Puzzle for January 3, 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
Add C to both sides of eq.4: C + D + C = B – C + C which becomes eq.4a) 2×C + D = B In eq.5, replace B with 2×C + D (from eq.4a): 2×C + D – D = A ÷ C which becomes 2×C = A ÷ C Multiply both sides of the equation above by C: 2×C × C = A ÷ C × C which makes 2×C² = A
Hint #2
In eq.2, replace B with 2×C + D (from eq.4a): 2×C + D + D = A + C which becomes 2×C + 2×D = A + C Subtract C from both sides of the equation above: 2×C + 2×D – C = A + C – C which becomes eq.2a) C + 2×D = A
Hint #3
In eq.3, substitute E + F for B (from eq.1): A + E + F – D = C + D + F In the equation above, add D to both sides, and subtract F from both sides: A + E + F – D + D – F = C + D + F + D – F which simplifies to eq.3a) A + E = C + 2×D
Hint #4
Substitute A for C + 2×D (from eq.2a) in eq.3a: A + E = A Subtract A from each side of the equation above: A + E – A = A – A which makes E = 0
Hint #5
Substitute 0 for E in eq.1: 0 + F = B which makes F = B
Hint #6
Substitute B for F in eq.6: D – C = B ÷ B which makes D – C = 1 Add C to both sides of the above equation: D – C + C = 1 + C which makes eq.6a) D = 1 + C
Hint #7
Substitute 1 + C for D (from eq.6a), and 2×C² for A in eq.2a: C + 2×(1 + C) = 2×C² which becomes C + 2×1 + 2×C = 2×C² which becomes 2 + 3×C = 2×C² Subtract 3×C and 2 from each side of the equation above: 2 + 3×C – 3×C – 2 = 2×C² – 3×C – 2 which becomes eq.2b) 0 = 2×C² – 3×C – 2
Hint #8
eq.2b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.2b yields: C = { (–1)×(–3) ± sq.rt.[(–3)² – (4 × 2 × (–2))] } ÷ (2 × 2) which becomes C = {3 ± sq.rt.(9 – (–16))} ÷ 4 which becomes C = {3 ± sq.rt.(25)} ÷ 4 which becomes C = (3 ± 5) ÷ 4 In the above equation, either C = (3 + 5) ÷ 4 = 8 ÷ 4 = 2 or C = (3 – 5) ÷ 4 = –2 ÷ 4 = –½ Since C must be a non-negative integer, then C ≠ –½ and therefore C = 2 making A = 2×C² = 2×2² = 2×4 = 8 D = 1 + C = 1 + 2 = 3 (from eq.6a)
Solution
Substitute 2 for C, and 3 for D in eq.4a: B = 2×2 + 3 = 4 + 3 = 7 F = B = 7 and ABCDEF = 872307