Puzzle for October 25, 2020 ( )
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.
* BC and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
eq.5 may be written as: 10×B + C – (10×D + E) = A + C + F which becomes 10×B + C – 10×D – E = A + C + F In the above equation, subtract C from both sides, and add E to both sides: 10×B + C – 10×D – E – C + E = A + C + F – C + E which becomes 10×B – 10×D = A + F + E which may be written as 10×B – 10×D = A + E + F Substitute C for E + F (from eq.2): eq.5a) 10×B – 10×D = A + C
Hint #2
Add A, B, and D to both sides of eq.4: C – A – B + A + B + D = B – D + A + B + D which becomes eq.4a) C + D = 2×B + A Add D to both sides of eq.5a: 10×B – 10×D + D = A + C + D which becomes eq.5b) 10×B – 9×D = A + C + D
Hint #3
Substitute 2×B + A for C + D (from eq.4a) in eq.5b: 10×B – 9×D = A + 2×B + A which becomes 10×B – 9×D = 2×A + 2×B Subtract 2×B from each side of the equation above: 10×B – 9×D – 2×B = 2×A + 2×B – 2×B which becomes 8×B – 9×D = 2×A Divide both sides by 2: (8×B – 9×D) ÷ 2 = 2×A ÷ 2 which becomes eq.5c) 4×B – 4½×D = A
Hint #4
Substitute 4×B – 4½×D for A (from eq.5c) in eq.5a: 10×B – 10×D = 4×B – 4½×D + C In the equation above, subtract 4×B from each side, and add 4½×D to each side: 10×B – 10×D – 4×B + 4½×D = 4×B – 4½×D + C – 4×B + 4½×D which becomes eq.5d) 6×B – 5½×D = C
Hint #5
Substitute (4×B – 4½×D) for A (from eq.5c), and 6×B – 5½×D for C (from eq.5d) in eq.3: D + F – (4×B – 4½×D) = (4×B – 4½×D) + 6×B – 5½×D which becomes D + F – 4×B + 4½×D = 10×B – 10×D which becomes 5½×D + F – 4×B = 10×B – 10×D In the equation above, add 4×B to both sides, and subtract 5½×D from both sides: 5½×D + F – 4×B + 4×B – 5½×D = 10×B – 10×D + 4×B – 5½×D which becomes eq.3a) F = 14×B – 15½×D
Hint #6
Substitute 14×B – 15½×D for F (from eq.3a), and 6×B – 5½×D for C (from eq.5d) in eq.2: E + 14×B – 15½×D = 6×B – 5½×D In the equation above, subtract 14×B from both sides, and add 15½×D to both sides: E + 14×B – 15½×D – 14×B + 15½×D = 6×B – 5½×D – 14×B + 15½×D which becomes E = –8×B + 10×D which may be written as eq.2a) E = 10×D – 8×B
Hint #7
Substitute (6×B – 5½×D) for C (from eq.5d), (4×B – 4½×D) for A (from eq.5c), and 14×B – 15½×D for F (from eq.3a) in eq.6: (6×B – 5½×D) × D = (4×B – 4½×D) × ((6×B – 5½×D) + 14×B – 15½×D) which becomes 6×B×D – 5½×D² = (4×B – 4½×D) × (20×B – 21×D) which becomes 6×B×D – 5½×D² = 80×B² – 84×B×D – 90×B×D + 94½×D² which becomes 6×B×D – 5½×D² = 80×B² – 174×B×D + 94½×D² In the above equation above, subtract 6×B×D from each side, and add 5½×D² to each side: 6×B×D – 5½×D² – 6×B×D + 5½×D² = 80×B² – 174×B×D + 94½×D² – 6×B×D + 5½×D² which becomes 0 = 80×B² – 180×B×D + 100×D² Divide both sides by 20: 0 ÷ 20 = (80×B² – 180×B×D + 100×D²) ÷ 20 which becomes 0 = 4×B² – 9×B×D + 5×D² which may be written as eq.6a) 0 = 4×B² – (9×D)×B + 5×D²
Hint #8
eq.6a is in the format of a quadratic equation in standard form. The independent variable is B. The co-efficents are 4, –9×D, and 5×D². Using the quadratic equation solution formula to solve for B in eq.6a yields: B = { (–1)×(–9×D) ± sq.rt.[(–9×D)² – (4 × 4 × 5×D²)] } ÷ (2 × 4) which becomes B = { 9×D ± sq.rt.[81×D² – (80×D²)] } ÷ 8 which becomes B = ( 9×D ± sq.rt.[D²] ) ÷ 8 which becomes B = ( 9×D ± D ) ÷ 8 In the above equation, either B = ( 9×D + D ) ÷ 8 = 10×D ÷ 8 = 1¼×D or B = ( 9×D – D ) ÷ 8 = 8×D ÷ 8 = D
Hint #9
Begin checking: B = D ... Substituting B for D in eq.5c would yield: 4×B – 4½×B = A which would become –½×B = A Since A and B are non-negative integers, the equation above would make: A = 0 and B = 0 and would also make D = B = 0 The above equation would make: C = 6×B – 5½×D = 6×0 – 5½×0 = 0 – 0 = 0 (from eq.5d) E = 10×D – 8×B = 10×0 – 8×0 = 0 – 0 = 0 (from eq.2a) F = 14×B – 15½×D = 14×0 – 15½×0 = 0 – 0 = 0 (from eq.3a)
Hint #10
Finish checking: B = D ... Substituting 0 for A, B, C, D, E, and F in eq.1 would make: 0 + 0 + 0 + 0 + 0 + 0 = 27 which would mean 0 = 27 Since 0 ≠ 27 then B ≠ D and therefore makes B = 1¼×D
Hint #11
Substitute (1¼×D) for B in eq.5c: 4×(1¼×D) – 4½×D = A which becomes 5×D – 4½×D = A which makes ½×D = A
Hint #12
Substitute (1¼×D) for B in eq.5d: 6×(1¼×D) – 5½×D = C which becomes 7½×D – 5½×D = C which makes 2×D = C
Hint #13
Substitute (1¼×D) for B in eq.2a: E = 10×D – 8×(1¼×D) which becomes E = 10×D – 10×D which makes E = 0
Hint #14
Substitute (1¼×D) for B in eq.3a: F = 14×(1¼×D) – 15½×D which becomes F = 17½×D – 15½×D which makes F = 2×D
Solution
Substitute ½×D for A, 1¼×D for B, 2×D for C and F, and 0 for E in eq.1: ½×D + 1¼×D + 2×D + D + 0 + 2×D = 27 which simplifies to 6¾×D = 27 Divide both sides of the above equation by 6¾: 6¾×D ÷ 6¾ = 27 ÷ 6¾ which means D = 4 making A = ½×D = ½ × 4 = 2 B = 1¼×D = 1¼ × 4 = 5 C = F = 2×D = 2 × 4 = 8 and ABCDEF = 258408