Puzzle for December 8, 2019 ( )
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.
(This is a very challenging puzzle. But it can be solved! Don't give up!)
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Hint #1
eq.1 may be written as: A + C + B + D + E + F = 41 In the above equation, replace A + C with D + E (from eq.2): D + E + B + D + E + F = 41 which may be written as B + 2×D + E + E + F = 41 Replace E + F with B + D (from eq.3): B + 2×D + E + B + D = 41 which becomes eq.1a) 2×B + 3×D + E = 41
Hint #2
eq.1 may also be written as: A + D + F + B + C + E = 41 In the above equation, replace A + D + F with B + C (from eq.5): B + C + B + C + E = 41 which becomes eq.1b) 2×B + 2×C + E = 41
Hint #3
In eq.1a, replace 41 with 2×B + 2×C + E (from eq.1b): 2×B + 3×D + E = 2×B + 2×C + E Subtract both 2×B and E from each side of the above equation: 2×B + 3×D + E – 2×B – E = 2×B + 2×C + E – 2×B – E which simplifies to 3×D = 2×C Divide each side by 2: 3×D ÷ 2 = 2×C ÷ 2 which makes 1½×D = C
Hint #4
Add B to both sides of eq.3: E + F + B = B + D + B which may be written as eq.3a) B + E + F = 2×B + D Add E to both sides of eq.4: C + E – B + E = B + F + E which may be written as C + 2×E – B = B + E + F Substitute C + 2×E – B for B + E + F in eq.3a: C + 2×E – B = 2×B + D Add B to both sides of the above equation: C + 2×E – B + B = 2×B + D + B which becomes eq.3b) C + 2×E = 3×B + D
Hint #5
Substitute 1½×D for C in eq.3b: 1½×D + 2×E = 3×B + D Subtract 1½×D from each side: 1½×D + 2×E – 1½×D = 3×B + D – 1½×D which makes eq.3c) 2×E = 3×B – ½×D
Hint #6
Multiply both sides of eq.1a by 2: 2×(2×B + 3×D + E) = 2×41 which becomes 4×B + 6×D + 2×E = 82 In the above equation, replace 2×E with 3×B – ½×D (from eq.3c): 4×B + 6×D + 3×B – ½×D = 82 which becomes 7×B + 5½×D = 82 Subtract 5½×D from each side: 7×B + 5½×D – 5½×D = 82 – 5½×D which becomes 7×B = 82 – 5½×D Divide both sides by 7: 7×B ÷ 7 = (82 – 5½×D) ÷ 7 which means eq.1c) B = (82 – 5½×D) ÷ 7
Hint #7
Substitute ((82 – 5½×D) ÷ 7) for B (from eq.1c), and 1½×D for C in eq.1b: 2×((82 – 5½×D) ÷ 7) + 2×(1½×D) + E = 41 which becomes ((164 – 11×D) ÷ 7) + 3×D + E = 41 Multiply both sides of the above equation by 7: 7×(((164 – 11×D) ÷ 7) + 3×D + E) = 7×41 which becomes 164 – 11×D + 21×D + 7×E = 287 which becomes 164 + 10×D + 7×E = 287 Subtract 10×D and 164 from both sides: 164 + 10×D + 7×E – 10×D – 164 = 287 – 10×D – 164 which becomes 7×E = 123 – 10×D Divide both sides by 7: 7×E ÷ 7 = (123 – 10×D) ÷ 7 which means eq.1d) E = (123 – 10×D) ÷ 7
Hint #8
Substitute ((123 – 10×D) ÷ 7) for E (from eq.1d), and 1½×D for C in eq.2: D + ((123 – 10×D) ÷ 7) = A + 1½×D Subtract 1½×D from each side of the above equation: D + ((123 – 10×D) ÷ 7) – 1½×D = A + 1½×D – 1½×D which becomes ((123 – 10×D) ÷ 7) – ½×D = A Multiply both sides of the above equation by 7: 7×(((123 – 10×D) ÷ 7) – ½×D) = 7×A which becomes 123 – 10×D – 3½×D = 7×A which becomes 123 – 13½×D = 7×A Divide both sides by 7: (123 – 13½×D) ÷ 7 = 7×A ÷ 7 which becomes eq.1e) (123 – 13½×D) ÷ 7 = A
Hint #9
Substitute ((123 – 10×D) ÷ 7) for E (from eq.1d), and ((82 – 5½×D) ÷ 7) for B (from eq.1c) in eq.3: ((123 – 10×D) ÷ 7) + F = ((82 – 5½×D) ÷ 7) + D which may be written as ((123 – 10×D) ÷ 7) + F = ((82 – 5½×D) ÷ 7) + (7×D ÷ 7) which becomes ((123 – 10×D) ÷ 7) + F = (82 – 5½×D + 7×D) ÷ 7 which becomes ((123 – 10×D) ÷ 7) + F = (82 + 1½×D) ÷ 7 Subtract ((123 – 10×D) ÷ 7) from both sides of the above equation: ((123 – 10×D) ÷ 7) + F – ((123 – 10×D) ÷ 7) = (82 + 1½×D) ÷ 7 – ((123 – 10×D) ÷ 7) which becomes F = (82 – 123 + 1½×D + 10×D) ÷ 7 which becomes F = (–41 + 11½×D) ÷ 7 which is the same as eq.3c) F = (11½×D – 41) ÷ 7
Hint #10
Substitute ((123 – 13½×D) ÷ 7) for A (from eq.1e), 1½×D for C, ((123 – 10×D) ÷ 7) for E (from eq.1d), and ((11½×D – 41) ÷ 7) for F (from eq.3c) in eq.6: ((123 – 13½×D) ÷ 7) + 1½×D + ((123 – 10×D) ÷ 7) = D × ((11½×D – 41) ÷ 7) Multiply both sides of the above equation by 7: 7×(((123 – 13½×D) ÷ 7) + 1½×D + ((123 – 10×D) ÷ 7)) = 7×(D × ((11½×D – 41) ÷ 7)) which becomes 123 – 13½×D + 7×(1½×D) + 123 – 10×D = D × (11½×D – 41) which becomes 246 – 13×D = 11½×D² – 41×D Add (13×D – 246) to both sides of the above equation: 246 – 13×D + (13×D – 246) = 11½×D² – 41×D + (13×D – 246) which becomes eq.6a) 0 = 11½×D² – 28×D – 246
Solution
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.6a yields: D = { (–1)×(–28) ± sq.rt.[(–28)² – (4 × 11½ × (–246))] } ÷ (2 × 11½) which becomes D = {28 ± sq.rt.[784 – (–11316)]} ÷ 23 which becomes D = {28 ± sq.rt.[12100]} ÷ 23 which becomes D = {28 ± 110} ÷ 23 In the above equation, either D = {28 + 110} ÷ 23 = 138 ÷ 23 = 6 or D = {28 – 110} ÷ 23 = –82 ÷ 23 = –3.5652173913 Since D must be a non–negative integer, then D ≠ –3.5652173913 Therefore, D = 6 making A = (123 – 13½×D) ÷ 7 = (123 – 13½ × 6) ÷ 7 = (123 – 81) ÷ 7 = 42 ÷ 7 = 6 (from eq.1e) B = (82 – 5½×D) ÷ 7 = (82 – 5½ × 6) ÷ 7 = (82 – 33) ÷ 7 = 49 ÷ 7 = 7 (from eq.1c) C = 1½×D = 1½ × 6 = 9 E = (123 – 10×D) ÷ 7 = (123 – 10 × 6) ÷ 7 = (123 – 60) ÷ 7 = 63 ÷ 7 = 9 (from eq.1d) F = (11½×D – 41) ÷ 7 = (11½ × 6 – 41) ÷ 7 = (69 – 41) ÷ 7 = 28 ÷ 7 = 4 (from eq.3c) and ABCDEF = 679694