Puzzle for April 30, 2023 ( )
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.
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Hint #1
eq.1 may be re-written as: eq.1a) B + C + D + A + E + F = 30 In eq.1a, replace B + C + D with A + E + F (from eq.5): A + E + F + A + E + F = 30 which becomes 2×(A + E + F) = 30 Divide both sides by 2: 2×(A + E + F) ÷ 2 = 30 ÷ 2 which becomes eq.1b) A + E + F = 15
Hint #2
In eq.1a, replace A + E + F with 15 (from eq.1b): B + C + D + 15 = 30 Subtract 15 from both sides of the equation above: B + C + D + 15 - 15 = 30 - 15 which becomes eq.1c) B + C + D = 15
Hint #3
Add F and E to both sides of eq.3: E - F + F + E = A - E + F + E which becomes eq.3a) 2×E = A + F eq.1b may be re-written as: eq.1d) A + F + E = 15
Hint #4
In eq.1d, replace A + F with 2×E (from eq.3a): 2×E + E = 15 which makes 3×E = 15 Divide both sides of the above equation by 3: 3×E ÷ 3 = 15 ÷ 3 which makes E = 5
Hint #5
Subtract F from each side of eq.4: B + E - F = C + D + F - F which becomes eq.4a) B + E - F = C + D
Hint #6
In eq.2, substitute B + E - F for C + D (from eq.4a): F = A + B + E - F Add 2×F to both sides of the above equation: F + 2×F = A + B + E - F + 2×F which becomes 3×F = A + B + E + F which may be written as eq.2a) 3×F = A + E + F + B
Hint #7
Substitute 15 for A + E + F (from eq.1b) in eq.2a: 3×F = 15 + B Subtract 15 from each side of the equation above: 3×F - 15 = 15 + B - 15 which becomes eq.2b) 3×F - 15 = B
Hint #8
eq.1 may be re-written as: B + A + C + D + E + F = 30 Substitute 3×F - 15 for B (from eq.2b), F for A + C + D (from eq.2), and 5 for E in the above equation: 3×F - 15 + F + 5 + F = 30 which becomes 5×F - 10 = 30 Add 10 to both sides: 5×F - 10 + 10 = 30 + 10 which makes 5×F = 40 Divide both sides by 5: 5×F ÷ 5 = 40 ÷ 5 which makes F = 8
Hint #9
Substitute 5 for E, and 8 for F in eq.1b: A + 5 + 8 = 15 which becomes A + 13 = 15 Subtract 13 from each side of the above equation: A + 13 - 13 = 15 - 13 which makes A = 2
Hint #10
Substitute 8 for F in eq.2b: 3×8 - 15 = B which becomes 24 - 15 = B which makes 9 = B
Hint #11
Substitute 8 for F, and 2 for A in eq.2: 8 = 2 + C + D Subtract 2 and C from both sides of the above equation: 8 - 2 - C = 2 + C + D - 2 - C which becomes eq.2c) 6 - C = D
Hint #12
Substitute 9 for B, 5 for E, (6 - C) for D (from eq.2c), 8 for F, and 2 for A in eq.6: 8 - (2 ÷ C) = 9 - (5 ÷ (6 - C)) Subtract 9 from both sides of the equation above: 8 - (2 ÷ C) - 9 = 9 - (5 ÷ (6 - C)) - 9 which becomes -1 - (2 ÷ C) = -(5 ÷ (6 - C)) Multiply by each side by both (-1) and C: (-1) × C × (-1 - (2 ÷ C)) = (-1) × C × (-(5 ÷ (6 - C))) which becomes eq.6a) C + 2 = 5×C ÷ (6 - C)
Hint #13
Multiply both sides of eq.6a by (6 - C): (C + 2) × (6 - C) = 5×C ÷ (6 - C) × (6 - C) which becomes 6×C - C² + 12 - 2×C = 5×C which becomes 4×C - C² + 12 = 5×C Subtract 5×C from each side of the equation above: 4×C - C² + 12 - 5×C = 5×C - 5×C which becomes -C - C² + 12 = 0 which may be written as -C² - C + 12 = 0 Multiply by both sides by (-1): (-1) × (-C² - C + 12) = (-1) × 0 which becomes eq.6b) C² + C - 12 = 0
Hint #14
eq.6b is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for C in eq.6b. However, we will instead factor eq.6b into the product of two expressions: (C + 4) × (C - 3) = 0 The above equation makes either: (C + 4) = 0 which means C = -4 or: (C - 3) = 0 which means C = 3 Since C must be non-negative, then: C ≠ -4 and therefore makes: C = 3
Solution
Substitute 3 for C in eq.2c: 6 - 3 = D which makes 3 = D and ABCDEF = 293358