# Euler problems/151 to 160

## 1 Problem 151

Paper sheets of standard sizes: an expected-value problem.

Solution:

```problem_151 = fun (1,1,1,1)

fun (0,0,0,1) = 0
fun (0,0,1,0) = fun (0,0,0,1) + 1
fun (0,1,0,0) = fun (0,0,1,1) + 1
fun (1,0,0,0) = fun (0,1,1,1) + 1
fun (a,b,c,d) =
(pickA + pickB + pickC + pickD) / (a + b + c + d)
where
pickA | a > 0 = a * fun (a-1,b+1,c+1,d+1)
| otherwise = 0
pickB | b > 0 = b * fun (a,b-1,c+1,d+1)
| otherwise = 0
pickC | c > 0 = c * fun (a,b,c-1,d+1)
| otherwise = 0
pickD | d > 0 = d * fun (a,b,c,d-1)
| otherwise = 0```

## 2 Problem 152

Writing 1/2 as a sum of inverse squares

Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.

Solution:

```import Data.Ratio
import Data.List

invSq n = 1 % (n * n)
sumInvSq = sum . map invSq

subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
subsets _      = [[]]

primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
all (\q -> p `mod` q /= 0) [3, 5, 7]]

-- All subsets whose sum of inverse squares,
-- when added to x, does not contain a factor of p
pfree s x p = [(y, t) | t <- subsets s, let y =  x + sumInvSq t,
denominator y `mod` p /= 0]

-- All pairs (x, s) where x is a rational number whose reduced
-- denominator is not divisible by any prime greater than 3;
-- and s is all sets of numbers up to 80 divisible
-- by a prime greater than 3, whose sum of inverse squares is x.
only23 = foldl fun [(0, [[]])] [13, 7, 5]
where
fun a p =
collect \$ [(y, u ++ v) |
(x, s) <- a,
(y, v) <- pfree (terms p) x p,
u <- s]
terms p =
[n * p |
n <- [1..80`div`p],
all (\q -> n `mod` q /= 0) \$
11 : takeWhile (>= p) [13, 7, 5]
]
collect =
map (\z -> (fst \$ head z, map snd z)) .
groupBy fstEq . sortBy cmpFst
fstEq  (x, _) (y, _) = x == y
cmpFst (x, _) (y, _) = compare x y

-- All subsets (of an ordered set) whose sum of inverse squares is x
findInvSq x y =
fun x \$ zip3 y (map invSq y) (map sumInvSq \$ init \$ tails y)
where
fun 0 _        = [[]]
fun x ((n, r, s):ns)
| r > x     = fun x ns
| s < x     = []
| otherwise = map (n :) (fun (x - r) ns) ++ fun x ns
fun _ _        = []

-- All numbers up to 80 that are divisible only by the primes
-- 2 and 3 and are not divisible by 32 or 27.
all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]

solutions =
[sort \$ u ++ v |
(x, s) <- only23,
u <- findInvSq (1%2 - x) all23,
v <- s
]

problem_152 = length solutions```

## 3 Problem 153

Investigating Gaussian Integers

Solution: This does not seem Haskell code to me. If the argument: Learning Haskell were valid pure Haskell code would have been given.

```#include <stdio.h>
#include <math.h>
typedef long long lolo;
static const lolo sumTo( lolo n ) { return n * ( n + 1 ) / 2; }

#define LL (1000)
lolo ssTab[ LL ];
int gcd(int a, int b) {
if (b==0) return a;
return gcd(b, a%b);
}

static const lolo sumSigma( lolo n ) {
lolo a, r, s;

if( n == 0 ) return 0;
if( n < LL ) { r = ssTab[ n ]; if( r ) return r; }
s = floor(sqrt( n ));
r = 0;
for( a = 1; a <= s; ++a ) r += a * ( n / a );
for( a = 1; a <= s; ++a ) r += ( sumTo( n / a ) - sumTo ( n / ( a + 1 ) ) ) * a;
if( n / s == s ) r -= s * s;
if( n < LL ) ssTab[ n ] = r;

return r;
}

int main() {
const lolo m = 100000000;
lolo t;
int a, b;
long ab;
t = sumSigma(m);
for( a = 1; a <=floor(sqrt(m)); ++a ) {
for( b = 1; b <= a && a * a + b * b <= m; ++b ) {
ab=(a*a+b*b);
if( ( a | b ) & 1 && gcd( a, b ) == 1 ) {
t += 2 * sumSigma( m / ab) * ( a == b ? a : a + b );
}
}
}
printf( "t = %lld\n", t );

return 1;
}
problem_153 = main```

## 4 Problem 154

Exploring Pascal's pyramid.

## 5 Problem 155

Counting Capacitor Circuits.

Solution:

```--http://www.research.att.com/~njas/sequences/A051389
a051389=
[1, 2, 4, 8, 20, 42,
102, 250, 610, 1486,
3710, 9228, 23050, 57718,
145288, 365820, 922194, 2327914
]
problem_155 = sum a051389```

## 6 Problem 156

Counting Digits

Solution: This was my code, published here without my permission nor any attribution, shame on whoever put it here. Daniel.is.fischer

## 7 Problem 157

Solving the diophantine equation 1/a+1/b= p/10n

Solution:

```-- Call (a,b,p) a primitive tuple of equation 1/a+1/b=p/10^n
-- a and b are divisors of 10^n, gcd a b == 1, a <= b and a*b <= 10^n
-- I noticed that the number of variants with a primitive tuple
-- is equal to the number of divisors of p.
-- So I produced all possible primitive tuples per 10^n and
-- summed all the number of divisors of every p

import Data.List
k `deelt` n = n `mod` k == 0

delers n
| n == 10 = [1,2,5,10]
| otherwise =
[ d |
d <- [1..n `div` 5],
d `deelt` n ]
++ [n `div` 4, n `div` 2,n]
fp n =
[ n*(a+b) `div` ab |
a <- ds,
b <- dropWhile (<a) ds,
gcd a b == 1,
let ab = a*b,
ab <= n
]
where
ds = delers n
numDivisors :: Integer -> Integer
numDivisors n = product [ toInteger (a+1) | (p,a) <- primePowerFactors n]
numVgln = sum . map numDivisors . fp

main = do
print . sum . map numVgln . takeWhile (<=10^9) . iterate (10*) \$ 10
primePowerFactors x = [(head a ,length a)|a<-group\$primeFactors x]
merge xs@(x:xt) ys@(y:yt) = case compare x y of
LT -> x : (merge xt ys)
EQ -> x : (merge xt yt)
GT -> y : (merge xs yt)

diff  xs@(x:xt) ys@(y:yt) = case compare x y of
LT -> x : (diff xt ys)
EQ -> diff xt yt
GT -> diff xs yt

primes, nonprimes :: [Integer]
primes    = [2,3,5] ++ (diff [7,9..] nonprimes)
nonprimes = foldr1 f . map g \$ tail primes
where f (x:xt) ys = x : (merge xt ys)
g p = [ n*p | n <- [p,p+2..]]
primeFactors n =
factor n primes
where
factor n (p:ps)
| p*p > n        = [n]
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
| otherwise      = factor n ps```

## 8 Problem 158

Exploring strings for which only one character comes lexicographically after its neighbour to the left.

Solution:

```factorial n = product [1..toInteger n]
fallingFactorial x n = product [x - fromInteger i | i <- [0..toInteger n - 1] ]
choose n k = fallingFactorial n k `div` factorial k
fun n=(2 ^ n - n - 1) * choose 26 n
problem_158=maximum\$map fun [1..26]```

## 9 Problem 159

Digital root sums of factorisations.

Solution:

```import Control.Monad
import Data.Array.ST
import qualified Data.Array.Unboxed as U
spfArray :: U.UArray Int Int
spfArray  = runSTUArray (do
arr <- newArray (0,m-1) 0
loop arr 2
forM_ [2 .. m - 1] \$ \ x ->
loop2 arr x 2
return arr
)
where
m=10^6
loop arr n
|n>=m=return ()
|otherwise=do writeArray arr n (n-9*(div (n-1) 9))
loop arr (n+1)
loop2 arr x n
|n*x>=m=return ()
|otherwise=do incArray arr x n
loop2 arr x (n+1)
incArray arr x n = do
when(ab<a+b) (writeArray arr (x*n) (a + b))
writ x=appendFile "p159.log"\$foldl (++) "" [show x,"\n"]
main=do
mapM_ writ \$U.elems spfArray
problem_159 = main

--at first ,make main to get file "p159.log"
--then ,add all num in the file```

## 10 Problem 160

Factorial trailing digits

We use the following two facts:

Fact 1:
(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0
Fact 2:
product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1

We really only need these two facts for the special case of

d == 5
, and we can verify that directly by

evaluating the above two Haskell expressions.

More generally:

Fact 1 follows from the fact that the group of invertible elements

of the ring of integers modulo
5^d
has
4*5^(d-1)
elements.

Fact 2 follows from the fact that the group of invertible elements

of the ring of integers modulo
10^d
is isomorphic to the product

of a cyclic group of order 2 and another cyclic group.

Solution:

```problem_160 = trailingFactorialDigits 5 (10^12)

trailingFactorialDigits d n = twos `times` odds
where
base = 10 ^ d
x `times` y = (x * y) `mod` base
multiply = foldl' times 1
x `toPower` k = multiply \$ genericReplicate n x
e = facFactors 2 n - facFactors 5 n
twos
| e <= d    = 2 `toPower` e
| otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1)))
odds = multiply [odd | a <- takeWhile (<= n) \$ iterate (* 2) 1,
b <- takeWhile (<= n) \$ iterate (* 5) a,
odd <- [3, 5 .. n `div` b `mod` base],
odd `mod` 5 /= 0]

-- The number of factors of the prime p in n!
facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) .

-- The digits of n in base b representation
radix p = map snd . takeWhile (/= (0, 0)) .
iterate ((`divMod` p) . fst) . (`divMod` p)```

it have another fast way to do this .

Solution:

```import Data.List
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c= (b * c) `rem` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
where
f x n y
| n == 1 = x `mul` y
| r == 0 = f x2 q y
| otherwise = f x2 q (x `mul` y)
where
(q,r) = quotRem n 2
x2 = sq x
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)

productMod =foldl (mulMod (10^5)) 1
hFacial 0=1
hFacial a
|gcd a 5==1=mod (a*hFacial(a-1)) (5^5)
|otherwise=hFacial(a-1)
fastFacial a= hFacial \$mod a 6250
numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
p160 x=mulMod t5 a b
where
t5=10^5
lst=numPrime x 5
a=powMod t5 1563 \$mod c 2500
b=productMod  c6
c=sum lst
c6=map fastFacial \$x:lst
problem_160 = p160 (10^12)```