Puzzle for December 31, 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.
* BC is a 2-digit number (not B×C).
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Hint #1
Add B to both sides of eq.3: B + C + B = D + F - B + B which becomes eq.3a) 2×B + C = D + F Add A and D to both sides of eq.2: F - A + A + D = C - D + A + D which becomes F + D = C + A which is the same as eq.2a) D + F = C + A
Hint #2
In eq.2a, replace D + F with 2×B + C (from eq.3a): 2×B + C = C + A Subtract C from each side of the equation above: 2×B + C - C = C + A - C which becomes 2×B = A
Hint #3
In eq.4, replace A with 2×B: 2×B + F = B - D + E In the above equation, subtract B from both sides, and add D to both sides: 2×B + F - B + D = B - D + E - B + D which becomes eq.4a) B + F + D = E
Hint #4
eq.5 may be written as: D + E = B + C + A - D In the above equation, substitute D + F for C + A (from eq.2a): D + E = B + D + F - D which becomes eq.5a) D + E = B + F
Hint #5
In eq.5a, substitute B + F + D for E (from eq.4a): D + B + F + D = B + F which becomes 2×D + B + F = B + F Subtract B and F from both sides of the above equation: 2×D + B + F - B - F = B + F - B - F which simplifies to 2×D = 0 which means D = 0
Hint #6
Substitute 0 for D in eq.3a: 2×B + C = 0 + F which becomes eq.3b) 2×B + C = F
Hint #7
Substitute 0 for D, and 2×B + C for F (from eq.3b) in eq.5a: 0 + E = B + 2×B + C which becomes eq.5b) E = 3×B + C
Hint #8
eq.6 may be written as: 10×B + C = A × F Substitute 2×B for A, and (2×B + C) for F (from eq.3b) in the above equation: 10×B + C = 2×B × (2×B + C) which becomes 10×B + C = 4×B² + 2×B×C Subtract C and 4×B² from each side of the equation above: 10×B + C - C - 4×B² = 4×B² + 2×B×C - C - 4×B² which becomes 10×B - 4×B² = 2×B×C - C which may be written as 10×B - 4×B² = C × (2×B - 1) Since 2×B - 1 ≠ 0 (B ≠ ½, since B is an integer), divide both sides by (2×B - 1): (10×B - 4×B²) ÷ (2×B - 1) = C × (2×B - 1) ÷ (2×B - 1) which becomes eq.6a) (10×B - 4×B²) ÷ (2×B - 1) = C
Hint #9
Substitute ((10×B - 4×B²) ÷ (2×B - 1)) for C (from eq.6a) into eq.3b: eq.3c) 2×B + ((10×B - 4×B²) ÷ (2×B - 1)) = F
Hint #10
Substitute ((10×B - 4×B²) ÷ (2×B - 1)) for C (from eq.6a) into eq.5b: eq.5c) E = 3×B + ((10×B - 4×B²) ÷ (2×B - 1))
Hint #11
In eq.1, substitute -- 2×B for A, ((10×B - 4×B²) ÷ (2×B - 1)) for C (from eq.6a), 0 for D, 3×B + ((10×B - 4×B²) ÷ (2×B - 1)) for E (from eq.5c), and 2×B + ((10×B - 4×B²) ÷ (2×B - 1)) for F (from eq.3c): 2×B + B + ((10×B - 4×B²) ÷ (2×B - 1)) + 0 + 3×B + ((10×B - 4×B²) ÷ (2×B - 1)) + 2×B + ((10×B - 4×B²) ÷ (2×B - 1)) = 26 which becomes eq.1a) 8×B + 3×(10×B - 4×B²) ÷ (2×B - 1) = 26
Hint #12
Multiply both sides of eq.1a by (2×B - 1): (2×B - 1) × (8×B + 3×(10×B - 4×B²) ÷ (2×B - 1)) = (2×B - 1) × 26 which becomes 16×B² - 8×B + 3×(10×B - 4×B²) = 52×B - 26 which becomes 16×B² - 8×B + 30×B - 12×B² = 52×B - 26 which becomes 4×B² + 22×B = 52×B - 26 In the equation above, subtract 52×B from both sides, and add 26 to both sides: 4×B² + 22×B - 52×B + 26 = 52×B - 26 - 52×B + 26 which becomes eq.1b) 4×B² - 30×B + 26 = 0
Hint #13
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.1b yields: B = { (-1)×(-30) ± sq.rt.[(-30)² - (4 × 4 × 26)] } ÷ (2 × 4) which becomes B = {30 ± sq.rt.(900 - 416)} ÷ 8 which becomes B = {30 ± sq.rt.(484)} ÷ 8 which becomes eq.1c) B = (30 ± 22) ÷ 8
Hint #14
In eq.1c, either B = (30 + 22) ÷ 8 = 52 ÷ 8 = 6½ or B = (30 - 22) ÷ 8 = 8 ÷ 8 = 1 Since B must be an integer, then B ≠ 6½ and therefore makes B = 1
Hint #15
Substitute 1 for B in eq.6a: (10×1 - 4×1²) ÷ (2×1 - 1) = C which becomes (10 - 4) ÷ (2 - 1) = C which becomes 6 ÷ 1 = C which makes 6 = C
Solution
Since B = 1 and C = 6, then: A = 2×1 = 2 E = 3×1 + 6 = 3 + 6 = 9 (from eq.5b) F = 2×1 + 6 = 2 + 6 = 8 (from eq.3b) and ABCDEF = 216098