Puzzle for October 2, 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.
* "D mod B" equals the remainder of (D ÷ B).
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Hint #1
eq.3 may be written as: B = (C + F) ÷ 2 Multiply both sides of the above equation by 2: 2 × B = 2 × (C + F) ÷ 2 which becomes eq.3a) 2×B = C + F
Hint #2
In eq.5, replace C + F with 2×B (from eq.3a): A ÷ (B – E) = 2×B Multiply both sides of the above equation by (B – E): A ÷ (B – E) × (B – E) = 2×B × (B – E) which becomes eq.5a) A = 2×B × (B – E)
Hint #3
Multiply both sides of eq.4 by E: E × B = E × A ÷ E which becomes E × B = A In eq.5a, replace A with E × B: eq.4a) E × B = 2×B × (B – E)
Hint #4
Since B ≠ 0 (from eq.6), divide both sides of eq.4a by B: (E × B) ÷ B = (2×B × (B – E)) ÷ B which becomes E = 2 × (B – E) which becomes E = 2×B – 2×E Add 2×E to both sides of the above equation: E + 2×E = 2×B – 2×E + 2×E which becomes 3×E = 2×B Divide both sides by 2: 3×E ÷ 2 = 2×B ÷ 2 which makes 1½×E = B
Hint #5
Substitute 1½×E for B in eq.4: 1½×E = A ÷ E Multiply both sides of the equation above by E: E × 1½×E = E × A ÷ E which makes 1½×E² = A
Hint #6
Substitute (1½×E) for B in eq.3a: 2×(1½×E) = C + F which becomes eq.3b) 3×E = C + F
Hint #7
In eq.2, add B and C and D to both sides, and subtract A from both sides: A – B – C + B + C + D – A = B – D + F + B + C + D – A which becomes D = 2×B + F + C – A which may be written as D = 2×B + C + F – A Substitute (1½×E) for B, 3×E for C + F (from eq.3b), and 1½×E² for A in the above equation: D = 2×(1½×E) + 3×E – 1½×E² which becomes D = 3×E + 3×E – 1½×E² which becomes eq.2a) D = 6×E – 1½×E²
Hint #8
eq.1 may be written as: A + B + C + F + D + E = 23 Substitute 1½×E² for A, 1½×E for B, 3×E for C + F (from eq.3b), and 6×E – 1½×E² for D (from eq.2a) in eq.1: 1½×E² + 1½×E + 3×E + 6×E – 1½×E² + E = 23 which becomes 11½×E = 23 Divide both sides of the above equation by 11½: 11½×E ÷ 11½ = 23 ÷ 11½ which makes E = 2 making A = 1½×E² = 1½ × 2² = 1½ × 4 = 6 B = 1½×E = 1½ × 2 = 3 D = 6×E – 1½×E² = (6 × 2) – (1½ × 2²) = 12 – (1½ × 4) = 12 – 6 = 6 (from eq.2a)
Hint #9
Substitute 6 for D, and 3 for B in eq.6: 6 mod 3 = F which means remainder of (6 ÷ 3) = F which means 0 = F
Solution
Substitute 2 for E, and 0 for F in eq.3b: 3×2 = C + 0 which makes 6 = C and makes ABCDEF = 636620