Puzzle for December 16, 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.
* AB, DE, and EF are 2-digit numbers (not A×B, D×E, or E×F).
Scratchpad
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Hint #1
Subtract B from each side of eq.2: A - B = B + C + E - B which becomes eq.2a) A - B = C + E Add C + F to both sides of eq.6: B - C + C + F = E - F + C + F which becomes B + F = E + C which is equivalent to eq.6a) B + F = C + E
Hint #2
Substitute B + F for C + E (from eq.6a) into eq.2: A = B + B + F which becomes eq.2b) A = 2×B + F Add (D - B) to each side of eq.4: A - D + (D - B) = B + C + D + F + (D - B) which becomes eq.4a) A - B = C + 2×D + F
Hint #3
In eq.4a, substitute C + E for A - B (from eq.2a): C + E = C + 2×D + F Subtract C from both sides: C + E - C = C + 2×D + F - C which becomes eq.4b) E = 2×D + F
Hint #4
Substitute 2×B + F for A (from eq.2b), and 2×D + F for E (from eq.4b) in eq.3: 2×B + F + B + D = 2×D + F + F which becomes 3×B + F + D = 2×D + 2×F Subtract both 2×D and F from both sides: 3×B + F + D - 2×D - F = 2×D + 2×F - 2×D - F which becomes eq.3a) 3×B - D = F
Hint #5
eq.5 can be re-stated as: 10×A + B = 10×D + E + 10×E + F which is the same as 10×A + B = 10×D + 11×E + F Substitute (2×B + F) for A (from eq.2b), and (2×D + F) for E (from eq.4b) in the equation above: 10×(2×B + F) + B = 10×D + 11×(2×D + F) + F which is the same as 20×B + 10×F + B = 10×D + 22×D + 11×F + F which becomes 21×B + 10×F = 32×D + 12×F Subtract 10×F from each side: 21×B + 10×F - 10×F = 32×D + 12×F - 10×F which simplifies to eq.5a) 21×B = 32×D + 2×F
Hint #6
In eq.5a, substitute (3×B - D) for F (from eq.3a): 21×B = 32×D + 2×(3×B - D) which is the same as 21×B = 32×D + 6×B - 2×D which becomes 21×B = 30×D + 6×B Subtract 6×B from each side: 21×B - 6×B = 30×D + 6×B - 6×B which simplifies to 15×B = 30×D Divide both sides by 15: 15×B ÷ 15 = 30×D ÷ 15 which means B = 2×D
Hint #7
Substitute (2×D) for B in eq.3a: 3×(2×D) - D = F which becomes 6×D - D = F which makes 5×D = F
Hint #8
Substitute (2×D) for B, and 5×D for F in eq.2b: A = 2×(2×D) + 5×D which becomes A = 4×D + 5×D which makes A = 9×D
Hint #9
Replace F with 5×D in eq.4b: E = 2×D + 5×D which means E = 7×D
Hint #10
Substitute 2×D for B, 7×D for E, and 5×D for F in eq.6a: 2×D + 5×D = C + 7×D which means 7×D = C + 7×D Subtract 7×D from each side: 7×D - 7×D = C + 7×D - 7×D which makes 0 = C
Solution
Substitute 9×D for A, 2×D for B, 0 for C, 7×D for E, and 5×D for F in eq.1: 9×D + 2×D + 0 + D + 7×D + 5×D = 24 which simplifies to 24×D = 24 Divide both sides by 24: 24×D ÷ 24 = 24 ÷ 24 which makes D = 1 making A = 9×D = 9 × 1 = 9 B = 2×D = 2 × 1 = 2 E = 7×D = 7 × 1 = 7 F = 5×D = 5 × 1 = 5 and ABCDEF = 920175