E + D - F = 2×C which is the same as eq.6a) A + D - F + E = 2×C
Hint #2
In eq.6a, replace A + D - F with E - 2×B (from eq.5a): E - 2×B + E = 2×C which becomes 2×E - 2×B = 2×C Divide both sides of the above equation by 2: (2×E - 2×B) ÷ 2 = 2×C ÷ 2 which becomes eq.6b) E - B = C
Hint #3
In eq.2, replace C with E - B (from eq.6b): D + E = B + E - B which becomes D + E = E Subtract E from each side of the equation above: D + E - E = E - E which makes D = 0
Hint #4
In eq.3, substitute (E - B) for C (from eq.6b), and 0 for D: B - (E - B) = (E - B) - 0 - E which becomes B - E + B = E - B - E which becomes 2×B - E = -B Add E and B to both sides of the equation above: 2×B - E + E + B = -B + E + B which becomes 3×B = E
Hint #5
Substitute 3×B for E in eq.6b: 3×B - B = C which makes 2×B = C
Hint #6
Substitute 2×B for C, and 3×B for E in eq.4: 2×B + 3×B = A + B + F which becomes 5×B = A + B + F Subtract B from each side of the above equation: 5×B - B = A + B + F - B which becomes eq.4a) 4×B = A + F
Hint #7
Substitute 3×B for E, and 0 for D in eq.5: 3×B + F - B = A + B + 0 which becomes 2×B + F = A + B Subtract B from each side of the above equation: 2×B + F - B = A + B - B which becomes eq.5b) B + F = A
Hint #8
Substitute B + F for A (from eq.5b) in eq.4a: 4×B = B + F + F which becomes 4×B = B + 2×F Subtract B from both sides of the above equation: 4×B - B = B + 2×F - B which makes 3×B = 2×F Divide both sides by 2: 3×B ÷ 2 = 2×F ÷ 2 which makes 1½×B = F
Hint #9
Substitute 1½×B for F in eq.5b: B + 1½×B = A which makes 2½×B = A
Solution
Substitute 2½×B for A, 2×B for C, 0 for D, 3×B for E, and 1½×B for F in eq.1: 2½×B + B + 2×B + 0 + 3×B + 1½×B = 20 which simplifies to 10×B = 20 Divide both sides of the above equation by 10: 10×B ÷ 10 = 20 ÷ 10 which means B = 2 making A = 2½×B = 2½ × 2 = 5 C = 2×B = 2 × 2 = 4 E = 3×B = 3 × 2 = 6 F = 1½×B = 1½ × 2 = 3 and ABCDEF = 524063