Puzzle for February 10, 2019 ( )
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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.
* AB, BC, CD, and EF are 2-digit numbers (not A×B, B×C, C×D, or E×F). BCD and CDE are 3-digit numbers (not B×C×D or C×D×E). EF is an angle expressed in degrees.
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Hint #1
The cosine of any angle is a number from -1 thru 1. Since A must be a one-digit non-negative integer, then A must be either 0 or 1 (from eq.6). Check: A = 0 ... In eq.6, replace A with 0: cosine (EF) = 0 which would make EF = 90
Hint #2
Check: A = 1 ... In eq.6, replace A with 1: cosine (EF) = 1 which would make EF = 00 Therefore, in either case, F = 0
Hint #3
In eq.3, replace F with 0: A + D + 0 = C - A Add A to both sides of the above equation: A + D + 0 + A = C - A + A which becomes eq.3a) 2×A + D = C Subtract 2×A from each side of eq.3a: 2×A + D - 2×A = C - 2×A which becomes eq.3b) D = C - 2×A
Hint #4
eq.4 may be written as: 10×B + C = A + 10×A + B - C + 10×C + D which becomes 10×B + C = 11×A + B + 9×C + D Subtract both B and C from each side of the above equation: 10×B + C - B - C = 11×A + B + 9×C + D - B - C which becomes 9×B = 11×A + 8×C + D In the above equation, substitute (2×A + D) for C (from eq.3a): 9×B = 11×A + 8×(2×A + D) + D which is the same as 9×B = 11×A + 16×A + 8×D + D which becomes 9×B = 27×A + 9×D Divide both sides by 9: 9×B ÷ 9 = (27×A + 9×D) ÷ 9 which becomes eq.4a) B = 3×A + D
Hint #5
In eq.4a, substitute C - 2×A for D (from eq.3b): B = 3×A + C - 2×A which becomes eq.4b) B = A + C In eq.2, substitute A + C for B: A + C - A = C + E which becomes C = C + E Subtract C from each side: C - C = C + E - C which makes 0 = E
Hint #6
eq.5 may be written as: 10×A + B + B + 10×B + C = 100×B + 10×C + D - (100×C + 10×D + E) which is equivalent to 10×A + 12×B + C = 100×B + 10×C + D - 100×C - 10×D - E which becomes 10×A + 12×B + C = 100×B - 90×C - 9×D - E Subtract both 12×B and C from each side of the above equation: 10×A + 12×B + C - 12×B - C = 100×B - 90×C - 9×D - E - 12×B - C which becomes 10×A = 88×B - 91×C - 9×D - E Substitute 0 for E, and (C - 2×A) for D (from eq.3b): 10×A = 88×B - 91×C - 9×(C - 2×A) - 0 which is the same as 10×A = 88×B - 91×C - 9×C + 18×A which becomes 10×A = 88×B - 100×C + 18×A Add 100×C - 10×A to both sides: 10×A + 100×C - 10×A = 88×B - 100×C + 18×A + 100×C - 10×A which simplifies to eq.5a) 100×C = 88×B + 8×A
Hint #7
Substitute (A + C) for B (from eq.4b) in eq.5a: 100×C = 88×(A + C) + 8×A which is equivalent to 100×C = 88×A + 88×C + 8×A Subtract 88×C from both sides of the above equation: 100×C - 88×C = 88×A + 88×C + 8×A - 88×C which becomes 12×C = 96×A Divide both sides by 12: 12×C ÷ 12 = 96×A ÷ 12 which means C = 8×A
Hint #8
Replace C with 8×A in eq.4b: B = A + 8×A which makes B = 9×A
Hint #9
Replace B with 9×A in eq.4a: 9×A = 3×A + D Subtract 3×A from each side: 9×A - 3×A = 3×A + D - 3×A which becomes 6×A = D
Solution
Substitute 9×A for B, 8×A for C, 6×A for D, and 0 for E and F in eq.1: A + 9×A + 8×A + 6×A + 0 + 0 = 24 which simplifies to 24×A = 24 Divide both sides by 24: 24×A ÷ 24 = 24 ÷ 24 which makes A = 1 making B = 9×A = 9 × 1 = 9 C = 8×A = 8 × 1 = 8 D = 6×A = 6 × 1 = 6 and ABCDEF = 198600