Puzzle for May 5, 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, DE, and EF are 2-digit numbers (not A×B, D×E, or E×F).
Scratchpad
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Hint #1
Subtract both B and C from each side of eq.2: B + E - B - C = C + F - B - C which becomes eq.2a) E - C = F - B Subtract both B and F from each side of eq.2: B + E - B - F = C + F - B - F which becomes eq.2b) E - F = C - B
Hint #2
eq.6 may be written as: 10×E + F - A - D = E + F Add A and D to each side, and subtract E and F from each side of the above equation: 10×E + F - A - D + A + D - E - F = E + F + A + D - E - F which simplifies to eq.6a) 9×E = A + D
Hint #3
Add B + C + D to both sides of eq.4: D - C + B + C + D = A - B - D + B + C + D which becomes eq.4a) B + 2×D = A + C Subtract A from both sides: B + 2×D - A = A + C - A which becomes eq.4b) B + 2×D - A = C
Hint #4
In eq.3, replace F - B with E - C (from eq.2a): A + B - D - E = E + E - C Add C + D + E to both sides of the equation above: A + B - D - E + C + D + E = E + E - C + C + D + E which becomes A + B + C = 3×E + D which may be written as A + C + B = 3×E + D In the above equation, replace A + C with B + 2×D (from eq.4a): B + 2×D + B = 3×E + D Subtract D from both sides: B + 2×D + B - D = 3×E + D - D which becomes eq.3a) 2×B + D = 3×E
Hint #5
Multiply both sides of eq.3a by 3: 3×(2×B + D) = 3×(3×E) which becomes 6×B + 3×D = 9×E In the above equation, substitute A + D for 9×E (from eq.6a): 6×B + 3×D = A + D Subtract D from both sides: 6×B + 3×D - D = A + D - D which becomes eq.3b) 6×B + 2×D = A
Hint #6
eq.5 may be written as: 10×A + B - (10×D + E) + F = 10×D + E - F which is the same as 10×A + B - 10×D - E + F = 10×D + E - F Add (10×D + E - F) to both sides of the equation above: 10×A + B - 10×D - E + F + 10×D + E - F = 10×D + E - F + 10×D + E - F which becomes 10×A + B = 20×D + 2×E - 2×F which may be written as eq.5a) 10×A + B = 20×D + 2×(E - F)
Hint #7
Substitute C - B for E - F (from eq.2b) in eq.5a: 10×A + B = 20×D + 2×(C - B) which becomes 10×A + B = 20×D + 2×C - 2×B Substitute (B + 2×D - A) for C (from eq.4b) in the above equation: 10×A + B = 20×D + 2×(B + 2×D - A) - 2×B which is equivalent to 10×A + B = 20×D + 2×B + 4×D - 2×A - 2×B which becomes 10×A + B = 24×D - 2×A Add 2×A to both sides: 10×A + B + 2×A = 24×D - 2×A + 2×A which becomes eq.5b) 12×A + B = 24×D
Hint #8
Substitute (6×B + 2×D) for A (from eq.3b) in eq.5b: 12×(6×B + 2×D) + B = 24×D which becomes 72×B + 24×D + B = 24×D Subtract 24×D from each side: 72×B + 24×D + B - 24×D = 24×D - 24×D which becomes 73×B = 0 which means B = 0
Hint #9
Substitute 0 for B in eq.3a: 2×0 + D = 3×E which makes D = 3×E
Hint #10
Substitute 0 for B, and 3×E for D in eq.3b: 6×0 + 2×3×E = A which means 6×E = A
Hint #11
Substitute 0 for B, 3×E for D, and 6×E for A in eq.4b: 0 + 2×3×E - 6×E = C which means 6×E - 6×E = C which makes 0 = C
Hint #12
In eq.2a, replace both B and C with 0: E - 0 = F - 0 which means E = F
Solution
Substitute 6×E for A, 0 for B and C, 3×E for D, and E for F in eq.1: 6×E + 0 + 0 + 3×E + E + E = 11 which becomes 11×E = 11 Divide both sides by 11: 11×E ÷ 11 = 11 ÷ 11 which makes E = 1 making A = 6×E = 6 × 1 = 6 D = 3×E = 3 × 1 = 3 F = E = 1 and ABCDEF = 600311