Puzzle for January 8, 2023 ( )
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
Help Area
Hint #1
Subtract E and B from both sides of eq.4: D + E - E - B = B + C - E - B which becomes eq.4a) D - B = C - E Subtract D and E from both sides of eq.5: C + D - D - E = B + E - D - E which becomes eq.5a) C - E = B - D
Hint #2
In eq.5a, replace C - E with D - B (from eq.4a): D - B = B - D Add B and D to both sides of the above equation: D - B + B + D = B - D + B + D which makes 2×D = 2×B Divide both sides by 2: 2×D ÷ 2 = 2×B ÷ 2 which makes D = B
Hint #3
In eq.4, substitute B for D: B + E = B + C Subtract B from each side of the equation above: B + E - B = B + C - B which makes E = C
Hint #4
In eq.6, substitute B for D, and C for E: B × B = A + C + C which becomes eq.6a) B² = A + 2×C
Hint #5
Subtract D from both sides of eq.2: E - D = D + F - D which becomes eq.2a) E - D = F Substitute E - D for F (from eq.2a) in eq.3: eq.3a) A = C + E - D
Hint #6
Substitute C for E, and B for D in eq.3a: A = C + C - B which becomes eq.3b) A = 2×C - B
Hint #7
Substitute 2×C - B for A (from eq.3b) into eq.6a: B² = 2×C - B + 2×C which becomes B² = 4×C - B Add B to both sides of the equation above: B² + B = 4×C - B + B which becomes B² + B = 4×C Divide both sides by 4: (B² + B) ÷ 4 = 4×C ÷ 4 which makes (B² + B) ÷ 4 = C and also makes eq.6b) (B² + B) ÷ 4 = C = E
Hint #8
Add B to both sides of eq.3b: A + B = 2×C - B + B which becomes eq.3c) A + B = 2×C
Hint #9
Substitute A + B for 2×C (from eq.3c) into eq.6a: B² = A + A + B which becomes B² = 2×A + B Subtract B from each side of the equation above: B² - B = 2×A + B - B which becomes B² - B = 2×A Divide both sides by 2: (B² - B) ÷ 2 = 2×A ÷ 2 which makes eq.6c) (B² - B) ÷ 2 = A
Hint #10
Substitute (B² + B) ÷ 4 for E (from eq.6b), and B for D in eq.2a: eq.2b) (B² + B) ÷ 4 - B = F
Hint #11
Substitute (B² - B) ÷ 2 for A (from eq.6c), (B² + B) ÷ 4 for C and E (from eq.6b), B for D, and (B² + B) ÷ 4 - B for F (from eq.2b) in eq.1: (B² - B) ÷ 2 + B + (B² + B) ÷ 4 + B + (B² + B) ÷ 4 + (B² + B) ÷ 4 - B = 25 Multiply both sides of the above equation by 4: 4 × ((B² - B) ÷ 2 + B + (B² + B) ÷ 4 + B + (B² + B) ÷ 4 + (B² + B) ÷ 4 - B) = 4 × 25 which becomes 2×B² - 2×B + 4×B + B² + B + 4×B + B² + B + B² + B - 4×B = 100 which becomes 5×B² + 5×B = 100 Divide both sides by 5: (5×B² - 5×B) ÷ 5 = 100 ÷ 5 which becomes B² + B = 20 Subtract 20 from both sides: B² + B - 20 = 20 - 20 which becomes eq.1a) B² + B - 20 = 0
Hint #12
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)×(1) ± sq.rt.[(1)² - (4 × (1) × (-20))] } ÷ (2 × (1)) which becomes B = {-1 ± sq.rt.(1 - (-80))} ÷ 2 which becomes B = {-1 ± sq.rt.(81)} ÷ 2 which becomes B = (-1 ± 9) ÷ 2 In the above equation, either B = (-1 + 9) ÷ 2 = 8 ÷ 2 = 4 or B = (-1 - 9) ÷ 2 = -10 ÷ 2 = -5 Since B must be a non-negative integer, then B ≠ -5 and therefore makes B = 4
Solution
Since B = 4, then: A = (B² - B) ÷ 2 = (4² - 4) ÷ 2 = (16 - 4) ÷ 2 = 12 ÷ 2 = 6 (from eq.6c) C = E = (B² + B) ÷ 4 = (4² + 4) ÷ 4 = (16 + 4) ÷ 4 = 20 ÷ 4 = 5 (from eq.6b) D = B = 4 F = (B² + B) ÷ 4 - B = (4² + 4) ÷ 4 - 4 = (16 + 4) ÷ 4 - 4 = 20 ÷ 4 - 4 = 5 - 4 = 1 (from eq.2b) and ABCDEF = 645451