Puzzle for June 11, 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.6 may be written as: F = (A + B + C + D + E) ÷ 5 Multiply both sides of the above equation by 5: 5 × F = 5 × (A + B + C + D + E) ÷ 5 which becomes eq.6a) 5×F = A + B + C + D + E
Hint #2
In eq.1, replace A + B + C + D + E with 5×F (from eq.6a): 5×F + F = 30 which makes 6×F = 30 Divide both sides of the above equation by 6: 6×F ÷ 6 = 30 ÷ 6 which makes F = 5
Hint #3
Subtract C from both sides of eq.2: E – C = A + C + D – C which becomes eq.2a) E – C = A + D eq.4 may be written as: eq.4a) C + F = A + D + B
Hint #4
In eq.4a, replace A + D with E – C (from eq.2a): C + F = E – C + B In the above equation, add C to both sides, and subtract F and E from both sides: C + F + C – F – E = E – C + B + C – F – E which simplifies to eq.4b) 2×C – E = B – F
Hint #5
Add the left and right sides of eq.3 to the left and right sides of eq.4b, respectively: 2×C – E + C + E = B – F + B + F which simplifies to 3×C = 2×B Divide both sides of the above equation by 2: 3×C ÷ 2 = 2×B ÷ 2 which makes 1½×C = B
Hint #6
In eq.3, substitute 1½×C for B, and 5 for F: C + E = 1½×C + 5 Subtract C from each side of the above equation: C + E – C = 1½×C + 5 – C which becomes eq.3a) E = ½×C + 5
Hint #7
Substitute ½×C + 5 for E (from eq.3a), 5 for F, and 1½×C for B in eq.5: C + ½×C + 5 – 5 = 1½×C – C + D + 5 which becomes 1½×C = ½×C + D + 5 Subtract ½×C and 5 from each side of the equation above: 1½×C – ½×C – 5 = ½×C + D + 5 – ½×C – 5 which becomes eq.5a) C – 5 = D
Hint #8
Substitute ½×C + 5 for E (from eq.3a), and C – 5 for D (from eq.5a) in eq.2: ½×C + 5 = A + C + C – 5 which becomes ½×C + 5 = A + 2×C – 5 In the above equation, subtract 2×C from both sides, and add 5 to both sides: ½×C + 5 – 2×C + 5 = A + 2×C – 5 – 2×C + 5 which becomes eq.2a) 10 – 1½×C = A
Solution
Substitute 10 – 1½×C for A (from eq.2a), 1½×C for B, C – 5 for D (from eq.5a), ½×C + 5 for E (from eq.3a), and 5 for F in eq.1: 10 – 1½×C + 1½×C + C + C – 5 + ½×C + 5 + 5 = 30 which simplifies to 15 + 2½×C = 30 Subtract 15 from both sides of the above equation: 15 + 2½×C – 15 = 30 – 15 which makes 2½×C = 15 Divide both sides by 2½: 2½×C ÷ 2½ = 15 ÷ 2½ which means C = 6 making A = 10 – 1½×C = 10 – (1½ × 6) = 10 – 9 = 1 (from eq.2a) B = 1½×C = 1½ × 6 = 9 D = C – 5 = 6 – 5 = 1 (from eq.5a) E = ½×C + 5 = (½ × 6) + 5 = 3 + 5 = 8 (from eq.3a) and ABCDEF = 196185