Puzzle for November 7, 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 B and E to both sides of eq.5: F – B + B + E = B + D – E + B + E which becomes F + E = 2×B + D which is the same as eq.5a) E + F = 2×B + D
Hint #2
In eq.4, replace E + F with 2×B + D (from eq.5a): A + B – C + D – E = 2×B + D Subtract B and D from each side of the above equation: A + B – C + D – E – B – D = 2×B + D – B – D which simplifies to A – C – E = B Add E to both sides: A – C – E + E = B + E which becomes eq.4a) A – C = B + E
Hint #3
In eq.2, replace B + E with A – C (from eq.4a): A – C = A + C In the above equation, subtract A from each side, and add C to each side: A – C – A + C = A + C – A + C which simplifies to 0 = 2×C which means 0 = C
Hint #4
eq.6 may be written as: E = (B + C + D) ÷ 3 Multiply both sides of the equation above by 3: 3 × E = 3 × (B + C + D) ÷ 3 which becomes eq.6a) 3×E = B + C + D
Hint #5
In eq.6a, substitute 0 for C: 3×E = B + 0 + D which becomes eq.6b) 3×E = B + D Subtract D from each side of eq.6b: 3×E – D = B + D – D which becomes eq.6c) 3×E – D = B
Hint #6
Substitute (3×E – D) for B (from eq.6c) in eq.5a: E + F = 2×(3×E – D) + D which becomes E + F = 6×E – 2×D + D which becomes E + F = 6×E – D Subtract E from each side of the above equation: E + F – E = 6×E – D – E which becomes eq.5b) F = 5×E – D
Hint #7
Substitute 5×E – D for F (from eq.5b) in eq.3: 5×E – D – A = A – D Add D and A to both sides of the above equation: 5×E – D – A + D + A = A – D + D + A which simplifies to 5×E = 2×A Divide both sides by 2: 5×E ÷ 2 = 2×A ÷ 2 which makes 2½×E = A
Hint #8
Substitute 2½×E for A, and 0 for C in eq.2: B + E = 2½×E + 0 which becomes B + E = 2½×E Subtract E from each side of the equation above: B + E – E = 2½×E – E which makes B = 1½×E
Hint #9
Substitute 1½×E for B in eq.6b: 3×E = 1½×E + D Subtract 1½×E from each side of the above equation: 3×E – 1½×E = 1½×E + D – 1½×E which makes 1½×E = D
Hint #10
Substitute 1½×E for D in eq.5b: F = 5×E – 1½×E which makes F = 3½×E
Solution
Substitute 2½×E for A, 1½×E for B and D, 0 for C, and 3½×E for F in eq.1: 2½×E + 1½×E + 0 + 1½×E + E + 3½×E = 20 which simplifies to 10×E = 20 Divide both sides of the above equation by 10: 10×E ÷ 10 = 20 ÷ 10 which means E = 2 making A = 2½×E = 2½ × 2 = 5 B = D = 1½×E = 1½ × 2 = 3 F = 3½×E = 3½ × 2 = 7 and ABCDEF = 530327