how to generate barcode in c# windows application Inside Microsoft SQL Server 2008: T-SQL Querying in Visual C#.NET

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Inside Microsoft SQL Server 2008: T-SQL Querying
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Reducing the number of days from 3/2 to 1 has a similar effect on the number of eggs; namely, you need to divide 1 (egg) by 3/2, giving you this equation: 1 chicken 1 day = 2/3 egg If you increase the number of days from 1 to 3, the effect on the number of eggs is a factor of 3 as well: 1 chicken 3 days = 2 eggs So the correct answer to the puzzle is that one chicken lays two eggs in three days. In a very similar manner, you can express the relationship between builders, houses, years, and tools with the following equation: 3/2 builders 3/2 years 3/2 tools = 3/2 house To reduce the number of builders, years, and tools to one each, you need to divide the number of houses by 3/2 three times; in other words, by (3/2)3: 1 builder 1 year 1 tool = 3/2 3/2 3/2 3/2 houses This gives you the following equation: 1 builder 1 year 1 tool = 4/9 house Thus, one builder with one tool will build four houses in nine years. To generalization the equation, you need to divide the right side of the equation by 3/2 n times for n elements in the left side of the equation. Or, if you want to express the calculation as a multiplication instead of division, multiply by (2/3)n. For example, take our last equation: 3/2 builders 3/2 years 3/2 tools = 3/2 houses The left side of the equation contains three elements; therefore, you get this equation: 1 builder 1 year 1 tool = 3/2 (2/3)3 houses This is equal to: 1 builder 1 year 1 tool = 4/9 house
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Puzzle 21: A Cat, a String, and the Earth
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As I said, although this puzzle is quite simple, I like it because it s so counterintuitive. It probably seems inconceivable that adding only 1 meter to such a large circumference would make any noticeable difference in the radius, let alone allow a cat to pass below the string in the space that was added. But if you do the math, you realize that the actual
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Appendix A
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Logic Puzzles
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circumference has no signi cance in determining how the radius would be affected when extending the circumference. Instead, only the addition is signi cant. The circumference can be expressed as C = 2 r (2 times times the radius). Hence, the original radius can be expressed as roriginal = C/(2 ). Adding 1 meter to the existing circumference would change the equation to C + 1 = 2 rnew. Isolating rnew , you get rnew = (C + 1)/(2 ). Expanding the parentheses, you get rnew = C/(2 ) + 1/(2 ). Because the original radius was C/(2 ), the new radius is 1/(2 ) greater, which is about 16 centimeters (a bit more than 6 inches) greater. That s enough for a cat to go under and move from one hemisphere to the other.
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Puzzle 22: Josephus Problem
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An easy way to nd a generic solution to this puzzle with any number of men is to rst solve it with very small numbers of men (1, 2, 3, and so on) and to look for a pattern in the results. If you solve the puzzle for small numbers, you get the results shown in Table A-1, where n is the number of men and p is the position of the only man left.
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TABLE A-1
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Results of the Josephus Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1
The pattern you can identify is that p is an increasing sequence of odd integers that restarts from 1 when n is a power of 2. You express n as 2a + b, where b >= 0 and b < 2a. That is, a is the highest power of 2 such that 2a is smaller than n, and b is n minus 2a. Then, p can be expressed as 2b + 1. For example, for n = 41, express n as 25 + 9. Since b = 9 and p = 2b + 1, you get p = 19.
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