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... We refrain from describing our initial guesses in these cases, believing instead that the reader is ready to see some results. 3. Results We wish to consider nontrivial sequences (an ) of integers that satisfy the recurrence relation an+1 = ban + an−1 for some positive integer b, but whose initial t ...
... We refrain from describing our initial guesses in these cases, believing instead that the reader is ready to see some results. 3. Results We wish to consider nontrivial sequences (an ) of integers that satisfy the recurrence relation an+1 = ban + an−1 for some positive integer b, but whose initial t ...
Solution
... expression and moving everything to the left we find that all summands but one are divisible by 42 ). Hence e = d = 0, z = 1, b = 2, x = 2 and a = 4, and we obtain the classical 24 + 32 = 42 + 32 = 52 . 10. How many positive integers satisfy the following three conditions: (i) All digits of the numb ...
... expression and moving everything to the left we find that all summands but one are divisible by 42 ). Hence e = d = 0, z = 1, b = 2, x = 2 and a = 4, and we obtain the classical 24 + 32 = 42 + 32 = 52 . 10. How many positive integers satisfy the following three conditions: (i) All digits of the numb ...
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... Please submit all new problem proposals and their solutions to the Problems Editor, DR. RUSS EULER, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468, or by email at [email protected]. All solutions to others’ proposals must ...
... Please submit all new problem proposals and their solutions to the Problems Editor, DR. RUSS EULER, Department of Mathematics and Statistics, Northwest Missouri State University, 800 University Drive, Maryville, MO 64468, or by email at [email protected]. All solutions to others’ proposals must ...
1 Introduction to Logic
... (2) We can put zero or more a-gram weights in the pan with the potato and one or more b-gram weights in the other. The equation for this is w+xa = yb. (3) We can put some or none of the b-gram weights with the spud and a-gram weights in the other pan, corresponding to the equation w+xb = ya. There i ...
... (2) We can put zero or more a-gram weights in the pan with the potato and one or more b-gram weights in the other. The equation for this is w+xa = yb. (3) We can put some or none of the b-gram weights with the spud and a-gram weights in the other pan, corresponding to the equation w+xb = ya. There i ...
Section 9.1
... conclude that if a sequence has an upper bound, it must have a least upper bound (an upper bound that is smaller than all other upper bounds for the sequence). ...
... conclude that if a sequence has an upper bound, it must have a least upper bound (an upper bound that is smaller than all other upper bounds for the sequence). ...
Classifying Real Numbers
... A rational number is a number that can be written as the ratio of two integers. An irrational number cannot be written as the ratio of two integers. • The square root of any whole number that is not a perfect square is irrational. The cube root of any integer that is not a perfect cube is irrational ...
... A rational number is a number that can be written as the ratio of two integers. An irrational number cannot be written as the ratio of two integers. • The square root of any whole number that is not a perfect square is irrational. The cube root of any integer that is not a perfect cube is irrational ...
Beyond the Fundamentals: Technical Analysis
... (with 34 petals, you will always finish with “she loves me not!”) ...
... (with 34 petals, you will always finish with “she loves me not!”) ...
Quotients of Fibonacci Numbers
... numbers. Instead of powers 10n with n running from some nonnegative N to −∞, we have powers pn with n running from N (potentially negative) to +∞. Further details on Q p can be found in [11, 20]. To prove our theorem, we require a few preliminary results. In what follows, we write a|b to denote that ...
... numbers. Instead of powers 10n with n running from some nonnegative N to −∞, we have powers pn with n running from N (potentially negative) to +∞. Further details on Q p can be found in [11, 20]. To prove our theorem, we require a few preliminary results. In what follows, we write a|b to denote that ...
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.