Puzzle for July 24, 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 positive integer.
Scratchpad
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Hint #1
Subtract the left and right sides of eq.2 from the left and right sides of eq.1, respectively: C – B = A + F – (D + F) which becomes C – B = A + F – D – F which becomes C – B = A – D Add B and D to both sides of the above equation: C – B + B + D = A – D + B + D which becomes eq.1a) C + D = A + B
Hint #2
In eq.3, replace C + D with A + B (from eq.1a): B + F = A + B + E Subtract B from each side of the equation above: B + F – B = A + B + E – B which becomes eq.3a) F = A + E
Hint #3
In eq.1, replace F with A + E (from eq.3a): C = A + A + E which becomes eq.1b) C = 2×A + E
Hint #4
In eq.4, substitute A + F for C (from eq.1): A + A + F – E = D + E + F which becomes 2×A + F – E = D + E + F Subtract F and E from both sides of the above equation: 2×A + F – E – F – E = D + E + F – F – E which simplifies to eq.4a) 2×A – 2×E = D
Hint #5
Substitute 2×A – 2×E for D (from eq.4a), and A + E for F (from eq.3a) in eq.2: B = 2×A – 2×E + A + E which becomes eq.2a) B = 3×A – E
Hint #6
Substitute 3×A – E for B (from eq.2a), 2×A + E for C (from eq.1b), and A + E for F (from eq.3a) in eq.5: 3×A – E + 2×A + E = A × (A + E) which becomes 5×A = A × (A + E) Divide both sides of the above equation by A: 5×A ÷ A = A × (A + E) ÷ A which becomes 5 = A + E Subtract A from both sides: 5 – A = A + E – A which makes eq.5a) 5 – A = E
Hint #7
Substitute 5 – A for E (from eq.5a) in eq.3a: F = A + 5 – A which makes F = 5
Hint #8
Substitute 5 for F in eq.1: eq.1c) C = A + 5
Hint #9
Substitute 5 – A for E (from eq.5a) in eq.4a: 2×A – 2×(5 – A) = D which becomes 2×A – 10 + 2×A = D which becomes eq.4b) 4×A – 10 = D
Hint #10
Substitute (5 – A) for E (from eq.5a) in eq.2a: B = 3×A – (5 – A) which becomes B = 3×A – 5 + A which becomes eq.2b) B = 4×A – 5
Hint #11
Substitute A + 5 for C (from eq.1c), 4×A – 10 for D (from eq.4b), 5 – A for E (from eq.5a), and 5 for F in eq.6: A + 5 + 4×A – 10 = (5 – A) × 5 which becomes 5×A – 5 = 25 – 5×A Add 5 and 5×A to both sides of the equation above: 5×A – 5 + 5 + 5×A = 25 – 5×A + 5 + 5×A which makes 10×A = 30 Divide both sides by 10: 10×A ÷ 10 = 30 ÷ 10 which makes A = 3
Solution
Since A = 3, then: B = 4×A – 5 = (4 × 3) – 5 = 12 – 5 = 7 (from eq.2b) C = A + 5 = 3 + 5 = 8 (from eq.1c) D = 4×A – 10 = (4 × 3) – 10 = 12 – 10 = 2 (from eq.4b) E = 5 – A = 5 – 3 = 2 (from eq.5a) and ABCDEF = 378225