Puzzle for January 12, 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 non-negative integer.
Scratchpad
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Hint #1
Add B to both sides of eq.3: F + B = A - B + B which becomes F + B = A In eq.2, replace A with F + B: F + B + F = B + D Subtract B from both sides of the above equation: F + B + F - B = B + D - B which simplifies to 2×F = D
Hint #2
Add (D - E) to each side of eq.4: A - D + (D - E) = C + E + (D - E) which becomes A - E = C + D In eq.5, substitute A - E for C + D: A - E = B - F Add E + F to both sides of the above equation: A - E + E + F = B - F + E + F which becomes eq.4a) A + F = B + E
Hint #3
In eq.2, replace A + F with B + E (from eq.4a): B + E = B + D Subtract B from both sides of the above equation: B + E - B = B + D - B which makes E = D and makes E = D = 2×F
Hint #4
In eq.6, substitute 2×F for both D and E: C = 2×F × 2×F which makes C = 4×F²
Hint #5
Substitute 4×F² for C, and 2×F for both D and E in eq.4: A - 2×F = 4×F² + 2×F Add 2×F to each side of the equation above: A - 2×F + 2×F = 4×F² + 2×F + 2×F which becomes eq.4b) A = 4×F² + 4×F
Hint #6
Substitute 4×F² for C, and 2×F for D in eq.5: 4×F² + 2×F = B - F Add F to each side: 4×F² + 2×F + F = B - F + F which makes eq.5a) 4×F² + 3×F = B
Hint #7
Substitute 4×F² + 4×F for A (from eq.4b), 4×F² + 3×F for B (from eq.5a), 4×F² for C, and 2×F for D and E in eq.1: 4×F² + 4×F + 4×F² + 3×F + 4×F² + 2×F + 2×F + F = 24 which simplifies to 12×F² + 12×F = 24 Divide both sides of the above equation by 12: (12×F² + 12×F) ÷ 12 = 24 ÷ 12 which becomes F² + F = 2 Subtract 2 from each side: F² + F - 2 = 2 - 2 which means eq.1a) F² + F - 2 = 0
Solution
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)×1 ± sq.rt.[1² - (4 × 1 × (-2))]} ÷ (2 × 1) which becomes F = {-1 ± sq.rt.[1 - (-8)]} ÷ 2 which is equivalent to F = {-1 ± sq.rt.[9]} ÷ 2 which becomes F = {-1 ± 3} ÷ 2 In the above equation, either: F = {-1 + 3} ÷ 2 = 2 ÷ 2 = 1 or: F = {-1 - 3} ÷ 2 = -4 ÷ 2 = -2 Since F must be a non-negative integer, then F ≠ -2 and therefore makes F = 1 making A = 4×F² + 4×F = 4×1² + 4×1 = 4 + 4 = 8 (from eq.4b) B = 4×F² + 3×F = 4×1² + 3×1 = 4 + 3 = 7 (from eq.5a) C = 4×F² = 4×1² = 4 D = E = 2×F = 2×1 = 2 and ABCDEF = 874221