Puzzle for April 16, 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
eq.5 may be written as: A = (B + C + D + E + F) ÷ 5 Multiply both sides of the above equation by 5: 5 × A = 5 × (B + C + D + E + F) ÷ 5 which becomes eq.5a) 5×A = B + C + D + E + F
Hint #2
In eq.1, replace B + C + D + E + F with 5×A (from eq.5a): A + 5×A = 18 which becomes 6×A = 18 Divide both sides of the above equation by 6: 6×A ÷ 6 = 18 ÷ 6 which makes A = 3
Hint #3
eq.6 may be written as: B = (A + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + D + F) ÷ 3 which becomes 3×B = A + D + F In the above equation, replace A with 3: eq.6a) 3×B = 3 + D + F
Hint #4
Subtract the left and right sides of eq.4 from the left and right sides of eq.3, respectively: C + D – (B + C) = A + B + F – (A – B + D) which becomes C + D – B – C = A + B + F – A + B – D which becomes D – B = 2×B + F – D In the above equation, subtract 2×B from both sides, and add D to both sides: D – B – 2×B + D = 2×B + F – D – 2×B + D which becomes eq.3a) 2×D – 3×B = F
Hint #5
In eq.6a, substitute 2×D – 3×B for F (from eq.3a): 3×B = 3 + D + 2×D – 3×B which becomes 3×B = 3 + 3×D – 3×B In the equation above, subtract 3 from both sides, and add 3×B to both sides: 3×B – 3 + 3×B = 3 + 3×D – 3×B – 3 + 3×B which becomes 6×B – 3 = 3×D Divide both sides by 3: (6×B – 3) ÷ 3 = 3×D ÷ 3 which becomes eq.6b) 2×B – 1 = D
Hint #6
Substitute 3 for A, and 2×B – 1 for D (from eq.6b) in eq.4: B + C = 3 – B + 2×B – 1 which becomes B + C = 2 + B Subtract B from each side of the equation above: B + C – B = 2 + B – B which makes C = 2
Hint #7
Substitute (2×B – 1) for D (from eq.6b) in eq.3a: 2×(2×B – 1) – 3×B = F which becomes 4×B – 2 – 3×B = F which becomes eq.3b) B – 2 = F
Hint #8
Substitute 2×B – 1 for D (from eq.6b), 3 for A, 2 for C, and B – 2 for F (from eq.3b) in eq.2: 2×B – 1 + E = 3 + 2 + B – 2 which becomes 2×B – 1 + E = 3 + B In the above equation, subtract 2×B from both sides, and add 1 to both sides: 2×B – 1 + E – 2×B + 1 = 3 + B – 2×B + 1 which becomes eq.2a) E = 4 – B
Solution
Substitute 3 for A, 2 for C, 2×B – 1 for D (from eq.6b), 4 – B for E (from eq.2a), and B – 2 for F (from eq.3b) in eq.1: 3 + B + 2 + 2×B – 1 + 4 – B + B – 2 = 18 which simplifies to 6 + 3×B = 18 Subtract 6 from both sides of the above equation: 6 + 3×B – 6 = 18 – 6 which makes 3×B = 12 Divide both sides by 3: 3×B ÷ 3 = 12 ÷ 3 which means B = 4 making D = 2×B – 1 = 2×4 – 1 = 8 – 1 = 7 (from eq.6b) E = 4 – B = 4 – 4 = 0 (from eq.2a) F = B – 2 = 4 – 2 = 2 (from eq.3b) and ABCDEF = 342702