# Euler problems/21 to 30

(Difference between revisions)

## 1 Problem 21

Evaluate the sum of all amicable pairs under 10000.

This is a little slow because of the naive method used to compute the divisors.

problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n]
where amicable m n = m < n && n < 10000 && divisorsSum ! n == m
divisorsSum = array (1,9999)
[(i, sum (divisors i)) | i <- [1..9999]]
divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]

Here is an alternative using a faster way of computing the sum of divisors.

problem_21_v2 = sum [n | n <- [2..9999], let m = d n,
m > 1, m < 10000, n == d m]
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group \$ primeFactors n, let p = head g
] - n
primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0    = p : pf ps q
| otherwise = pf ps' n
where (q, r) = n `divMod` p
primes = 2 : filter (null . tail . primeFactors) [3,5..]

## 2 Problem 22

What is the total of all the name scores in the file of first names?

Solution:

import Data.List
import Data.Char
problem_22 =
do input <- readFile "names.txt"
let names = sort \$ read\$"["++ input++"]"
let scores = zipWith score names [1..]
print . show . sum \$ scores
where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) \$ w

## 3 Problem 23

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution:

--http://www.research.att.com/~njas/sequences/A048242
import Data.Array
n = 28124
abundant n = eulerTotient n - n > n
abunds_array = listArray (1,n) \$ map abundant [1..n]
abunds = filter (abunds_array !) [1..n]

rests x = map (x-) \$ takeWhile (<= x `div` 2) abunds
isSum = any (abunds_array !) . rests

problem_23 = putStrLn . show . foldl1 (+) . filter (not . isSum) \$ [1..n]

## 4 Problem 24

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Solution:

import Data.List

fac 0 = 1
fac n = n * fac (n - 1)
perms [] _= []
perms xs n= x : perms (delete x xs) (mod n m)
where m = fac \$ length xs - 1
y = div n m
x = xs!!y

problem_24 = perms "0123456789" 999999

## 5 Problem 25

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution:

valid ( i, n ) = length ( show n ) == 1000

problem_25 = fst . head . filter valid . zip [ 1 .. ] \$ fibs
where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs )

## 6 Problem 26

Find the value of d < 1000 for which 1/d contains the longest recurring cycle.

Solution:

problem_26 = fst \$ maximumBy (\a b -> snd a `compare` snd b)
[(n,recurringCycle n) | n <- [1..999]]
where  recurringCycle d = remainders d 10 []
remainders d 0 rs = 0
remainders d r rs = let r' = r `mod` d
in case findIndex (== r') rs of
Just i  -> i + 1
Nothing -> remainders d (10*r') (r':rs)

## 7 Problem 27

Find a quadratic formula that produces the maximum number of primes for consecutive values of n.

Solution:

problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head \$ filter (\x->x^2-x+41>n) [1..]
a = m-1

## 8 Problem 28

What is the sum of both diagonals in a 1001 by 1001 spiral?

Solution:

problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1

## 9 Problem 29

How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Solution:

problem_29 = length . group . sort \$ liftM2 (^) [2..100] [2..100]

## 10 Problem 30

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Solution:

--http://www.research.att.com/~njas/sequences/A052464
problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979]

I'm sorry, but I find the solution to problem 30 very unsatisfying. I'm using the Euler problems to learn Haskell, so looking up the answer and adding the terms isn't really that helpful. I would like to present the following as a clearer solution that perhaps gives a little more insight into the problem and programming in Haskell. -- Henry Laxen, Feb 20, 2008

problem_30  = sum \$ map listToInt (drop 2 ans)
-- we drop 2 because the first two members of the ans are 0 and 1,
-- which are considered "trivial" solutions and should not count in the sum
where       maxFirstDigit = (6*9^5 `div` 10^5) + 1
-- The largest number that can be the sum of fifth powers
-- is 6*9^5 = 354294, which has 6 digits
listToInt n = foldl (\x y -> 10*x+y) 0 n
isSumOfPowers p n = (sum \$ map (\x -> x^p) n) == listToInt n
ans = filter (isSumOfPowers 5) [ [a,b,c,d,e,f] |
a <- [0..maxFirstDigit],
b <- [0..9],
c <- [0..9],
d <- [0..9],
e <- [0..9],
f <- [0..9] ]