Puzzle for October 9, 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 non-negative integer.
Scratchpad
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Hint #1
Add B and E to both sides of eq.3: E - B + B + E = A - E + B + E which becomes eq.3a) 2×E = A + B Add B and D to both sides of eq.4: D - B + B + D = A - D + E + B + D which becomes 2×D = A + E + B which may be written as eq.4a) 2×D = A + B + E
Hint #2
In eq.4a, replace A + B with 2×E (from eq.3a): 2×D = 2×E + E which makes 2×D = 3×E Divide both sides of the above equation by 2: 2×D ÷ 2 = 3×E ÷ 2 which makes D = 1½×E
Hint #3
In eq.6, replace D with 1½×E: 1½×E - E = A ÷ E which becomes ½×E = A ÷ E Multiply both sides of the above equation by E: E × ½×E = E × (A ÷ E) which makes ½×E² = A
Hint #4
In eq.3a, substitute ½×E² for A: 2×E = ½×E² + B Subtract ½×E² from each side of the equation above: 2×E - ½×E² = ½×E² + B - ½×E² which becomes eq.3b) 2×E - ½×E² = B
Hint #5
eq.5 may be written as: A - D = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × (A - D) = 3 × (B + C + D) ÷ 3 which becomes 3×A - 3×D = B + C + D Add 3×D to both sides of the above equation: 3×A - 3×D + 3×D = B + C + D + 3×D which becomes eq.5a) 3×A = B + C + 4×D
Hint #6
In eq.5a, substitute (½×E²) for A, 2×E - ½×E² for B (from eq.3b), and (1½×E) for D: 3×(½×E²) = 2×E - ½×E² + C + 4×(1½×E) which becomes 1½×E² = 2×E - ½×E² + C + 6×E which becomes 1½×E² = 8×E - ½×E² + C In the above equation, subtract 8×E from both sides, and add ½×E² to both sides: 1½×E² - 8×E + ½×E² = 8×E - ½×E² + C - 8×E + ½×E² which simplifies to eq.5b) 2×E² - 8×E = C
Hint #7
Substitute 1½×E for D, and ½×E² for A in eq.2: F - 1½×E = ½×E² - F Add 1½×E and F to both sides of the equation above: F - 1½×E + 1½×E + F = ½×E² - F + 1½×E + F which becomes 2×F = ½×E² + 1½×E Divide both sides by 2: 2×F ÷ 2 = (½×E² + 1½×E) ÷ 2 which becomes eq.2a) F = ¼×E² + ¾×E
Hint #8
Substitute ½×E² for A, 2×E - ½×E² for B (from eq.3b), 2×E² - 8×E for C (from eq.5b), 1½×E for D, and ¼×E² + ¾×E for F (from eq.2a) in eq.1: ½×E² + 2×E - ½×E² + 2×E² - 8×E + 1½×E + E + ¼×E² + ¾×E = 25 which becomes -2¾×E + 2¼×E² = 25 Subtract 25 from each side of the above equation: -2¾×E + 2¼×E² - 25 = 25 - 25 which becomes -2¾×E + 2¼×E² - 25 = 0 which may be written as 2¼×E² - 2¾×E - 25 = 0 To eliminate fractions (and make a little less difficult to solve), multiply both sides by 4: 4 × (2¼×E² - 2¾×E - 25) = 4 × 0 which becomes eq.1a) 9×E² - 11×E - 100 = 0
Hint #9
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.1a yields: E = { (-1)×(-11) ± sq.rt.[(-11)² - (4 × 9 × (-100))] } ÷ (2 × 9) which becomes E = {11 ± sq.rt.(121 - (-3600))} ÷ 18 which becomes E = {11 ± sq.rt.(3721)} ÷ 18 which becomes E = (11 ± 61) ÷ 18 In the above equation, either E = (11 + 61) ÷ 18 = 72 ÷ 18 = 4 or E = (11 - 61) ÷ 18 = -50 ÷ 18 = -2.777777777 Since E must be a non-negative integer, then E ≠ -2.777777777 and therefore makes E = 4
Solution
Since E = 4, then: A = ½×E² = ½ × 4² = ½ × 16 = 8 B = 2×E - ½×E² = (2 × 4) - (½ × 4²) = 8 - (½ × 16) = 8 - 8 = 0 (from eq.3b) C = 2×E² - 8×E = (2 × 4²) - (8 × 4) = (2 × 16) - 32 = 32 - 32 = 0 (from eq.5b) D = 1½×E = 1½ × 4 = 6 F = ¼×E² + ¾×E = (¼ × 4²) + (¾ × 4) = (¼ × 16) + 3 = 4 + 3 = 7 (from eq.2a) and ABCDEF = 800647