Puzzle for November 21, 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
Multiply both sides of eq.5 by B: D × B = ((D + E + F) ÷ B) × B which becomes D × B = D + E + F which may be written as eq.5a) B×D = D + E + F eq.3 may be written as: B = (D + E + F) ÷ 3 Multiply both sides of the above equation by 3: B × 3 = ((D + E + F) ÷ 3) × 3 which becomes B × 3 = D + E + F which may be written as eq.3a) 3×B = D + E + F
Hint #2
In eq.3a, replace D + E + F with B×D (from eq.5a): 3×B = B×D Since B ≠ 0 (from eq.5), divide both sides of the above equation by B: 3×B ÷ B = B×D ÷ B which makes 3 = D
Hint #3
Multiply both sides of eq.6 by D: A × D = ((A + B – E) ÷ D) × D which becomes A×D = A + B – E Add E to both sides of the equation above: A×D + E = A + B – E + E which becomes eq.6a) A×D + E = A + B
Hint #4
In eq.1, replace A + B with A×D + E (from eq.6a): A×D + E + D – E = E + F which becomes eq.1a) A×D + D = E + F
Hint #5
In eq.5a, substitute A×D + D for E + F (from eq.1a): B×D = D + A×D + D which becomes B×D = 2×D + A×D which may be written as B×D = D×(2 + A) Divide both sides of the above equation by D (D ≠ 0, since D = 3): B×D ÷ D = D×(2 + A) ÷ D which makes eq.5b) B = 2 + A
Hint #6
Substitute 3 for D, and 2 + A for B (from eq.5b) in eq.6a: A×3 + E = A + 2 + A which becomes 3×A + E = 2×A + 2 Subtract 3×A from each side of the equation above: 3×A + E – 3×A = 2×A + 2 – 3×A which makes eq.6b) E = 2 – A
Hint #7
Substitute 2 + A for B (from eq.5b), and 2 – A for E (from eq.6b) in eq.4: C = (2 + A + 2 – A) ÷ A which makes eq.4a) C = 4 ÷ A
Hint #8
eq.2 may be written as: C = (A + B + E) ÷ 3 Multiply both sides of the above equation by 3: C × 3 = ((A + B + E) ÷ 3) × 3 which becomes eq.2a) 3×C = A + B + E
Hint #9
Substitute (4 ÷ A) for C (from eq.4a), 2 + A for B (from eq.5b), and 2 – A for E (from eq.6b) in eq.2a: 3×(4 ÷ A) = A + 2 + A + 2 – A which becomes 12 ÷ A = A + 4 Multiply both sides of the above equation by A: (12 ÷ A) × A = (A + 4) × A which becomes 12 = A² + 4×A Subtract 12 from both sides: 12 – 12 = A² + 4×A – 12 which becomes eq.2b) 0 = A² + 4×A – 12
Hint #10
eq.2b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.2b yields: A = { (–1)×4 ± sq.rt.[4² – (4 × 1 × (–12))] } ÷ (2 × 1) which becomes A = {–4 ± sq.rt.(16 + 48)} ÷ 2 which becomes A = {–4 ± sq.rt.(64)} ÷ 2 which becomes A = (–4 ± 8) ÷ 2 In the above equation, either A = (–4 + 8) ÷ 2 = 4 ÷ 2 = 2 or A = (–4 – 8) ÷ 2 = –12 ÷ 2 = –6 Since A must be non-negative, then A ≠ –6 and therefore makes A = 2
Hint #11
Since A = 2, then: B = 2 + A = 2 + 2 = 4 (from eq.5b) C = 4 ÷ A = 4 ÷ 2 = 2 (from eq.4a) E = 2 – A = 2 – 2 = 0 (from eq.6b)
Solution
Substitute 2 for A, 3 for D, and 0 for E in eq.1a: 2×3 + 3 = 0 + F which makes 9 = F and makes ABCDEF = 242309