# euler.lisp @ 2e707232cee0

Problem 5

author | Steve Losh <steve@stevelosh.com> |
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date | Sun, 10 Apr 2016 01:27:54 +0000 |

parents | 5dc80bffbecc |

children | 37bd082d9946 |

(in-package #:euler) ;;;; (defun digits (n) "Return how many digits `n` has in base 10." (values (truncate (1+ (log n 10))))) (defun definitely-palindrome-p (n) "Return whether `n` is a palindrome (in base 10), the slow-but-sure way." (let ((s (format nil "~D" n))) (string= s (reverse s)))) (defun palindrome-p (n) "Return whether `n` is a palindrome (in base 10)." (assert (>= n 0) (n) "~A must be a non-negative integer" n) ;; All even-length base-10 palindromes are divisible by 11, so we can shortcut ;; the awful string comparison. E.g.: ;; ;; abccba = ;; 100001 * a + ;; 010010 * b + ;; 001100 * c (cond ((zerop n) t) ((and (evenp (digits n)) (not (dividesp n 11))) nil) (t (definitely-palindrome-p n)))) (defun range (from below) (loop :for i :from from :below below :collect i)) ;;;; Problems (defun problem-1 () ;; If we list all the natural numbers below 10 that are multiples of 3 or 5, ;; we get 3, 5, 6 and 9. The sum of these multiples is 23. ;; ;; Find the sum of all the multiples of 3 or 5 below 1000. (loop :for i :from 1 :below 1000 :when (or (dividesp i 3) (dividesp i 5)) :sum i)) (defun problem-2 () ;; Each new term in the Fibonacci sequence is generated by adding the previous ;; two terms. By starting with 1 and 2, the first 10 terms will be: ;; ;; 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... ;; ;; By considering the terms in the Fibonacci sequence whose values do not ;; exceed four million, find the sum of the even-valued terms. (loop :with p = 0 :with n = 1 :while (<= n 4000000) :when (evenp n) :sum n :do (psetf p n n (+ p n)))) (defun problem-3 () ;; The prime factors of 13195 are 5, 7, 13 and 29. ;; ;; What is the largest prime factor of the number 600851475143 ? (apply #'max (prime-factorization 600851475143))) (defun problem-4 () ;; A palindromic number reads the same both ways. The largest palindrome made ;; from the product of two 2-digit numbers is 9009 = 91 × 99. ;; ;; Find the largest palindrome made from the product of two 3-digit numbers. (let ((result (list))) (loop :for i :from 0 :to 999 :do (loop :for j :from 0 :to 999 :for product = (* i j) :when (palindrome-p product) :do (push product result))) (apply #'max result))) (defun problem-5 () ;; 2520 is the smallest number that can be divided by each of the numbers from ;; 1 to 10 without any remainder. ;; ;; What is the smallest positive number that is evenly divisible by all of the ;; numbers from 1 to 20? (let ((divisors (range 11 21))) ;; all numbers are divisible by 1 and we can skip checking everything <= 10 ;; because: ;; ;; anything divisible by 12 is automatically divisible by 2 ;; anything divisible by 12 is automatically divisible by 3 ;; anything divisible by 12 is automatically divisible by 4 ;; anything divisible by 15 is automatically divisible by 5 ;; anything divisible by 12 is automatically divisible by 6 ;; anything divisible by 14 is automatically divisible by 7 ;; anything divisible by 16 is automatically divisible by 8 ;; anything divisible by 18 is automatically divisible by 9 ;; anything divisible by 20 is automatically divisible by 10 (loop :for i :from 20 :by 20 ; it must be divisible by 20 :when (every (lambda (n) (dividesp i n)) divisors) :return i)))