Puzzle for September 25, 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 to both sides of eq.5: D + E – F + B = A – B + C + B which may be written as D + B + E – F = A + C In the equation above, replace B + E with C + D + F (from eq.3): D + C + D + F – F = A + C which becomes C + 2×D = A + C Subtract C from both sides: C + 2×D – C = A + C – C which makes 2×D = A
Hint #2
In eq.4, replace A with 2×D: eq.4a) 2×D + B = C + E Add D to both sides of eq.2: B + C – D + D = D + E + D which becomes B + C = 2×D + E which may be written as eq.2a) 2×D + E = B + C
Hint #3
Subtract the left and right sides of eq.2a from the left and right sides of eq.4a, respectively: 2×D + B – (2×D + E) = C + E – (B + C) which is equivalent to 2×D + B – 2×D – E = C + E – B – C which becomes B – E = E – B Add E and B to both sides of the equation above: B – E + E + B = E – B + E + B which makes 2×B = 2×E Divide both sides by 2: 2×B ÷ 2 = 2×E ÷ 2 which makes B = E
Hint #4
In eq.2a, substitute B for E: 2×D + B = B + C Subtract B from both sides of the above equation: 2×D + B – B = B + C – B which makes 2×D = C
Hint #5
eq.6 may be written as: C = (B + E + F) ÷ 3 Multiply both sides of the equation above by 3: 3 × C = 3 × (B + E + F) ÷ 3 which becomes eq.6a) 3×C = B + E + F
Hint #6
Substitute C + D + F for B + E (from eq.3) in eq.6a: 3×C = C + D + F + F which becomes 3×C = C + D + 2×F Subtract C from each side of the equation above: 3×C – C = C + D + 2×F – C which becomes eq.6b) 2×C = D + 2×F
Hint #7
Substitute (2×D) for C in eq.6b: 2×(2×D) = D + 2×F which becomes 4×D = D + 2×F Subtract D from each side of the equation above: 4×D – D = D + 2×F – D which becomes 3×D = 2×F Divide both sides by 2: 3×D ÷ 2 = 2×F ÷ 2 which makes 1½×D = F
Hint #8
Substitute 2×D for C, 1½×D for F, and B for E in eq.3: 2×D + D + 1½×D = B + B which becomes 4½×D = 2×B Divide both sides of the equation above by 2: 4½×D ÷ 2 = 2×B ÷ 2 which makes 2¼×D = B and also makes E = B = 2¼×D
Solution
Substitute 2×D for A and C, 2¼×D for B and E, and 1½×D for F in eq.1: 2×D + 2¼×D + 2×D + D + 2¼×D + 1½×D = 44 which simplifies to 11×D = 44 Divide both sides of the equation above by 11: 11×D ÷ 11 = 44 ÷ 11 which means D = 4 making A = C = 2×D = 2 × 4 = 8 B = E = 2¼×D = 2¼ × 4 = 9 F = 1½×D = 1½ × 4 = 6 and ACBDEF = 898496