Puzzle for April 27, 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.
* AB, BC, DE, and EF are 2-digit numbers (not A×B, B×C, D×E, or E×F).
Scratchpad
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Hint #1
Subtract the left and right sides of eq.2 from the left and right sides of eq.3, respectively: A + B - (A + C) = E + F - (B + F) which is equivalent to A + B - A - C = E + F - B - F which becomes B - C = E - B Add B to each side of the equation above: B - C + B = E - B + B which becomes eq.3a) 2×B - C = E
Hint #2
eq.5 may be written as: 10×B + C = A + E In the above equation, replace E with 2×B - C (from eq.3a): 10×B + C = A + 2×B - C Add C to both sides, and subtract 2×B from each side of the equation above: 10×B + C + C - 2×B = A + 2×B - C + C - 2×B which simplifies to eq.5a) 8×B + 2×C = A
Hint #3
In eq.3, substitute 8×B + 2C for A (from eq.5a), and 2×B - C for E (from eq.3a): 8×B + 2×C + B = 2×B - C + F Add (C - 2×B) to both sides of the above equation: 8×B + 2×C + B + (C - 2×B) = 2×B - C + F + (C - 2×B) which simplifies to eq.3b) 7×B + 3×C = F
Hint #4
Substitute 7×B + 3×C for F (from eq.3b) in eq.4: C + 7×B + 3×C - B = B + D which becomes 6×B + 4×C = B + D Subtract B from both sides: 6×B + 4×C - B = B + D - B which becomes eq.4a) 5×B + 4×C = D
Hint #5
eq.6 may be written as: 10×A + B - E = 10×D + E + 10×E + F Add E to both sides of the above equation: 10×A + B - E + E = 10×D + E + 10×E + F + E which becomes eq.6a) 10×A + B = 10×D + 12×E + F
Hint #6
Substitute (8×B + 2×C) for A (from eq.5a), (5×B + 4×C) for D (from eq.4a), (2×B - C) for E (from eq.3a), and 7×B + 3×C for F (from eq.3b) in eq.6a: 10×(8×B + 2×C) + B = 10×(5×B + 4×C) + 12×(2×B - C) + 7×B + 3×C which becomes 80×B + 20×C + B = 50×B + 40×C + 24×B - 12×C + 7×B + 3×C which simplifies to 81×B + 20×C = 81×B + 31×C Subtract 81×B and 20×C from both sides: 81×B + 20×C - 81×B - 20×C = 81×B + 31×C - 81×B - 20×C which makes 0 = 11×C which means 0 = C
Hint #7
Substitute 0 for C in eq.5a: 8×B + 2×0 = A which makes 8×B = A
Hint #8
Substitute 0 for C in eq.4a: 5×B + 4×0 = D which makes 5×B = D
Hint #9
Substitute 0 for C in eq.3a: 2×B - 0 = E which means 2×B = E
Hint #10
Substitute 0 for C in eq.3b: 7×B + 3×0 = F which makes 7×B = F
Solution
Substitute 8×B for A, 0 for C, 5×B for D, 2×B for E, and 7×B for F in eq.1: 8×B + B + 0 + 5×B + 2×B + 7×B = 23 which simplifies to 23×B = 23 Divide both sides of the above equation by 23: 23×B ÷ 23 = 23 ÷ 23 which means B = 1 making A = 8×B = 8 × 1 = 8 D = 5×B = 5 × 1 = 5 E = 2×B = 2 × 1 = 2 F = 7×B = 7 × 1 = 7 and ABCDEF = 810527