Puzzle for February 21, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
Subtract the left and right sides of eq.2 from the left and right sides of eq.4, respectively: C + F - (D + F) = A + B + D - (A + B - D) which is equivalent to C + F - D - F = A + B + D - A - B + D which becomes C - D = 2×D Add D to both sides of the above equation: C - D + D = 2×D + D which makes C = 3×D
Hint #2
In eq.4, replace F with D + E (from eq.1): C + D + E = A + B + D Subtract D from each side of the equation above: C + D + E - D = A + B + D - D which becomes eq.4a) C + E = A + B
Hint #3
In eq.3, replace C with 3×D: A = B + 3×D + D which becomes eq.3a) A = B + 4×D
Hint #4
In eq.6, substitute B + 4×D for A (from eq.3a), 3×D for C, and D + E for F (from eq.1): (B + 4×D) × B = B + 3×D + E + D + E which becomes eq.6a) B² + 4×D×B = B + 4×D + 2×E
Hint #5
In eq.4a, substitute 3×D for C, and B + 4×D for A (from eq.3a): 3×D + E = B + 4×D + B which becomes 3×D + E = 2×B + 4×D Subtract 3×D from each side of the above equation: 3×D + E - 3×D = 2×B + 4×D - 3×D which becomes eq.4b) E = 2×B + D
Hint #6
Substitute 3×D for C, and D + E for F (from eq.1) in eq.5: B × 3×D = D + D + E which becomes eq.5a) B × 3×D = 2×D + E
Hint #7
Substitute 2×B + D for E (from eq.4b) in eq.5a: B × 3×D = 2×D + 2×B + D which becomes eq.5b) 3×D×B = 3×D + 2×B
Hint #8
Substitute 2×B + D for E (from eq.4b) in eq.6a: B² + 4×D×B = B + 4×D + 2×(2×B + D) which becomes B² + 4×D×B = B + 4×D + 4×B + 2×D which becomes eq.4c) B² + 4×D×B = 5×B + 6×D
Hint #9
Subtract the left and right sides of eq.5b from the left and right sides of eq.4c, respectively: B² + 4×D×B - 3×D×B = 5×B + 6×D - (3×D + 2×B) which becomes B² + D×B = 3×B + 3×D which may be written as B×(B + D) = 3×(B + D) Divide both sides of the above equation by (B + D): B×(B + D) ÷ (B + D) = 3×(B + D) ÷ (B + D) which becomes B = 3
Solution
Substitute 3 for B in eq.5b: 3×D×3 = 3×D + 2×3 which becomes 9×D = 3×D + 6 Subtract 3×D from each side of the above equation: 9×D - 3×D = 3×D + 6 - 3×D which becomes 6×D = 6 Divide both sides by 6: 6×D ÷ 6 = 6 ÷ 6 which means D = 1 making A = B + 4×D = 3 + 4×1 = 3 + 4 = 7 (from eq.3a) C = 3×D = 3×1 = 3 E = 2×B + D = 2×3 + 1 = 6 + 1 = 7 (from eq.4b) F = D + E = 1 + 7 = 8 (from eq.1) and ABCDEF = 733178