Puzzle for April 28, 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
In eq.2, replace D + E with B + C (from eq.3): A + C = B + C + F Subtract C from both sides of the equation above: A + C - C = B + C + F - C which becomes eq.2a) A = B + F
Hint #2
In eq.2a, replace A with B + D (from eq.5): B + D = B + F Subtract B from both sides of the above equation: B + D - B = B + F - B which makes D = F
Hint #3
In eq.4, replace A with B + D (from eq.5): C = B + D + B which becomes C = 2×B + D Substitute 2×B + D for C in eq.3: D + E = B + 2×B + D which becomes D + E = 3×B + D Subtract D from each side: D + E - D = 3×B + D - D which makes E = 3×B
Hint #4
Substitute 3×B for E in eq.6: A + D = B × 3×B which may be expressed as A + D = 3×B² In the equation above, substitute B + D for A (from eq.5): B + D + D = 3×B² Subtract B from each side: B + D + D - B = 3×B² - B which becomes 2×D = 3×B² - B Divide both sides by 2: 2×D ÷ 2 = (3×B² - B) ÷ 2 which becomes D = 1½×B² - ½×B which also makes eq.6a) F = D = 1½×B² - ½×B
Hint #5
Substitute 1½×B² - ½×B for F (from eq.6a) in eq.2a: A = B + 1½×B² - ½×B which becomes eq.2b) A = 1½×B² + ½×B
Hint #6
Substitute 1½×B² + ½×B for A (from eq.2b) in eq.4: C = 1½×B² + ½×B + B which becomes eq.4a) C = 1½×B² + 1½×B
Hint #7
In eq.1, substitute 1½×B² + ½×B for A (from eq.2b), 1½×B² + 1½×B for C (from eq.4a), 1½×B² - ½×B for both D and F (from eq.6a), and 3×B for E: 1½×B² + ½×B + B + 1½×B² + 1½×B + 1½×B² - ½×B + 3×B + 1½×B² - ½×B = 34 which simplifies to 6×B² + 5×B = 34 Subtract 34 from each side of the above equation: 6×B² + 5×B - 34 = 34 - 34 which makes eq.1a) 6×B² + 5×B - 34 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.1a yields: B = {(-1)×5 ± sq.rt.[5² - (4 × 6 × (-34))]} ÷ (2 × 6) which becomes B = {-5 ± sq.rt.[25 - (-816)]} ÷ 12 which is equivalent to B = {-5 ± sq.rt.[841]} ÷ 12 which becomes B = {-5 ± 29} ÷ 12 In the above equation, either: B = {-5 + 29} ÷ 12 = 24 ÷ 12 = 2 or: B = {-5 - 29} ÷ 12 = -34 ÷ 12 = -2⅚ Since B must be a non-negative integer, then B ≠ -2⅚ and therefore B = 2 making A = 1½×B² + ½×B = 1½ × 2² + ½ × 2 = 6 + 1 = 7 (from eq.2b) C = 1½×B² + 1½×B = 1½ × 2² + 1½ × 2 = 6 + 3 = 9 (from eq.4a) D = F = 1½×B² - ½×B = 1½ × 2² - ½ × 2 = 6 - 1 = 5 (from eq.6a) E = 3×B = 3×2 = 6 and ABCDEF = 729565