Puzzle for January 17, 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 positive integer.
Scratchpad
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Hint #1
In eq.1, replace A + B + C + D + E with A×E (from eq.6): eq.1a) A×E + F = 33
Hint #2
In eq.2, replace D + E with A (from eq.3): B + C = A + A which becomes eq.2a) B + C = 2×A
Hint #3
In eq.1, replace B + C with 2×A (from eq.2a), and D + E with A (from eq.3): A + 2×A + A + F = 33 which becomes eq.1b) 4×A + F = 33
Hint #4
In eq.1a, substitute 4×A + F for 33 (from eq.1b): A×E + F = 4×A + F Subtract F from both sides of the above equation: A×E + F – F = 4×A + F – F which makes A×E = 4×A Divide both sides of the above equation by A: A×E ÷ A = 4×A ÷ A which makes E = 4
Hint #5
Substitute 4 for E in eq.3: D + 4 = A Subtract 4 from both sides of the equation above: D + 4 – 4 = A – 4 which becomes eq.3a) D = A – 4
Hint #6
Subtract 4×A from both sides of eq.1b: 4×A + F – 4×A = 33 – 4×A which becomes eq.1c) F = 33 – 4×A Substitute 33 – 4×A for F in eq.4 (from eq.1c): 33 – 4×A = A + B Subtract A from both sides of the equation above: 33 – 4×A – A = A + B – A which becomes eq.1d) 33 – 5×A = B
Hint #7
Subtract B from both sides of eq.2a: B + C – B = 2×A – B which becomes C = 2×A – B Substitute (33 – 5×A) for B (from eq.1d) in the equation above: C = 2×A – (33 – 5×A) which is equivalent to C = 2×A – 33 + 5×A which becomes eq.2b) C = 7×A – 33
Hint #8
Substitute (33 – 5×A) for B (from eq.1d), and (A – 4) for D (from eq.3a) in eq.5: (33 – 5×A)×(A – 4) = A which becomes 33×A – 132 – 5×A² + 20×A = A which becomes 53×A – 132 – 5×A² = A Subtract A from both sides of the above equation: 53×A – 132 – 5×A² – A = A – A which becomes 52×A – 132 – 5×A² = 0 which may be written as eq.5a) –5×A² + 52×A – 132 = 0
Solution
eq.5a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.5a yields: A = { (–1)×52 ± sq.rt.[52² – (4 × (–5) × (–132))] } ÷ (2 × (–5)) which becomes A = {(–52) ± sq.rt.(2704 – 2640} ÷ (–10) which becomes A = {(–52) ± sq.rt.(64)} ÷ (–10) which becomes A = ((–52) ± 8) ÷ (–10) In the above equation, either: A = ((–52) + 8) ÷ (–10) = –44 ÷ (–10) = 4.4 or A = ((–52) – 8) ÷ (–10) = –60 ÷ (–10) = 6 Since A must be an integer, then A ≠ 4.4 and therefore means A = 6 making B = 33 – 5×A = 33 – 5×6 = 33 – 30 = 3 (from eq.1d) C = 7×A – 33 = 7×6 – 33 = 42 – 33 = 9 (from eq.2b) D = A – 4 = 6 – 4 = 2 (from eq.3a) F = 33 – 4×A = 33 – 4×6 = 33 – 24 = 9 (from eq.1c) and ABCDEF = 639249