Puzzle for June 6, 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
eq.6 may be written as: A = (B + D) ÷ 2 Multiply both sides of the equation above by 2: A × 2 = (B + D) ÷ 2 × 2 which becomes eq.6a) 2×A = B + D Add A and F to both sides of eq.2: D – A + A + F = A + C – F + A + F which becomes eq.2a) D + F = 2×A + C
Hint #2
In eq.2a, replace 2×A with B + D (from eq.6a): D + F = B + D + C Subtract D from each side of the equation above: D + F – D = B + D + C – D which becomes eq.2b) F = B + C
Hint #3
In eq.1, substitute B + C for F (from eq.2b): B + B + C = C – B which becomes 2×B + C = C – B In the above equation, subtract C from both sides, and add B to both sides: 2×B + C – C + B = C – B – C + B which becomes 3×B = 0 which means B = 0
Hint #4
Substitute 0 for B in eq.2b: F = 0 + C which makes F = C
Hint #5
Substitute C for F in eq.2a: D + C = 2×A + C Subtract C from both sides of the above equation: D + C – C = 2×A + C – C which makes D = 2×A
Hint #6
Substitute C for F in eq.5: C ÷ C = E which makes 1 = E
Hint #7
Substitute 2×A for D, 1 for E, and C for F in eq.4: A × 2×A = A + C + 1 + C which becomes 2×A² = A + 2×C + 1 Subtract A and 1 from both sides of the above equation: 2×A² – A – 1 = A + 2×C + 1 – A – 1 which becomes eq.4a) 2×A² – A – 1 = 2×C
Hint #8
Substitute C for F, 0 for B, 2×A for D, and 1 for E in eq.3: C = 0 + 2×A + 1 which becomes C = 2×A + 1 and also makes eq.3a) F = C = 2×A + 1
Hint #9
Substitute (2×A + 1) for C (from eq.3a) in eq.4a: 2×A² – A – 1 = 2×(2×A + 1) which becomes 2×A² – A – 1 = 4×A + 2 Subtract 4×A and 2 from both sides of the equation above: 2×A² – A – 1 – 4×A – 2 = 4×A + 2 – 4×A – 2 which becomes eq.4b) 2×A² – 5×A – 3 = 0
Solution
eq.4b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.4b yields: A = { (–1)×(–5) ± sq.rt.[(–5)² – (4 × 2 × (–3))] } ÷ (2 × 2) which becomes A = {5 ± sq.rt.(25 – (–24))} ÷ 4 which becomes A = {5 ± sq.rt.(49)} ÷ 4 which becomes A = {5 ± 7} ÷ 4 In the above equation, either A = (5 + 7) ÷ 4 = 12 ÷ 4 = 3 or A = (5 – 7) ÷ 4 = –2 ÷ 4 = –½ Since A must be a non-negative integer, then A ≠ –½ and therefore means A = 3 making D = 2×A = 2×3 = 6 F = C = 2×A + 1 = 2×3 + 1 = 6 + 1 = 7 (from eq.3a) and ABCDEF = 307617