Puzzle for August 7, 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.
Scratchpad
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Hint #1
In eq.4, add C to both sides, and subtract (B × E) from both sides: (A × F) – C + C – (B × E) = (B × E) + D + C – (B × E) which becomes (A × F) – (B × E) = D + C which may be written as eq.4a) (A × F) – (B × E) = C + D
Hint #2
Subtract (E ÷ F) and C from both sides of eq.3: (E ÷ F) – D – (E ÷ F) – C = (A ÷ B) + C – (E ÷ F) – C which becomes –D – C = (A ÷ B) – (E ÷ F) which may be written as eq.3a) (–1) × (C + D) = (A ÷ B) – (E ÷ F)
Hint #3
Multiply both sides of eq.3a by B and F: B × F × (–1) × (C + D) = B × F × ((A ÷ B) – (E ÷ F)) which becomes B × F × (–1) × (C + D) = (B × F × (A ÷ B)) – (B × F × (E ÷ F)) which becomes B × F × (–1) × (C + D) = (F × A) – (B × E) which may be written as eq.3b) (–1) × B × F × (C + D) = (A × F) – (B × E)
Hint #4
In eq.3b, replace (A × F) – (B × E) with C + D (from eq.4a): (–1) × B × F × (C + D) = C + D Add (B × F × (C + D)) to both sides of the equation above: ((–1) × B × F × (C + D)) + (B × F × (C + D)) = C + D + (B × F × (C + D)) which becomes 0 = C + D + (B × F × (C + D)) which is the same as 0 = (1 × (C + D)) + (B × F × (C + D)) which may be written as eq.3c) 0 = (1 + (B × F)) × (C + D)
Hint #5
Since B and F are ≥ 0, then in eq.3c: 1 + (B × F) ≥ 1 To make eq.3c true, the above inequality requires that: C + D = 0 Since C and D are ≥ 0, then the above equation makes: C = 0 and D = 0
Hint #6
In eq.1, replace C and D with 0: 0 + E – A = A + 0 – F which becomes E – A = A – F Add A and F to both sides of the above equation: E – A + A + F = A – F + A + F which becomes eq.1a) E + F = 2×A
Hint #7
In eq.2, substitute 2×A for E + F (from eq.1a), and 0 for C and D: 0 + 2×A – B = A + B + 0 which becomes 2×A – B = A + B In the above equation, add B to both sides, and subtract A from both sides: 2×A – B + B – A = A + B + B – A which makes eq.2a) A = 2×B
Hint #8
In eq.3, substitute 0 for C and D, and 2×B for A: (E ÷ F) – 0 = (2×B ÷ B) + 0 which becomes E ÷ F = 2 Multiply both sides of the above equation by F: (E ÷ F) × F = 2 × F which makes E = 2×F
Hint #9
Substitute 2×B for A, and 2×F for E in eq.5: (2×B × 2×F) ÷ F = B × F which becomes (2×B × 2) = B × F which becomes 4×B = B × F Since B ≠ 0 (from eq.3), divide both sides of the above equation by B: 4×B ÷ B = (B × F) ÷ B which makes 4 = F and makes E = 2×F = 2 × 4 = 8
Hint #10
Substitute 8 for E, and 4 for F in eq.1a: 8 + 4 = 2×A which makes 12 = 2×A Divide both sides of the above equation by 2: 12 ÷ 2 = 2×A ÷ 2 which makes 6 = A
Solution
Substitute 6 for A in eq.2a: 6 = 2×B Divide both sides of the above equation by 2: 6 ÷ 2 = 2×B ÷ 2 which makes 3 = B and makes ABCDEF = 630084