Puzzle for December 24, 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
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Hint #1
eq.1 may be written as: A + B + F + C + D + E = 26 In the above equation, replace C + D + E with A + B + F (from eq.5): A + B + F + A + B + F = 26 which becomes 2×(A + B + F) = 26 Divide both sides by 2: 2×(A + B + F) ÷ 2 = 26 ÷ 2 which becomes eq.1a) A + B + F = 13
Hint #2
In eq.1a, replace A + B + F with C + D + E (from eq.5): eq.1b) C + D + E = 13
Hint #3
Add F and D to both sides of eq.3: D - F + F + D = B - D + F + D which becomes eq.3a) 2×D = B + F
Hint #4
In eq.5, substitute C + D for A (from eq.2), and 2×D for B + F (from eq.3a): C + D + E = C + D + 2×D Subtract C and D from each side of the equation above: C + D + E - C - D = C + D + 2×D - C - D which makes E = 2×D
Hint #5
Substitute 2×D for E in eq.4: B + D = C + 2×D Subtract D from each side of the above equation: B + D - D = C + 2×D - D which becomes eq.4a) B = C + D
Hint #6
Substitute B for C + D (from eq.4a) in eq.2: A = B
Hint #7
Substitute 2×D for E in eq.1b: C + D + 2×D = 13 which becomes C + 3×D = 13 Subtract 3×D from each side of the above equation: C + 3×D - 3×D = 13 - 3×D which becomes eq.1c) C = 13 - 3×D
Hint #8
Substitute 13 - 3×D for C (from eq.1c) into eq.4a: B = 13 - 3×D + D which becomes B = 13 - 2×D and also makes eq.4b) A = B = 13 - 2×D
Hint #9
Substitute 13 - 2×D for B (from eq.4b) into eq.3a: 2×D = 13 - 2×D + F In the above equation, subtract 13 from both sides, and add 2×D to both sides: 2×D - 13 + 2×D = 13 - 2×D + F - 13 + 2×D which becomes eq.3b) 4×D - 13 = F
Hint #10
Substitute (13 - 2×D) for A and B (from eq.4b), 13 - 3×D for C (from eq.1c), 2×D for E, and (4×D - 13) for F (from eq.3b) in eq.6: (13 - 2×D) × (13 - 2×D) = 13 - 3×D + (2×D × (4×D - 13)) which becomes 169 - 26×D - 26×D + 4×D² = 13 - 3×D + 8×D² - 26×D which becomes 169 - 52×D + 4×D² = 13 - 29×D + 8×D² In the above equation, subtract 169 and 4×D² from both sides, and add 52×D to both sides: 169 - 52×D + 4×D² - 169 - 4×D² + 52×D = 13 - 29×D + 8×D² - 169 - 4×D² + 52×D which becomes 0 = -156 + 23×D + 4×D² which may be written as eq.6a) 0 = 4×D² + 23×D - 156
Hint #11
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.6a yields: D = { (-1)×23 ± sq.rt.[23² - (4 × 4 × (-156))] } ÷ (2 × 4) which becomes D = {-23 ± sq.rt.(529 + 2496)} ÷ 8 which becomes D = {-23 ± sq.rt.(3025)} ÷ 8 which becomes eq.6b) D = (-23 ± 55) ÷ 8
Hint #12
In eq.6b, either: D = (-23 + 55) ÷ 8 = 32 ÷ 8 = 4 or: D = (-23 - 55) ÷ 8 = -78 ÷ 8 = -9¾ Since D must be a non-negative integer, then D ≠ -9¾ and therefore makes D = 4
Solution
Since D = 4, then: A = B = 13 - 2×D = 13 - 2×4 = 13 - 8 = 5 (from eq.4b) C = 13 - 3×D = 13 - 3×4 = 13 - 12 = 1 (from eq.1c) E = 2×D = 2×4 = 8 F = 4×D - 13 = 4×4 - 13 = 16 - 13 = 3 (from eq.3b) and ABCDEF = 551483