Puzzle for September 20, 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.
* CD is a 2-digit number (not C×D).
Scratchpad
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Hint #1
Add A and E to both sides of eq.2: F – A + A + E = A – E + A + E which becomes eq.2a) F + E = 2×A Add A to both sides of eq.5: D – A + A = A + B – E – F + A which becomes D = 2×A + B – E – F In the above equation, replace 2×A with F + E (from eq.2a): D = F + E + B – E – F which makes D = B
Hint #2
In eq.4, replace B with D: D – D = D – F which becomes 0 = D – F Add F to both sides of the equation above: 0 + F = D – F + F which makes F = D and also makes F = D = B
Hint #3
eq.6 may be written as: 10×C + D = (B×F) + C In the above equation, substitute B for D and F: 10×C + B = (B×B) + C Subtract C and B from each side: 10×C + B – C – B = (B×B) + C – C – B which becomes 9×C = B² – B Divide both sides by 9: 9×C ÷ 9 = (B² – B) ÷ 9 which makes eq.6a) C = (B² – B) ÷ 9
Hint #4
In eq.3, add C and A to both sides, and subtract D from both sides: D – C + C + A – D = C – A + C + A – D which becomes A = 2×C – D Substitute ((B² – B) ÷ 9) for C (from eq.6a), and B for D in the equation above: A = 2×((B² – B) ÷ 9) – B which may be written as eq.3a) A = ((2×B² – 2×B) ÷ 9) – B
Hint #5
Subtract F from both sides of eq.2a: F + E – F = 2×A – F which becomes E = 2×A – F In the equation above, substitute (((2×B² – 2×B) ÷ 9) – B) for A (from eq.3a), and B for F: E = 2×(((2×B² – 2×B) ÷ 9) – B) – B which may be written as E = ((4×B² – 4×B) ÷ 9) – 2×B – B which becomes eq.2b) E = ((4×B² – 4×B) ÷ 9) – 3×B
Hint #6
In eq.1, substitute ((2×B² – 2×B) ÷ 9) – B for A (from eq.3a), (B² – B) ÷ 9 for C (from eq.6a), B for D and F, and ((4×B² – 4×B) ÷ 9) – 3×B for E (from eq.2b): ((2×B² – 2×B) ÷ 9) – B + B + (B² – B) ÷ 9 + B + ((4×B² – 4×B) ÷ 9) – 3×B + B = 47 which is equivalent to (2×B² – 2×B + B² – B + 4×B² – 4×B) ÷ 9) – B = 47 which becomes ((7×B² – 7×B) ÷ 9) – B = 47 Add B to both sides of the above equation: ((7×B² – 7×B) ÷ 9) – B + B = 47 + B which becomes (7×B² – 7×B) ÷ 9 = 47 + B Multiply both sides by 9: 9 × ((7×B² – 7×B) ÷ 9) = 9 × (47 + B) which becomes 7×B² – 7×B = 423 + 9×B Subtract 423 and 9×B from both sides: 7×B² – 7×B – 423 – 9×B = 423 + 9×B – 423 – 9×B which becomes eq.1a) 7×B² – 16×B – 423 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.1a yields: B = {(–1)×(–16) ± sq.rt.((–16)² – (4 × 7 × (–423))} ÷ (2 × 7) which becomes B = {16 ± sq.rt.(256 – (–11844))} ÷ 14 which becomes B = {16 ± sq.rt.(12100)} ÷ 14 which becomes B = (16 ± 110) ÷ 14 In the equation above, either B = (16 + 110) ÷ 14 = 126 ÷ 14 = 9 or B = (16 – 110) ÷ 14 = –94 ÷ 14 = –6.7142857142857 Since B must be a non-negative integer, then B ≠ –6.7142857142857 and therefore makes B = 9 making A = ((2×B² – 2×B) ÷ 9) – B = ((2×9² – 2×9) ÷ 9) – 9 = ((2×81 – 18) ÷ 9) – 9 = ((162 – 18) ÷ 9) – 9 = (144 ÷ 9) – 9 = 16 – 9 = 7 (from eq.3a) C = (B² – B) ÷ 9 = (9² – 9) ÷ 9 = (81 – 9) ÷ 9 = 72 ÷ 9 = 8 (from eq.6a) D = F = B = 9 E = ((4×B² – 4×B) ÷ 9) – 3×B = ((4×9² – 4×9) ÷ 9) – 3×9 = ((4×81 – 36) ÷ 9) – 27 = ((324 – 36) ÷ 9) – 27 = (288 ÷ 9) – 27 = 32 – 27 = 5 (from eq.2a) and ABCDEF = 798959