Puzzle for December 10, 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
eq.6 may be written as: C = (D + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (D + E + F) ÷ 3 which becomes eq.6a) 3×C = D + E + F
Hint #2
In eq.6a, replace E + F with C + D (from eq.2): 3×C = D + C + D which becomes 3×C = 2×D + C Subtract C from both sides of the above equation: 3×C – C = 2×D + C – C which makes 2×C = 2×D Divide both sides by 2: 2×C ÷ 2 = 2×D ÷ 2 which makes C = D
Hint #3
Subtract the left and right sides of eq.4 from the left and right sides of eq.3, respectively: B + D – (D + E) = A + E – (A + B – E) which is equivalent to B + D – D – E = A + E – A – B + E which becomes B – E = 2×E – B Add E and B to both sides of the equation above: B – E + E + B = 2×E – B + E + B which makes 2×B = 3×E Divide both sides by 2: 2×B ÷ 2 = 3×E ÷ 2 which makes eq.3a) B = 1½×E
Hint #4
In eq.2, replace C with D: E + F = D + D which becomes E + F = 2×D Divide both sides of the equation above by 2: (E + F) ÷ 2 = 2×D ÷ 2 which becomes eq.2a) ½×E + ½×F = D
Hint #5
In eq.5, substitute 1½×E for B, and ½×E + ½×F for D (from eq.2a): 1½×E = ½×E + ½×F + F which becomes 1½×E = ½×E + 1½×F Subtract ½×E from each side of the above equation: 1½×E – ½×E = ½×E + 1½×F – ½×E which makes E = 1½×F
Hint #6
Substitute (1½×F) for E in eq.2a: ½×(1½×F) + ½×F = D which becomes ¾×F + ½×F = D which makes 1¼×F = D and also makes C = D = 1¼×F
Hint #7
Substitute (1½×F) for E in eq.3a: B = 1½×(1½×F) which makes B = 2¼×F
Hint #8
Substitute 2¼×F for B, 1¼×F for D, and 1½×F for E in eq.3: 2¼×F + 1¼×F = A + 1½×F which becomes 3½×F = A + 1½×F Subtract 1½×F from both sides of the equation above: 3½×F – 1½×F = A + 1½×F – 1½×F which makes 2×F = A
Solution
Substitute 2×F for A, 2¼×F for B, 1¼×F for C and D, and 1½×F for E in eq.1: 2×F + 2¼×F + 1¼×F + 1¼×F + 1½×F + F = 37 which simplifies to 9¼×F = 37 Divide both sides of the equation above by 9¼: 9¼×F ÷ 9¼ = 37 ÷ 9¼ which means F = 4 making A = 2×F = 2 × 4 = 8 B = 2¼×F = 2¼ × 4 = 9 C = D = 1¼×F = 1¼ × 4 = 5 E = 1½×F = 1½ × 4 = 6 and ABCDEF = 895564