Puzzle for April 23, 2022 ( )
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 F" equals the remainder of B ÷ F.
** "E!" means E-factorial.
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Hint #1
Add C, E, and F to both sides of eq.2: A – C + F + C + E + F = C + D – E – F + C + E + F which becomes eq.2a) A + 2×F + E = 2×C + D Add C to both sides of eq.1: C + D + F + C = A + B – C + E + C which becomes eq.1a) 2×C + D + F = A + B + E
Hint #2
In eq.1a, replace 2×C + D with A + 2×F + E (from eq.2a): A + 2×F + E + F = A + B + E which becomes A + 3×F + E = A + B + E Subtract A and E from each side of the equation above: A + 3×F + E – A – E = A + B + E – A – E which simplifies to 3×F = B
Hint #3
In eq.5, replace B with 3×F: 3×F mod F = C which means remainder of (3×F ÷ F) = C which becomes remainder of (3) = C which means 0 = C
Hint #4
Substitute 3×F for B, and 0 for C in eq.3: 3×F ÷ E = F – 0 which becomes 3×F ÷ E = F Multiply both sides of the above equation by E: E × (3×F ÷ E) = E × F which becomes 3×F = E × F Since F ≠ 0 (from eq.5), divide both sides by F: 3×F ÷ F = (E × F) ÷ F which makes 3 = E
Hint #5
Substitute 3 for E, and 0 for C in eq.4: 3 ÷ A = F + 0 which makes 3 ÷ A = F Multiply both sides of the above equation by A: A × (3 ÷ A) = A × F which makes eq.4a) 3 = A × F
Hint #6
Substitute 3 for E, 3×F for B, and 0 for C in eq.6: 3! = A + 3×F + 0 which becomes 6 = A + 3×F which may be written as eq.6a) 2×3 = A + 3×F
Hint #7
Substitute (A × F) for 3 (from eq.4a) into eq.6a: 2 × (A × F) = A + ((A × F) × F) Since A ≠ 0 (from eq.4), divide both sides of the above equation by A: (2 × (A × F)) ÷ A = (A + ((A × F) × F)) ÷ A which becomes 2 × F = 1 + (F × F) which becomes eq.6b) 2×F = 1 + F²
Hint #8
Subtract 2×F from each side of eq.6b: 2×F – 2×F = 1 + F² – 2×F which becomes 0 = 1 + F² – 2×F which may be written as 0 = F² – 2×F + 1 which may be re-written as 0 = (F – 1)² which makes F = 1 and makes B = 3×F = 3×1 = 3
Hint #9
Substitute 1 for F in eq.4a: 3 = A × 1 which makes 3 = A
Solution
Substitute 3 for A and E, 1 for F, and 0 for C in eq.2a: 3 + 2×1 + 3 = 2×0 + D which becomes 3 + 2 + 3 = 0 + D which makes 8 = D and makes ABCDEF = 330831