Puzzle for July 11, 2021 ( )
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.
* "B mod A" equals the remainder of (B ÷ A).
Scratchpad
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Hint #1
Multiply both sides of eq.5 by C: A × C = B ÷ C × C which becomes eq.5a) A × C = B In eq.6, substitute (A × C) for B (from eq.5a): F = (A × C) mod A which is the same as eq.6a) F = remainder of ((A × C) ÷ A)
Hint #2
In eq.6a: (A × C) ÷ A = C which may be written as (A × C) ÷ A = C ÷ 1 In eq.6a, substitute C ÷ 1 for (A × C) ÷ A: F = remainder of (C ÷ 1) which makes F = 0
Hint #3
In eq.3, replace F with 0: B - A = A + C - 0 which becomes B - A = A + C Add A to both sides of the above equation: B - A + A = A + C + A which becomes eq.3a) B = 2×A + C
Hint #4
In eq.3a, replace B with A × C (from eq.5a): eq.3b) A × C = 2×A + C Subtract 2×A from each side of eq.3b: A × C - 2×A = 2×A + C - 2×A which becomes A × C - 2×A = C which may be written as A × (C - 2) = C Divide both sides of the equation above by (C - 2): A × (C - 2) ÷ (C - 2) = C ÷ (C - 2) which becomes eq.3c) A = C ÷ (C - 2) (assumes C ≠ 2)
Hint #5
Check: C ≠ 2 ... If C = 2, then substituting 2 for C in eq.3b would yield: A × 2 = 2×A + 2 which may be written as 2×A = 2×A + 2 Subtracting 2×A from both sides of the above equation would yield: 2×A - 2×A = 2×A + 2 - 2×A which would make 0 = 2 Since 0 ≠ 2, then C ≠ 2
Hint #6
In eq.5, substitute C ÷ (C - 2) for A (from eq.3c): C ÷ (C - 2) = B ÷ C Multiply both sides of the equation above by C: (C ÷ (C - 2)) × C = (B ÷ C) × C which becomes eq.5b) C² ÷ (C - 2) = B
Hint #7
Substitute 0 for F in eq.4: C - E = A - D + 0 which becomes eq.4a) C - E = A - D
Hint #8
Substitute 0 for F in eq.2: D + E - A = A + B + C - 0 which becomes D + E - A = A + B + C Add A to both sides of the equation above: D + E - A + A = A + B + C + A which becomes eq.2a) D + E = 2×A + B + C
Hint #9
Add the left and right sides of eq.4a to the left and right sides of eq.2a, respectively: D + E + C - E = 2×A + B + C + A - D which becomes D + C = 3×A + B + C - D In the above equation, add D to both sides, and subtract C from both sides: D + C + D - C = 3×A + B + C - D + D - C which becomes eq.2b) 2×D = 3×A + B
Hint #10
Substitute (C ÷ (C - 2)) for A (from eq.3c), and (C² ÷ (C - 2)) for B (from eq.5b) in eq.2b: 2×D = 3×(C ÷ (C - 2)) + (C² ÷ (C - 2)) which may be written as 2×D = (3×C + C²) ÷ (C - 2) Divide both sides of the above equation by 2: 2×D ÷ 2 = ((3×C + C²) ÷ (C - 2)) ÷ 2 which becomes eq.2c) D = (1½×C + ½×C²) ÷ (C - 2)
Hint #11
Substitute (C ÷ (C - 2)) for A (from eq.3c), and ((1½×C + ½×C²) ÷ (C - 2)) for D (from eq.2c) in eq.4a: C - E = (C ÷ (C - 2)) - ((1½×C + ½×C²) ÷ (C - 2)) which becomes C - E = (C - 1½×C - ½×C²) ÷ (C - 2) which becomes C - E = (-½×C - ½×C²) ÷ (C - 2) Subtract C from each side of the equation above: C - E - C = (-½×C - ½×C²) ÷ (C - 2) - C which becomes eq.4b) -E = (-½×C - ½×C²) ÷ (C - 2) - C
Hint #12
To give terms a common denominator, multiply the last C in eq.4b by ((C - 2) ÷ (C - 2)): -E = (-½×C - ½×C²) ÷ (C - 2) - (C × ((C - 2) ÷ (C - 2))) which becomes -E = (-½×C - ½×C²) ÷ (C - 2) - (C² - 2×C) ÷ (C - 2) which becomes -E = (-½×C - ½×C² - C² + 2×C) ÷ (C - 2) which becomes -E = (-1½×C² + 1½×C) ÷ (C - 2) Multiply both sides by (-1): -E × (-1) = (-1½×C² + 1½×C) ÷ (C - 2) × (-1) which becomes eq.4c) E = (1½×C² - 1½×C) ÷ (C - 2)
Hint #13
In eq.1, substitute -- C ÷ (C - 2) for A (from eq.3c), C² ÷ (C - 2) for B (from eq.5b), (1½×C + ½×C²) ÷ (C - 2) for D (from eq.2c), (1½×C² - 1½×C) ÷ (C - 2) for E (from eq.4c), and 0 for F: C ÷ (C - 2) + C² ÷ (C - 2) + C + (1½×C + ½×C²) ÷ (C - 2) + (1½×C² - 1½×C) ÷ (C - 2) + 0 = 30 which becomes (C + C² + 1½×C + ½×C² + 1½×C² - 1½×C) ÷ (C - 2) + C = 30 which becomes eq.1a) (3×C² + C) ÷ (C - 2) + C = 30
Hint #14
Subtract C from both sides of eq.1a: (3×C² + C) ÷ (C - 2) + C - C = 30 - C which becomes (3×C² + C) ÷ (C - 2) = 30 - C Multiply both sides of the equation above by (C - 2): (3×C² + C) ÷ (C - 2) × (C - 2) = (30 - C) × (C - 2) which becomes 3×C² + C = 30×C - C² - 60 + 2×C which becomes 3×C² + C = 32×C - C² - 60 In the equation above, subtract 32×C from both sides, and add C² and 60 to both sides: 3×C² + C - 32×C + C² + 60 = 32×C - C² - 60 - 32×C + C² + 60 which makes eq.1b) 4×C² - 31×C + 60 = 0
Solution
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.1b yields: C = { (-1)×(-31) ± sq.rt.[(-31)² - (4 × (4) × (60))] } ÷ (2 × (4)) which becomes C = {31 ± sq.rt.[961 - (4 × (4) × (60))]} ÷ 8) which becomes C = {31 ± sq.rt.[961 - 960]} ÷ 8) which becomes C = {31 ± sq.rt.(1)} ÷ 8) which becomes C = (31 ± 1) ÷ 8) In the above equation, either C = (31 + 1) ÷ 8 = 32 ÷ 8 = 4 or C = (31 - 1) ÷ 8 = 30 ÷ 8 = 3¾ Since C must be an integer, then C ≠ 3¾ and therefore makes C = 4 making A = C ÷ (C - 2) = 4 ÷ (4 - 2) = 4 ÷ 2 = 2 (from eq.3c) B = C² ÷ (C - 2) = 4² ÷ (4 - 2) = 16 ÷ 2 = 8 (from eq.5b) D = (1½×C + ½×C²) ÷ (C - 2) = (1½×4 + ½×4²) ÷ (4 - 2) = (6 + 8) ÷ 2 = 14 ÷ 2 = 7 (from eq.2c) E = (1½×C² - 1½×C) ÷ (C - 2) = (1½×4² - 1½×4) ÷ (4 - 2) = (24 - 6) ÷ 2 = 18 ÷ 2 = 9 (from eq.4c) and ABCDEF = 284790