Puzzle for March 14, 2020 ( )
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
Add C and E to both sides of eq.4: B – E + C + E = D – B – C + C + E which becomes B + C = D – B + E In eq.2, replace B + C with D – B + E: A + D = D – B + E In the above equation, add B to both sides, and subtract D from each side: A + D + B – D = D – B + E + B – D which becomes eq.2a) A + B = E
Hint #2
In eq.3, replace E with A + B (from eq.2a): A + B + F = A + B Subtract A and B from each side of the above equation: A + B + F – A – B = A + B – A – B which simplifies to F = 0
Hint #3
In eq.5, replace F with 0: B + E + 0 = C + D which means B + E = C + D Add the left and right sides of the above equation to the left and right sides of eq.4, respectively: B – E + (B + E) = D – B – C + (C + D) which becomes 2×B = 2×D – B Add B to each side of the equation above: 2×B + B = 2×D – B + B which becomes 3×B = 2×D Divide both sides by 2: 3×B ÷ 2 = 2×D ÷ 2 which means 1½×B = D
Hint #4
In eq.4, substitute 1½×B for D: B – E = 1½×B – B – C which becomes B – E = ½×B – C In the above equation, add C and E to both sides, and subtract ½×B from each side: B – E + C + E – ½×B = ½×B – C + C + E – ½×B which simplifies to eq.4a) ½×B + C = E
Hint #5
In eq.2a, substitute ½×B + C for E (from eq.4a): A + B = ½×B + C Subtract B from each side of the above equation: A + B – B = ½×B + C – B which becomes eq.2b) A = C – ½×B
Hint #6
Substitute 1½×B for D, ½×B + C for E (from eq.4a), and C – ½×B for A (from eq.2b) in eq.6: 1½×B + ½×B + C = C – ½×B + C – 1½×B which becomes 2×B + C = 2×C – 2×B In the above equation, subtract C from both sides, and add 2×B to each side: 2×B + C – C + 2×B = 2×C – 2×B – C + 2×B which makes 4×B = C
Hint #7
Substitute 4×B for C in eq.2b: A = 4×B – ½×B which makes A = 3½×B
Hint #8
Substitute 4×B for C in eq.4a: ½×B + 4×B = E which makes 4½×B = E
Solution
Substitute 3½×B for A, 4×B for C, 1½×B for D, 4½×B for E, and 0 for F in eq.1: 3½×B + B + 4×B + 1½×B + 4½×B + 0 = 29 which simplifies to 14½×B = 29 Divide both sides of the equation above by 14½: 14½×B ÷ 14½ = 29 ÷ 14½ which means E = 2 making A = 3½×B = 3½ × 2 = 7 C = 4×B = 4 × 2 = 8 D = 1½×B = 1½ × 2 = 3 E = 4½×B = 4½ × 2 = 9 and ABCDEF = 728390