Puzzle for October 23, 2020 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
* A! is A-factorial. BC and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
In eq.3, replace A + C with D + E (from eq.1): D + E – F = D + F In the equation above, subtract D from both sides, and add F to both sides: D + E – F – D + F = D + F – D + F which makes E = 2×F
Hint #2
In eq.4, replace A + C with D + E (from eq.1): B + D = D + E + F Subtract D from each side of the above equation: B + D – D = D + E + F – D which becomes eq.4a) B = E + F
Hint #3
In eq.4a, substitute 2×F for E: B = 2×F + F which makes B = 3×F
Hint #4
Subtract C from both sides of eq.1: D + E – C = A + C – C which becomes D + E – C = A Substitute 2×F for E in the above equation: eq.1a) D + 2×F – C = A
Hint #5
Substitute 2×F for E, D + 2×F – C for A (from eq.1a), and 3×F for B in eq.2: C + D + 2×F = D + 2×F – C + 3×F + F which becomes C + D + 2×F = D – C + 6×F Subtract D and 2×F from both sides of the above equation: C + D + 2×F – D – 2×F = D – C + 6×F – D – 2×F which becomes C = –C + 4×F Add C to both sides: C + C = –C + 4×F + C which becomes 2×C = 4×F Divide both sides by 2: 2×C ÷ 2 = 4×F ÷ 2 which makes C = 2×F
Hint #6
Substitute 2×F for C in eq.1a: D + 2×F – 2×F = A which makes D = A
Hint #7
Substitute D for A, and 2×F for both C and E in eq.5: 2×F + 2×F + F = D + D which makes 5×F = 2×D Divide both sides of the equation above by 2: 5×F ÷ 2 = 2×D ÷ 2 which makes eq.5a) 2½×F = D
Hint #8
eq.6 may be written as: A! = 10×B + C + 10×D + E + F Substitute 3×F for B, 2×F for C and E, and 2½×F for D in the equation above: A! = 10×(3×F) + 2×F + 10×(2½×F) + 2×F + F which is equivalent to A! = 30×F + 2×F + 25×F + 2×F + F which becomes eq.6a) A! = 60×F
Hint #9
In eq.5a, substitute A for D: 2½×F = A Divide both sides of the above equation by 2½: 2½×F ÷ 2½ = A ÷ 2½ which makes F = ⅖×A
Solution
Substitute (⅖×A) for F in eq.6a: A! = 60×(⅖×A) which becomes A! = 24×A Divide both sides of the above equation by A: A! ÷ A = 24×A ÷ A which becomes (A – 1)! = 24 which means A – 1 = 4 Add 1 to both sides: A – 1 + 1 = 4 + 1 which makes A = 5 making D = A = 5 F = ⅖×A = ⅖ × 5 = 2 B = 3×F = 3 × 2 = 6 C = E = 2×F = 2 × 2 = 4 and 564542