Not A Data Scientist

The Problem: If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? NOTE:  Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage. The Solution: This problem requires you to have some knowledge about a library "num2words"; specifically created to convert numbers into words into python. Convert numbers into words in Python >>> from num2words import num2words >>> num2words(3) 'three' >>> num2words(200) 'two hundred' # The Program   >>> count = 0 >>> for i in range(1, 1001): y = num2words(i) ...

The Problem: The following iterative sequence is defined for the set of positive integers: n → n /2 ( n is even) n → 3 n + 1 ( n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1   It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? NOTE: Once the chain starts the terms are allowed to go above one million. The Solution: We can divide the solution in 3 task. To form the chain of sequence. To feed the data to the first function. Third will be sort the values.  To form the chain of sequence: def chain(n):        a = []  # will hold the sequence     a.append(n)     while n...

This problem seems difficult at first, but once you understand it; this will prove out to be easiest of all the problems in Project Euler: I guess you will have seen the problem till now, if not then visit Project Euler 13 else scroll down for this problem. The Solution: The number mentions in the problem is of 100 line and in one line there are 50 numbers. So basically you just have to add all the 100 numbers (each number consist of 50 digits).  you will need str.split() function to do the job. num = "37107287533902102798797998220837590246510135740250 \n \ 46376937677490009712648124896970078050417018260538 \n \ 74324986199524741059474233309513058123726617309629 \n \ 91942213363574161572522430563301811072406154908250 \n \ 23067588207539346171171980310421047513778063246676 \n \ 89261670696623633820136378418383684178734361726757 \n \ 28112879812849979408065481931592621691275889832738 \n \ 44274228917432520321923589422876796487670272189318 \n \ 474514457360013064...

You guys can get this programming ask from Project Euler. This  is the 11th question  of this series. The solution of this question needs you to be good in matrix. You can find the solution below. The Solution: You have to find the greatest product of the four adjacent number and the number can should be in the same direction. So basically, if we start by first number of the series 08; then we can find only 3 adjacent number in the same direction and that will be: 1>  08 02 22 97(going in positive direction) 2>  08 49 81 52 (Going in down direction) 3> 08 49 31 23 (Going in positive diagonal direction) but a number can have maximum 8 products in the same direction; for better understanding we have marked it in the red and you can also check the below image for better understanding. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67...

You guys can get this programming question from Project Euler.   The code for this ask is :   for a in range(1,1000):     for b in range(1,1000):         c = math.sqrt(a*a + b*b)         if a + b + c == 1000.0:             print a*b*c You guys will get the answer once you execute the code.

We have already wrote a similar program on Prime Number:  So, before explaining anything about prime number we will directly jump in program.  Programme for What is the 10001st prime number? n = 2 a = [] while n < n+1: sum = 0 for i in range(1, n+1): if n%i == 0: sum += 1 if sum == 2: a.append(n) if len(a) == 10001: break n = n + 1 This shall give us all the prime number till 10001st. Type a[len(a) - 1] into the console to get the output. Keep reading Python an hour a day to learn simple Python tricks, tips and program.