Puzzle for June 27, 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.3, replace E with A + D + F (from eq.2): A + D + F – A = A + B + C + F which becomes D + F = A + B + C + F Subtract F from each side of the equation above: D + F – F = A + B + C + F – F which becomes eq.3a) D = A + B + C
Hint #2
In eq.3a, replace A + B with C + D (from eq.1): D = C + D + C which becomes D = 2×C + D Subtract D from both sides of the above equation: D – D = 2×C + D – D which makes 0 = 2×C which means 0 = C
Hint #3
In eq.5, substitute 0 for C: 0 × E = B – F which makes 0 = B – F Add F to both sides: 0 + F = B – F + F which makes F = B
Hint #4
Substitute B for F in eq.6: B ÷ B = A which makes 1 = A
Hint #5
Substitute 0 for C in eq.3a: D = A + B + 0 which becomes D = A + B Subtract A from both sides: D – A = A + B – A which becomes eq.3b) D – A = B
Hint #6
Substitute B for D – A (from eq.3b), B for F, and 1 for A in eq.4: B × B = 1 + E which may be written as eq.4a) B² = 1 + E
Hint #7
Substitute 1 for A, 0 for C, and B for F in eq.3: E – 1 = 1 + B + 0 + B which becomes E – 1 = 1 + 2×B Add 1 to each side of the equation above: E – 1 + 1 = 1 + 2×B + 1 which becomes eq.3c) E = 2 + 2×B
Hint #8
Substitute 2 + 2×B for E (from eq.3c) in eq.4a: B² = 1 + 2 + 2×B which becomes B² = 3 + 2×B Subtract 2×B and 3 from both sides of the above equation: B² – 2×B – 3 = 3 + 2×B – 2×B – 3 which becomes eq.3d) B² – 2×B – 3 = 0
Solution
eq.3d is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.3d yields: B = {(–1)×(–2) ± sq.rt.[(–2)² – (4 × (1) × (–3))]} ÷ (2 × (1)) which becomes B = {2 ± sq.rt.[4 – (–12)]} ÷ 2 which becomes B = {2 ± sq.rt.(16)} ÷ 2 which becomes B = (2 ± 4) ÷ 2 In the above equation, either: B = (2 + 4) ÷ 2 = 6 ÷ 2 = 3 or: B = (2 – 4) ÷ 2 = –2 ÷ 2 = –1 Since B must be non-negative, then B ≠ –1 which therefore means B = 3 making D = A + B + C = 1 + 3 + 0 = 4 (from eq.3a) E = 2 + 2×B = 2 + 2×3 = 2 + 6 = 8 (from eq.3c) F = B = 3 and ABCDEF = 130483