Puzzle for September 18, 2022 ( )
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.3, replace F with A - C (from eq.2): A + D = B + C + A - C which becomes A + D = B + A Subtract A from each side of the equation above: A + D - A = B + A - A which makes D = B
Hint #2
eq.4 may be written as: D - E + F = (A + B + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × (D - E + F) = 3 × (A + B + E) ÷ 3 which becomes 3×D - 3×E + 3×F = A + B + E Add 3×E to both sides: 3×D - 3×E + 3×F + 3×E = A + B + E + 3×E which becomes eq.4a) 3×D + 3×F = A + B + 4×E
Hint #3
In eq.4a, replace D with B: 3×B + 3×F = A + B + 4×E Subtract B from each side of the above equation: 3×B + 3×F - B = A + B + 4×E - B which becomes eq.4b) 2×B + 3×F = A + 4×E
Hint #4
eq.5 may be written as: A + B - E = ((C + D) ÷ 2) + E Subtract E from both sides of the above equation: A + B - E - E = ((C + D) ÷ 2) + E - E which becomes A + B - 2×E = (C + D) ÷ 2 Multiply both sides by 2: 2 × (A + B - 2×E) = 2 × (C + D) ÷ 2 which becomes eq.5a) 2×A + 2×B - 4×E = C + D
Hint #5
In eq.5a, substitute B for D: 2×A + 2×B - 4×E = C + B In the above equation, add 4×E to both sides, and subtract C and B from both sides: 2×A + 2×B - 4×E + 4×E - C - B = C + B + 4×E - C - B which becomes eq.5b) 2×A + B - C = 4×E
Hint #6
Substitute 2×A + B - C for 4×E (from eq.5b) into eq.4b: 2×B + 3×F = A + 2×A + B - C which becomes 2×B + 3×F = 3×A + B - C Subtract B from each side of the equation above: 2×B + 3×F - B = 3×A + B - C - B which becomes eq.4c) B + 3×F = 3×A - C
Hint #7
Substitute (A - C) for F (from eq.2) into eq.4c: B + 3×(A - C) = 3×A - C which becomes B + 3×A - 3×C = 3×A - C In the above equation, subtract 3×A from both sides, and add 3×C to both sides: B + 3×A - 3×C - 3×A + 3×C = 3×A - C - 3×A + 3×C which makes B = 2×C and also makes eq.4d) D = B = 2×C
Hint #8
Substitute 2×C for B (from eq.4d) in eq.6: (A × 2×C) ÷ F = C × F In the equation above, multiply both sides by F, and divide both sides by C: (((A × 2×C) ÷ F) × F) ÷ C = ((C × F) × F) ÷ C which becomes (A × 2) = F × F which may be written as 2×A = F² Divide both sides by 2: (2×A) ÷ 2 = (F²) ÷ 2 which makes A = ½×F²
Hint #9
Substitute ½×F² for A in eq.2: F = ½×F² - C In the equation above, add C to both sides, and subtract F from both sides: F + C - F = ½×F² - C + C - F which becomes eq.2a) C = ½×F² - F
Hint #10
Substitute ½×F² - F for C (from eq.2a) in eq.4d: D = B = 2×(½×F² - F) which becomes eq.4e) D = B = F² - 2×F
Hint #11
Substitute (½×F²) for A, F² - 2×F for B (from eq,4e), and (½×F² - F) for C (from eq.2a) in eq.5b: 2×(½×F²) + F² - 2×F - (½×F² - F) = 4×E which becomes F² + F² - 2×F - ½×F² + F = 4×E which becomes 1½×F² - F = 4×E Divide both sides of the above equation by 4: (1½×F² - F) ÷ 4 = (4×E) ÷ 4 which becomes eq.5c) ⅜×F² - ¼×F = E
Hint #12
Substitute ½×F² for A, F² - 2×F for B and D (from eq.4e), ½×F² - F for C (from eq.2a), and ⅜×F² - ¼×F for E (from eq.5c) in eq.1: ½×F² + F² - 2×F + ½×F² - F + F² - 2×F + ⅜×F² - ¼×F + F = 37 which becomes 3⅜×F² - 4¼×F = 37 Subtract 37 from each side of the equation above: 3⅜×F² - 4¼×F - 37 = 37 - 37 which becomes 3⅜×F² - 4¼×F - 37 = 0 To eliminate fractions (and make a little less difficult to solve), multiply both sides by 8: 8 × (3⅜×F² - 4¼×F - 37) = 8 × 0 which becomes eq.1a) 27×F² - 34×F - 296 = 0
Hint #13
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for F in eq.1a yields: F = { (-1)×(-34) ± sq.rt.[(-34)² - (4 × 27 × (-296))] } ÷ (2 × 27) which becomes F = {34 ± sq.rt.(1156 - (-31968))} ÷ 54 which becomes F = {34 ± sq.rt.(33124)} ÷ 54 which becomes F = (34 ± 182) ÷ 54 In the above equation, either F = (34 + 182) ÷ 54 = 216 ÷ 54 = 4 or F = (34 - 182) ÷ 54 = -148 ÷ 54 = -2.74074074 Since F must be a positive integer, then F ≠ -2.74074074 and therefore makes F = 4
Solution
Since F = 4, then: A = ½×F² = ½ × 4² = ½ × 16 = 8 B = D = F² - 2×F = 4² - 2×4 = 16 - 8 = 8 (from eq.4e) C = ½×F² - F = (½ × 4²) - 4 = (½ × 16) - 4 = 8 - 4 = 4 (from eq.2a) E = ⅜×F² - ¼×F = (⅜ × 4²) - (¼ × 4) = (⅜ × 16) - 1 = 6 - 1 = 5 (from eq.5c) and ABCDEF = 884854