Puzzle for July 7, 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, CD, and DE are 2-digit numbers (not A×B, C×D, or D×E).
Scratchpad
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Hint #1
Add C to both sides of eq.4: C + E + F + C = B - C + C which becomes 2×C + E + F = B In eq.3, replace B with 2×C + E + F: A = 2×C + E + F + C - F which makes eq.3a) A = 3×C + E
Hint #2
Add B and E to both sides of eq.5: A - E + B + E = D - B + B + E which becomes A + B = D + E Subtract the left and right sides of eq.2 from the left and right sides of the above equation, respectively: A + B - (B + E) = D + E - (C + D) which is the same as A + B - B - E = D + E - C - D which becomes A - E = E - C Add E to both sides of the equation above: A - E + E = E - C + E which becomes eq.5a) A = 2×E - C
Hint #3
In eq.5a, replace A with 3×C + E (from eq.3a): 3×C + E = 2×E - C In the above equation, subtract E from both sides, and add C to both sides: 3×C + E - E + C = 2×E - C - E + C which simplifies to 4×C = E
Hint #4
In eq.3a, substitute 4×C for E: A = 3×C + 4×C which makes A = 7×C
Hint #5
eq.6 may be written as: C + 10×D + E + F = 10×A + B + 10×C + D Subtract C and D from both sides of the above equation: C + 10×D + E + F - C - D = 10×A + B + 10×C + D - C - D which becomes eq.6a) 9×D + E + F = 10×A + B + 9×C
Hint #6
Subtract C from both sides of eq.2: B + E - C = C + D - C which makes eq.2a) B + E - C = D
Hint #7
Subtract C from both sides of eq.4: C + E + F - C = B - C - C which makes eq.4a) E + F = B - 2×C
Hint #8
Substitute (B + E - C) for D (from eq.2a), and B - 2×C for E + F (from eq.4a) in eq.6a: 9×(B + E - C) + B - 2×C = 10×A + B + 9×C which is equivalent to 9×B + 9×E - 9×C + B - 2×C = 10×A + B + 9×C which becomes 10×B + 9×E - 11×C = 10×A + B + 9×C In the above equation, add 11×C to both sides, and subtract B from both sides: 10×B + 9×E - 11×C + 11×C - B = 10×A + B + 9×C + 11×C - B which becomes eq.6b) 9×B + 9×E = 10×A + 20×C
Hint #9
Substitute (4×C) for E, and (7×C) for A in eq.6b: 9×B + 9×(4×C) = 10×(7×C) + 20×C which becomes 9×B + 36×C = 70×C + 20×C Subtract 36×C from both sides: 9×B + 36×C - 36×C = 70×C + 20×C - 36×C which simplifies to 9×B = 54×C Divide both sides by 9: 9×B ÷ 9 = 54×C ÷ 9 which makes B = 6×C
Hint #10
Substitute 6×C for B, and 4×C for E in eq.2a: 6×C + 4×C - C = D which makes 9×C = D
Hint #11
Substitute 4×C for E, and 6×C for B in eq.4a: 4×C + F = 6×C - 2×C which becomes 4×C + F = 4×C Subtract 4×C from both sides: 4×C + F - 4×C = 4×C - 4×C which means F = 0
Solution
Substitute 7×C for A, 6×C for B, 9×C for D, 4×C for E, and 0 for F in eq.1: 7×C + 6×C + C + 9×C + 4×C + 0 = 27 which simplifies to 27×C = 27 Divide both sides by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 7×C = 7 × 1 = 7 B = 6×C = 6 × 1 = 6 D = 9×C = 9 × 1 = 9 E = 4×C = 4 × 1 = 4 and ABCDEF = 761940