New Year's Resolution 2019

This year I will consume 52 more books than I usually read or listen to a year; in other words, 52 books. Knowing myself, almost all books will be non-fiction, and that is OK.

I will also design, code and publish a mobile application.

Books

[X] Contact, Carl Sagan, 1985
[X] Bad Blood, John Carreyrou, 2018
[X] Educated, Tara Westover, 2018
[X] Skunk Works, Ben Rich, Leo Janos, 1994
[X] Siddhartha, Hermann Hesse, 1922
[ ] Heavy, Kiese Laymon, 2018

Other

[ ] Mobile Application

Project Euler 22 - Names Scores - Solved with JavaScript

The 22th Project Euler problem - Names Scores - is stated as follows.

Using names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.

What is the total of all the name scores in the file? Project Euler, Names Scores, Problem 22

const fs = require('fs');

// Read the textfile into a string.
const names = fs.readFileSync('p022_names.txt', 'utf8')
    // Split the string on every comma and remove the quotes
    .split(',').map(name => name.replace(/"/g, ''))
    // Sort the names.
    .sort();

// Get the alphabetical value by adding the chars position in
// the alphabet. A = Char Code 65 = Alphabet Position 1.
const getAlphabeticalValue = name => name.split('')
    .reduce((memo, item) => memo + item.charCodeAt(0) - 64, 0);

const result = names.reduce((memo, item, index) =>
    memo + getAlphabeticalValue(item) * (index + 1), 0);

Project Euler 21 - Amicable Numbers - Solved with JavaScript

The 21th Project Euler problem - Amicable Numbers - is stated as follows. Evaluate the sum of all the amicable numbers under 10000.

const getProperDivisors = (n) => {
  const divisors = [];
  const root = Math.floor(Math.sqrt(n));
  for (let i = 1; i <= root; i++) {
    if (n % i === 0) {
      divisors.push(i);
      if (Math.pow(i, 2) !== n) {
        divisors.push(n / i);
      }
    }
  }
  return divisors.sort(function (a, b) {
    return a - b;
  }).filter(x => x !== n);
};

// Let d(n) be defined as the sum of proper divisors of
// n (numbers less than n which divide evenly into n).
const d = (n) => {
  const divisors = getProperDivisors(n);
  return divisors.reduce((a,b) => a + b, 0);
};

// Iterate over all numbers from 0 to 9999.
const result = Array.from({length: 9999}).map((_, n) => {
    const a = d(n);
    const b = d(a);
    // If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
    // pair and each of a and b are called amicable numbers.
    return n !== a && n === b ? a : 0;
}).reduce((a,b) => a + b);

Project Euler 20 - Factorial Digit Sum - Solved with JavaScript

The 20th Project Euler problem - Factorial Digit Sum - is stated as follows. Find the sum of the digits in the number 100!

// Recursive function using BigInt to get n factorial.
const factorial = (n) => n <= 1 ? BigInt(n) : BigInt(n) * BigInt(factorial(--n));

// Convert the number to a string.
const sumDigits = n => n.toString()
    // Split the string on every char.
    .split('')
    // Convert all chars to ints.
    .map(x => parseInt(x))
    // Sum all ints.
    .reduce((a,b) => a + b);

// Get 100 factorial.
const f100 = factorial(100);

// Sum the digits.
const result = sumDigits(f100);

Project Euler 19 - Counting Sundays - Solved with JavaScript

The 19th Project Euler problem - Counting Sundays - is stated as follows. How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

// Get all the months between Jan 1901 to Dec 2000.
// We add 12 since we want the months in 2000 as well.
const months = (2000 - 1901) * 12 + 12;

// Iterate one month at a time.
const sundays = Array.from({length:months})
  .filter((_,month) => {
    const currentYear = 1901 + Math.floor(month/12);
    const currentMonth = month % 12;
    // Check the weekday of the 1 in the current year and month.
    // 0 equals Sunday.
    return new Date(currentYear, currentMonth, 1).getDay() === 0
  }).length;

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