Puzzle for July 23, 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.
* AB is a 2-digit number (not A×B).
Scratchpad
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Hint #1
Add D and C to both sides of eq.3: C - D + D + C = A - C + D + C which becomes eq.3a) 2×C = A + D
Hint #2
In eq.1, replace A + D with 2×C (from eq.3a): 2×C = B + C + E Subtract C from each side of the above equation: 2×C - C = B + C + E - C which becomes eq.1a) C = B + E
Hint #3
In eq.3a, substitute (B + E) for C (from eq.1a): 2×(B + E) = A + D which becomes eq.3b) 2×B + 2×E = A + D
Hint #4
eq.2 may be written as: E + F - B = A + D + B In the above equation, replace A + D with 2×B + 2×E (from eq.3b): E + F - B = 2×B + 2×E + B which becomes E + F - B = 3×B + 2×E Subtract E from both sides, and add B to both sides: E + F - B - E + B = 3×B + 2×E - E + B which becomes eq.2a) F = 4×B + E
Hint #5
eq.4 may be written as: E = (A + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + C + D + F) ÷ 4 which becomes 4×E = A + C + D + F which may be written as eq.4a) 4×E = A + D + C + F
Hint #6
In eq.4a, substitute 2×C for A + D (from eq.3a): 4×E = 2×C + C + F which becomes eq.4b) 4×E = 3×C + F
Hint #7
Substitute (B + E) for C (from eq.1a), and 4×B + E for F (from eq.2a) into eq.4b: 4×E = 3×(B + E) + 4×B + E which becomes 4×E = 3×B + 3×E + 4×B + E which becomes 4×E = 7×B + 4×E Subtract 4×E from both sides of the equation above: 4×E - 4×E = 7×B + 4×E - 4×E which makes 0 = 7×B which means 0 = B
Hint #8
Substitute 0 for B in eq.2a: F = 4×0 + E which becomes F = 0 + E which makes F = E
Hint #9
Substitute 0 for B in eq.1a: C = 0 + E which makes C = E
Hint #10
Substitute E for F in eq.6: A - (E ÷ E) = E - A which becomes A - 1 = E - A Add 1 and A to both sides of the above equation: A - 1 + 1 + A = E - A + 1 + A which becomes 2×A = E + 1 Divide both sides by 2: 2×A ÷ 2 = (E + 1) ÷ 2 which becomes eq.6a) A = ½×E + ½
Hint #11
Substitute E for C, and ½×E + ½ for A (from eq.6a) into eq.3a: 2×E = ½×E + ½ + D Subtract ½×E and ½ from each side of the equation above: 2×E - ½×E - ½ = ½×E + ½ + D - ½×E - ½ which becomes eq.3c) 1½×E - ½ = D
Hint #12
eq.5 may be written as: D × E = 10×A + B + C Substitute (1½×E - ½) for D (from eq.3c), ½×E + ½ for A (from eq.6a), 0 for B, and E for C in the above equation: (1½×E - ½) × E = 10×(½×E + ½) + 0 + E which becomes 1½×E² - ½×E = 5×E + 5 + E which becomes eq.5a) 1½×E² - ½×E = 6×E + 5
Hint #13
Subtract 6×E and 5 from both sides of eq.5a: 1½×E² - ½×E - 6×E - 5 = 6×E + 5 - 6×E - 5 which becomes 1½×E² - 6½×E - 5 = 0 Multiply both sides of the above equation by 2 (to eliminate fractions and make easier to solve): 2×(1½×E² - 6½×E - 5) = 0 which becomes eq.5b) 3×E² - 13×E - 10 = 0
Hint #14
eq.5b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.5b yields: E = { (-1)×(-13) ± sq.rt.[(-13)² - (4 × 3 × (-10))] } ÷ (2 × 3) which becomes E = {13 ± sq.rt.(169 - (-120))} ÷ 6 which becomes E = {13 ± sq.rt.(289)} ÷ 6 which becomes E = (13 ± 17) ÷ 6 In the above equation, either E = (13 + 17) ÷ 6 = 30 ÷ 6 = 5 or E = (13 - 17) ÷ 6 = (-4) ÷ 6 = -⅔ Since E must be a non-negative integer, then E ≠ -⅔ and therefore makes E = 5
Solution
Since E = 5, then: A = ½×E + ½ = ½×5 + ½ = 2½ + ½ = 3 (from eq.6a) C = F = E = 5 D = 1½×E - ½ = 1½×5 - ½ = 7½ - ½ = 7 (from eq.3c) and ABCDEF = 305755