Problem Set 1 Solutions

... tn (x) + tn (y) − 1/10n < x + y < tn (x) + tn (y) + 1/10n . Taking the truncation gives 1. tn (x + y) = z + w − 1/10n or z + w, if z + w ≥ 1/10n , 2. tn (x + y) = z + w or z + w + 1/10n , if z + w ≤ −1/10n , 3. tn (x + y) = z + w, if −1/10n < z + w < 1/10n (or equivalently, z + w = 0). Of course, we ...

... tn (x) + tn (y) − 1/10n < x + y < tn (x) + tn (y) + 1/10n . Taking the truncation gives 1. tn (x + y) = z + w − 1/10n or z + w, if z + w ≥ 1/10n , 2. tn (x + y) = z + w or z + w + 1/10n , if z + w ≤ −1/10n , 3. tn (x + y) = z + w, if −1/10n < z + w < 1/10n (or equivalently, z + w = 0). Of course, we ...

Explore/Investigate: Products of Proper Factors #1

... no type number a natural number that is not a deficient number, abundant number, or perfect number. The one-and-only ‘no type’ number is 1. 1 has no proper factors. A deficient number is not a number with a personality defect. Deficient numbers can be very mathemagical. For example, ALL prime number ...

... no type number a natural number that is not a deficient number, abundant number, or perfect number. The one-and-only ‘no type’ number is 1. 1 has no proper factors. A deficient number is not a number with a personality defect. Deficient numbers can be very mathemagical. For example, ALL prime number ...

Class VI Chapter 3 – Playing with Numbers Maths Exercise 3.1

... To make the number divisible by 3, the sum of its digits should be divisible by 3. The smallest multiple of 3 which comes after 19 is 21. Therefore, smallest number = 21 − 19 = 2 Now, 2 + 3 + 3 = 8 However, 2 + 3 + 3 + 3 = 11 If we put 8, then the sum of the digits will be 27 and as 27 is divisible ...

... To make the number divisible by 3, the sum of its digits should be divisible by 3. The smallest multiple of 3 which comes after 19 is 21. Therefore, smallest number = 21 − 19 = 2 Now, 2 + 3 + 3 = 8 However, 2 + 3 + 3 + 3 = 11 If we put 8, then the sum of the digits will be 27 and as 27 is divisible ...

Running Time of Euclidean Algorithm

... . Factoring large numbers is a surprisingly difficult problem. No-one knows how to do this in polynomial time, except on theoretical Quantum Computers. ...

... . Factoring large numbers is a surprisingly difficult problem. No-one knows how to do this in polynomial time, except on theoretical Quantum Computers. ...

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.