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... fi v for any two elements a f , a" of A (that i s , subsets A such that integers i and i + v do not both appear in A for any i = 1, 2, • • • , n - p) ? Church 1 s problem is then recovered from the above formulation on taking v = 2. For the solution of the general problem, we let n = m + r with m an ...
... fi v for any two elements a f , a" of A (that i s , subsets A such that integers i and i + v do not both appear in A for any i = 1, 2, • • • , n - p) ? Church 1 s problem is then recovered from the above formulation on taking v = 2. For the solution of the general problem, we let n = m + r with m an ...
Algebraic Proof
... Although most of his work is now lost his texts dealt with solving algebraic equations which included the proof that the quadratic formula could solve any real quadratic equation. ...
... Although most of his work is now lost his texts dealt with solving algebraic equations which included the proof that the quadratic formula could solve any real quadratic equation. ...
DAVID ESSNER EXAM 1999-2000
... 22. In a circle of radius r a line drawn from the center is perpendicular to a chord of length 4. If the line intersects the chord at point P and intersects the circle at a point Q which is 1 unit from P then r = (a) 5/2 (b) 7/2 (c) 3 (d) 9/4 (e) 4/3 23. A man invests $P in the stock market. Each o ...
... 22. In a circle of radius r a line drawn from the center is perpendicular to a chord of length 4. If the line intersects the chord at point P and intersects the circle at a point Q which is 1 unit from P then r = (a) 5/2 (b) 7/2 (c) 3 (d) 9/4 (e) 4/3 23. A man invests $P in the stock market. Each o ...
Conjecture
... Conjecture: Notice that 6 is 3 × 2 and 9 is 3 × 3. The next figure will increase by 3 × 4 or 12 segments. So, the next figure will have 18 + 12 or 30 segments. Answer: 30 segments ...
... Conjecture: Notice that 6 is 3 × 2 and 9 is 3 × 3. The next figure will increase by 3 × 4 or 12 segments. So, the next figure will have 18 + 12 or 30 segments. Answer: 30 segments ...
3. Number theory
... 4. Show that for every positive integer n, there is an integer N > n such that the number 5n appears as the last few digits of 5N . For example, if n = 3, we have 53 = 125, and 55 = 3125, so N = 5 would work. 5. Prove that the product of 3 consecutive integers is never a perfect power (i.e., a perfe ...
... 4. Show that for every positive integer n, there is an integer N > n such that the number 5n appears as the last few digits of 5N . For example, if n = 3, we have 53 = 125, and 55 = 3125, so N = 5 would work. 5. Prove that the product of 3 consecutive integers is never a perfect power (i.e., a perfe ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.