Puzzle for January 13, 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 and EF are 2-digit numbers (not A×B or E×F).
Scratchpad
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Hint #1
In eq.5, replace B with C + D (from eq.2): E + F = A + C + D + C - D which becomes E + F = A + 2×C Subtract both E and 2×C from each side of the above equation: E + F - E - 2×C = A + 2×C - E - 2×C which becomes F - 2×C = A - E In eq.3, substitute F - 2×C for A - E: F - 2×C = B - F Add F to both sides: F - 2×C + F = B - F + F which makes eq.3a) 2×F - 2×C = B
Hint #2
In eq.3a, replace B with C + D (from eq.2): 2×F - 2×C = C + D Subtract C from both sides of the above equation: 2×F - 2×C - C = C + D - C which becomes eq.3b) 2×F - 3×C = D
Hint #3
From eq.3b, substitute (2×F - 3×C) for D in eq.4: F - (2×F - 3×C) = A - C + (2×F - 3×C) which is equivalent to F - 2×F + 3×C = A - 4×C + 2×F Add (4×C - 2×F) to both sides of the equation above: F - 2×F + 3×C + (4×C - 2×F) = A - 4×C + 2×F + (4×C - 2×F) which simplifies to -3×F + 7×C = A which may also be written as eq.4a) 7×C - 3×F = A
Hint #4
Substitute 7×C - 3×F for A (from eq.4a), and 2×F - 2×C for B (from eq.3a) into eq.3: 7×C - 3×F - E = 2×F - 2×C - F Add E to each side of the above equation: 7×C - 3×F - E + E = 2×F - 2×C - F + E which becomes 7×C - 3×F = F - 2×C + E Add (2×C - F) to both sides: 7×C - 3×F + (2×C - F) = F - 2×C + E + (2×C - F) which simplifies to eq.3c) 9×C - 4×F = E
Hint #5
eq.6 may be written as: 10×A + B + C + D = 10×E + F - D Add D to both sides of the equation above: 10×A + B + C + D + D = 10×E + F - D + D which becomes 10×A + B + C + 2×D = 10×E + F Substitute 7×C - 3×F for A (from eq.4a), 2×F - 2×C for B (from eq.3a), 2×F - 3×C for D (from eq.3b), and 9×C - 4×F for E (from eq.3c): 10×(7×C - 3×F) + 2×F - 2×C + C + 2×(2×F - 3×C) = 10×(9×C - 4×F) + F which is the same as 70×C - 30×F + 2×F - 2×C + C + 4×F - 6×C = 90×C - 40×F + F which becomes 63×C - 24×F = 90×C - 39×F Add (39×F - 63×C) to both sides: 63×C - 24×F + (39×F - 63×C) = 90×C - 39×F + (39×F - 63×C) which simplifies to 15×F = 27×C Divide both sides of the above equation by 15: 15×F ÷ 15 = 27×C ÷ 15 which makes F = 1⅘×C
Hint #6
In eq.4a, replace F with (1⅘×C): 7×C - 3×(1⅘×C) = A which becomes 7×C - 5⅖×C = A which makes 1⅗×C = A
Hint #7
In eq.3a, replace F with (1⅘×C): 2×(1⅘×C) - 2×C = B which becomes 3⅗×C - 2×C = B which makes 1⅗×C = B
Hint #8
In eq.3b, replace F with (1⅘×C): 2×(1⅘×C) - 3×C = D which becomes 3⅗×C - 3×C = D which makes ⅗×C = D
Hint #9
In eq.3c, replace F with (1⅘×C): 9×C - 4×(1⅘×C) = E which becomes 9×C - 7⅕×C = E which makes 1⅘×C = E
Solution
Substitute 1⅗×C for A and B, ⅗×C for D, and 1⅘×C for E and F in eq.1: 1⅗×C + 1⅗×C + C + ⅗×C + 1⅘×C + 1⅘×C = 42 which simplifies to 8⅖×C = 42 Divide both sides by 8⅖: 8⅖×C ÷ 8⅖ = 42 ÷ 8⅖ which makes C = 5 making A = B = 1⅗×C = 1⅗ × 5 = 8 D = ⅗×C = ⅗ × 5 = 3 E = F = 1⅘×C = 1⅘ × 5 = 9 and ABCDEF = 885399