Puzzle for July 30, 2023 ( )
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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.
Our thanks go out to frequent contributor, Judah S (age 16), for sending us one of the most challenging puzzles we have seen so far! Thank you so much, Judah!
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Hint #1
Add C and B to both sides of eq.4: A + B - C + C + B = C + F - B + C + B which becomes eq.4a) A + 2×B = 2×C + F Add C to both sides of eq.6: C + E + F + C = (A × B) - C + C which becomes eq.6a) 2×C + E + F = A × B which may be written as eq.6b) 2×C + F + E = A × B
Hint #2
In eq.6a, replace E + F with B + C (from eq.2): 2×C + B + C = A × B which becomes eq.6c) 3×C + B = A × B
Hint #3
In eq.6b, replace 2×C + F with A + 2×B (from eq.4a), and A × B with 3×C + B (from eq.6c): A + 2×B + E = 3×C + B Subtract B from each side of the equation above: A + 2×B + E - B = 3×C + B - B which becomes eq.6d) A + B + E = 3×C
Hint #4
eq.1 may be written as: A + B + E + C + D + F = 15 In the above equation, substitute 3×C for A + B + E (from eq.6d): 3×C + C + D + F = 15 which becomes eq.1a) 4×C + D + F = 15
Hint #5
Substitute E + F for B + C (from eq.2) in eq.1: A + E + F + D + E + F = 15 which becomes eq.1b) A + 2×E + 2×F + D = 15
Hint #6
eq.1 may be written as: B + D + A + C + E + F = 15 Substitute A - D + F for B + D (from eq.3) in the above equation: A - D + F + A + C + E + F = 15 which becomes eq.1c) 2×A - D + 2×F + C + E = 15
Hint #7
Substitute A + 2×E + 2×F + D for 15 (from eq.1b) in eq.1c: 2×A - D + 2×F + C + E = A + 2×E + 2×F + D Subtract 2×F, E, and A from both sides of the above equation: 2×A - D + 2×F + C + E - 2×F - E - A = A + 2×E + 2×F + D - 2×F - E - A which simplifies to A - D + C = E + D which may be written as eq.1d) A - D + C = D + E
Hint #8
Add the left and right sides of eq.1 to the left and right sides of eq.4, respectively: A + B - C + A + B + C + D + E + F = C + F - B + 15 which becomes 2×A + 2×B + D + E + F = C + F - B + 15 In the above equation, subtract C and F from both sides, and add B to both sides: 2×A + 2×B + D + E + F - C - F + B = C + F - B + 15 - C - F + B which becomes eq.4b) 2×A + 3×B + D + E - C = 15
Hint #9
Substitute A - D + C for D + E (from eq.1d) into eq.4b: 2×A + 3×B + A - D + C - C = 15 which becomes eq.4c) 3×A + 3×B - D = 15
Hint #10
eq.5 may be written as: D - E = (A + B + C + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (D - E) = 4 × (A + B + C + F) ÷ 4 which becomes eq.5a) 4×D - 4×E = A + B + C + F
Hint #11
eq.1 may be written as: A + B + C + F + D + E = 15 Substitute 4×D - 4×E for A + B + C + F (from eq.5a) into the above equation: 4×D - 4×E + D + E = 15 which becomes eq.1e) 5×D - 3×E = 15
Hint #12
Substitute 5×D - 3×E for 15 (from eq.1e) in eq.4c: 3×A + 3×B - D = 5×D - 3×E Add D and 3×E to both sides of the equation above: 3×A + 3×B - D + D + 3×E = 5×D - 3×E + D + 3×E which becomes 3×A + 3×B + 3×E = 6×D Divide both sides by 3: (3×A + 3×B + 3×E) ÷ 3 = 6×D ÷ 3 which becomes eq.4d) A + B + E = 2×D
Hint #13
Substitute 3×C for A + B + E (from eq.6d) in eq.4d: 3×C = 2×D Divide both sides of the above equation by 2: 3×C ÷ 2 = 2×D ÷ 2 which makes 1½×C = D
Hint #14
Substitute (1½×C) for D in eq.1e: 5×(1½×C) - 3×E = 15 which becomes 7½×C - 3×E = 15 In the above equation, add 3×E to both sides, and subtract 15 from both sides: 7½×C - 15 = 3×E Divide both sides by 3: (7½×C - 15) ÷ 3 = 3×E ÷ 3 which becomes eq.1f) 2½×C - 5 = E
Hint #15
Substitute 1½×C for D in eq.1a: 4×C + 1½×C + F = 15 which becomes 5½×C + F = 15 Subtract 5½×C from each side of the equation above: 5½×C + F - 5½×C = 15 - 5½×C which becomes eq.1g) F = 15 - 5½×C
Hint #16
Substitute 2½×C - 5 for E (from eq.1f), and 15 - 5½×C for F (from eq.1g) in eq.2: 2½×C - 5 + 15 - 5½×C = B + C which becomes -3×C + 10 = B + C Subtract C from each side of the above equation: -3×C + 10 - C = B + C - C which becomes -4×C + 10 = B which is the same as eq.2a) 10 - 4×C = B
Hint #17
Substitute 10 - 4×C for B (from eq.2a), and 2½×C - 5 for E (from eq.1f) in eq.6d: A + 10 - 4×C + 2½×C - 5 = 3×C which becomes A + 5 - 1½×C = 3×C In the equation above, subtract 5 from both sides, and add 1½×C to both sides: A + 5 - 1½×C - 5 + 1½×C = 3×C - 5 + 1½×C which becomes eq.6e) A = 4½×C - 5
Hint #18
Substitute (10 - 4×C) for B (from eq.2a), and (4½×C - 5) for A (from eq.6e) in eq.6c: 3×C + (10 - 4×C) = (4½×C - 5) × (10 - 4×C) which becomes 10 - C = 45×C - 18×C² - 50 + 20×C which becomes 10 - C = 65×C - 18×C² - 50 In the equation above, subtract 10 from both sides, and add C to both sides: 10 - C - 10 + C = 65×C - 18×C² - 50 - 10 + C which becomes 0 = 66×C - 18×C² - 60 Divide both sides by 6: 0 ÷ 6 = (66×C - 18×C² - 60) ÷ 6 which becomes 0 = 11×C - 3×C² - 10 which may be written as eq.6f) 0 = -3×C² + 11×C - 10
Hint #19
eq.6f is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.6f yields: C = { (-1)×(11) ± sq.rt.[(11)² - (4 × (-3) × (-10))] } ÷ (2 × (-3)) which becomes C = {-11 ± sq.rt.(121 - 120)} ÷ (-6) which becomes C = {-11 ± sq.rt.(1)} ÷ (-6) which becomes eq.6g) C = (-11 ± 1) ÷ (-6)
Hint #20
In eq.6g, either: C = (-11 + 1) ÷ (-6) = -10 ÷ (-6) = 1⅔ or: C = (-11 - 1) ÷ (-6) = -12 ÷ (-6) = 2 Since C must be an integer, then C ≠ 1⅔ and therefore makes C = 2
Solution
Since C = 2, then: A = 4½×C - 5 = (4½ × 2) - 5 = 9 - 5 = 4 (from eq.6e) B = 10 - 4×C = 10 - (4 × 2) = 10 - 8 = 2 (from eq.2a) D = 1½×C = 1½ × 2 = 3 E = 2½×C - 5 = (2½ × 2) - 5 = 5 - 5 = 0 (from eq.1f) F = 15 - 5½×C = 15 - (5½ × 2) = 15 - 11 = 4 (from eq.1g) and ABCDEF = 422304