Puzzle for December 22, 2021 ( )
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.
Scratchpad
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Hint #1
In eq.2, replace C + F with D + E – F (from eq.4): B + D = A + D + E – F In the above equation, subtract D from both sides, and add F to both sides: B + D – D + F = A + D + E – F – D + F which becomes eq.2a) B + F = A + E
Hint #2
In eq.3, replace A + E with B + F (from eq.2a): D + F = B + F Subtract F from each side of the equation above: D + F – F = B + F – F which makes D = B
Hint #3
In eq.5, replace B with D: D + C = D + F Subtract D from each side of the above equation: D + C – D = D + F – D which makes C = F
Hint #4
In eq.6, substitute C for F: E – D = C ÷ C which becomes E – D = 1 Add D to both sides of the equation above: E – D + D = 1 + D which makes eq.6a) E = 1 + D
Hint #5
Substitute C for F, and 1 + D for E (from eq.6a) in eq.4: C + C = D + 1 + D – C which becomes 2×C = 2×D + 1 – C Add C to both sides of the above equation: 2×C + C = 2×D + 1 – C + C which becomes 3×C = 2×D + 1 Divide both sides by 3: 3×C ÷ 3 = (2×D + 1) ÷ 3 which makes C = ⅔×D + ⅓ and also makes eq.4a) F = C = ⅔×D + ⅓
Hint #6
Substitute ⅔×D + ⅓ for F (from eq.4a), and 1 + D for E (from eq.6a) in eq.3: D + ⅔×D + ⅓ = A + 1 + D which becomes 1⅔×D + ⅓ = A + 1 + D Subtract 1 and D from each side of the equation above: 1⅔×D + ⅓ – 1 – D = A + 1 + D – 1 – D which makes eq.3a) ⅔×D – ⅔ = A
Solution
Substitute ⅔×D – ⅔ for A (from eq.3a), D for B, ⅔×D + ⅓ for C and F (from eq.4a), and 1 + D for E (from eq.6a) in eq.1: ⅔×D – ⅔ + D + ⅔×D + ⅓ + D + 1 + D + ⅔×D + ⅓ = 36 which simplifies to 5×D + 1 = 36 Subtract 1 from each side of the equation above: 5×D + 1 – 1 = 36 – 1 which makes 5×D = 35 Divide both sides by 5: 5×D ÷ 5 = 35 ÷ 5 which means D = 7 making A = ⅔×D – ⅔ = ⅔×7 – ⅔ = 4⅔ – ⅔ = 4 (from eq.3a) B = D = 7 C = F = ⅔×D + ⅓ = ⅔×7 + ⅓ = 4⅔ + ⅓ = 5 (from eq.4a) E = 1 + D = 1 + 7 = 8 (from eq.6a) and ABCDEF = 475785