# Euler problems/151 to 160

## 1 Problem 151

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

Solution:

`problem_151 = undefined`

## 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]

-- Verify that we need not consider terms divisible by 11, or by any
-- prime greater than 13. Nor need we consider any term divisible
-- by 25, 27, 32, or 49.
verify = all (\p -> null \$ tail \$ pfree [p, 2*p..85] 0 p) \$
11 : dropWhile (< 17) primes ++ [25, 27, 32, 49]

-- 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 f [(0, [[]])] [13, 7, 5]
where
f 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 = f x \$ zip3 y (map invSq y) (map sumInvSq \$ init \$ tails y)
where
f 0 _        = [[]]
f x ((n, r, s):ns)
| r > x     = f x ns
| s < x     = []
| otherwise = map (n :) (f (x - r) ns) ++ f x ns
f _ _        = []

-- 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 = if verify
then [sort \$ u ++ v | (x, s) <- only23,
u <- findInvSq (1%2 - x) all23,
v <- s]
else undefined

problem_152 = length solutions```

## 3 Problem 153

Investigating Gaussian Integers

Solution:

```#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.

Solution:

```#include <stdio.h>
int main(){
int bound = 200000;
long long sum = 0;
int val2[bound+1], val5[bound+1]; // number of factors 2/5 in i!
int v2 = 0, v5  = 0;
int i;
int n;
for(n=0;n<=bound;n++)
{val5[n]=n/5+n/25+n/125+n/625+n/3125+n/15625+n/78125;
val2[n]=n/2+n/4+n/8+n/16+n/32+n/64+n/128+n/256+n/512+n/1024
+n/2048+n/4096+n/8192+n/16384+n/32768+n/65536+n/131072;}

v2 =val2[bound]- 11;
v5 = val5[bound]-11;
int j,k,vi2,vi5;
for(i = 2; i < 65625; i++){
if (!(i&1023)){
// look how many we got so far
printf("%d:\t%lld\n",i,sum);
}
vi5 = val5[i];
vi2 = val2[i];
int jb = ((bound - i) >> 1)+1;
// I want i <= j <= k
// by carry analysis, I know that if i < 4*5^5+2, then
// j must be at least 2*5^6+2
for(j = (i < 12502) ? 31252 : i; j < jb; j++){
k = bound - i - j;
if (vi5 + val5[j] + val5[k] < v5
&& vi2 + val2[j] + val2[k] < v2){
if (j == k || i == j){
sum += 3;
} else {
sum += 6;
}
}
}
}
printf("Total:\t%lld\n",sum);
return 0;
}
problem_154 = main```

## 5 Problem 155

Counting Capacitor Circuits.

Solution:

`problem_155 = undefined`

## 6 Problem 156

Counting Digits

Solution:

```digits =reverse.digits'
where
digits' n
|n<10=[n]
|otherwise= y:digits' x
where
(x,y)=divMod n 10
digitsToNum n=foldl dmm 0  n
where
dmm=(\x y->x*10+y)
countA :: Int -> Integer
countA 0 = 0
countA k = fromIntegral k * (10^(k-1))

countFun :: Integer -> Integer -> Integer
countFun _ 0 = 0
countFun d n = countL ds k
where
ds = digits n
k = length ds - 1
countL [a] _
| a < d     = 0
| otherwise = 1
countL (a:tl) m
| a < d     = a*countA m + countL tl (m-1)
| a == d    = a*countA m + digitsToNum tl + 1 + countL tl (m-1)
| otherwise = a*countA m + 10^m + countL tl (m-1)

fixedPoints :: Integer -> [Integer]
fixedPoints d
= [a*10^10+b | a <- [0 .. d-1], b <- findFrom 0 (10^10-1)]
where
fun = countFun d
good r = r == fun r
findFrom lo hi
| hi < lo   = []
| good lo   = lgs ++ findFrom (last lgs + 2) hi
| good hi   = findFrom lo (last hgs - 2) ++ reverse hgs
| h1 < l1   = []
| l1 == h1  = if good l1 then [l1] else []
| m0 == m1  = findFrom l1 (head mgs - 2) ++ mgs
++ findFrom (last mgs + 2) h1
| m0 < m1   = findFrom l1 (m0-1) ++ findFrom (goUp h1 m1) h1
| otherwise = findFrom l1 (goDown l1 m1) ++ findFrom (m0+1) h1
where
l1 = goUp hi lo
h1 = goDown l1 hi
goUp bd k
| k < k1 && k < bd  = goUp bd k1
| otherwise         = k
where
k1 = fun k
goDown bd k
| k1 < k && bd < k  = goDown bd k1
| otherwise         = k
where
k1 = fun k
m0 = (l1 + h1) `div` 2
m1 = fun m0
lgs = takeWhile good [lo .. hi]
hgs = takeWhile good [hi,hi-1 .. lo]
mgs = reverse (takeWhile good [m0,m0-1 .. l1])
++ takeWhile good [m0+1 .. h1]
problem_156=sum[sum \$fixedPoints a|a<-[1..9]]```

## 7 Problem 157

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

Solution:

`problem_157 = undefined`

## 8 Problem 158

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

Solution:

`problem_158 = undefined`

## 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)```