Puzzle for May 4, 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 positive integer.
Scratchpad
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Hint #1
Add D to both sides of eq.2: A + B + C + D = E + F + D which may be written as eq.2a) A + B + C + D = D + E + F Add F to each side of eq.5: eq.5a) D + E + F = A + C + F + F
Hint #2
In eq.2a, replace D + E + F with A + C + F + F (from eq.5a): A + B + C + D = A + C + F + F Subtract both A and C from each side of the above equation: A + B + C + D - A - C = A + C + F + F - A - C which simplifies to B + D = 2×F In eq.4, replace B + D with 2×F: 2×F = C + F Subtract F from each side: 2×F - F = C + F - F which makes F = C
Hint #3
In eq.2, substitute C for F: A + B + C = E + C Subtract C from both sides of the above equation: A + B + C - C = E + C - C which becomes eq.2b) A + B = E
Hint #4
In eq.1, substitute E for A + B (from eq.2b) and for C + D (from eq.3): E + E + E + F = 25 which becomes 3×E + F = 25 Subtract 3×E from both sides of the above equation: 3×E + F - 3×E = 25 - 3×E which makes F = 25 - 3×E which also makes eq.1a) C = F = 25 - 3×E
Hint #5
Subtract C from both sides of eq.3: E - C = C + D - C which becomes E - C = D Substitute (25 - 3×E) for C (from eq.1a) in the equation above: E - (25 - 3×E) = D which is equivalent to E - 25 + 3×E = D which makes eq.3a) 4×E - 25 = D
Hint #6
Subtract D from both sides of eq.4: B + D - D = C + F - D which becomes B = C + F - D Substitute 25 - 3×E for both C and F (from eq.1a), and (4×E - 25) for D (from eq.3a) in the equation above: B = 25 - 3×E + 25 - 3×E - (4×E - 25) which becomes B = 50 - 6×E - 4×E + 25 which makes eq.4a) B = 75 - 10×E
Hint #7
Substitute 75 - 10×E for B (from eq.4a) in eq.2b: A + 75 - 10×E = E Add (10×E - 75) to each side: A + 75 - 10×E + (10×E - 75) = E + (10×E - 75) which becomes eq.2c) A = 11×E - 75
Hint #8
In eq.6, substitute 25 - 3×E for both C and F (from eq.1a), (75 - 10×E) for B (from eq.4a), and (4×E - 25) for D (from eq.3a): 25 - 3×E + E + 25 - 3×E = (75 - 10×E) × (4×E - 25) which becomes 50 - 5×E = 300×E - 40×E² - 1875 + 250×E Divide both sides of the above equation by 5: (50 - 5×E) ÷ 5 = (300×E - 40×E² - 1875 + 250×E) ÷ 5 which becomes 10 - E = 60×E - 8×E² - 375 + 50×E Add (E - 10) to both sides: 10 - E + (E - 10) = 60×E - 8×E² - 375 + 50×E + (E - 10) which becomes eq.6a) 0 = -8×E² + 111×E - 385
Solution
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.6a yields: E = {(-1)×111 ± sq.rt.[111² - (4 × (-8) × (-385))]} ÷ (2 × (-8)) which becomes E = {-111 ± sq.rt.[12321 - 12320]} ÷ (-16) which becomes E = {-111 ± sq.rt.[1]} ÷ (-16) which becomes E = {-111 ± 1} ÷ (-16) In the above equation, either: E = {-111 + 1} ÷ (-16) = -110 ÷ (-16) = 6⅞ or: E = {-111 - 1} ÷ (-16) = -112 ÷ (-16) = 7 Since E must be a non-negative integer, then E ≠ 6⅞ and therefore E = 7 making A = 11×E - 75 = 11×7 - 75 = 77 - 75 = 2 (from eq.2c) B = 75 - 10×E = 75 - 10×7 = 75 - 70 = 5 (from eq.4a) C = F = 25 - 3×E = 25 - 3×7 = 25 - 21 = 4 (from eq.1a) D = 4×E - 25 = 4×7 - 25 = 28 - 25 = 3 (from eq.3a) and ABCDEF = 254374