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Euler problems/151 to 160

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Do them on your own!
+
== [http://projecteuler.net/index.php?section=problems&id=151 Problem 151] ==
  +
Paper sheets of standard sizes: an expected-value problem.
  +
  +
Solution:
  +
<haskell>
  +
problem_151 = undefined
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=152 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:
  +
<haskell>
  +
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
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=153 Problem 153] ==
  +
Investigating Gaussian Integers
  +
  +
Solution:
  +
<haskell>
  +
problem_153 = undefined
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=154 Problem 154] ==
  +
Exploring Pascal's pyramid.
  +
  +
Solution:
  +
<haskell>
  +
#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
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=155 Problem 155] ==
  +
Counting Capacitor Circuits.
  +
  +
Solution:
  +
<haskell>
  +
problem_155 = undefined
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=156 Problem 156] ==
  +
Counting Digits
  +
  +
Solution:
  +
<haskell>
  +
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]]
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=157 Problem 157] ==
  +
Solving the diophantine equation 1/a+1/b= p/10n
  +
  +
Solution:
  +
<haskell>
  +
problem_157 = undefined
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=158 Problem 158] ==
  +
Exploring strings for which only one character comes lexicographically after its neighbour to the left.
  +
  +
Solution:
  +
<haskell>
  +
problem_158 = undefined
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=159 Problem 159] ==
  +
Digital root sums of factorisations.
  +
  +
Solution:
  +
<haskell>
  +
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
  +
a <- readArray arr x
  +
b <- readArray arr n
  +
ab <- readArray arr (x*n)
  +
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
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=160 Problem 160] ==
  +
Factorial trailing digits
  +
  +
We use the following two facts:
  +
  +
Fact 1: <hask>(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0</hask>
  +
  +
Fact 2: <hask>product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1</hask>
  +
  +
We really only need these two facts for the special case of
  +
<hask>d == 5</hask>, 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 <hask>5^d</hask> has
  +
<hask>4*5^(d-1)</hask> elements.
  +
  +
Fact 2 follows from the fact that the group of invertible elements
  +
of the ring of integers modulo <hask>10^d</hask> is isomorphic to the product
  +
of a cyclic group of order 2 and another cyclic group.
  +
  +
Solution:
  +
<haskell>
  +
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) .
  +
tail . radix p
  +
  +
-- The digits of n in base b representation
  +
radix p = map snd . takeWhile (/= (0, 0)) .
  +
iterate ((`divMod` p) . fst) . (`divMod` p)
  +
</haskell>
  +
it have another fast way to do this .
  +
  +
Solution:
  +
<haskell>
  +
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)
  +
  +
</haskell>

Revision as of 04:59, 30 January 2008

Contents

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:

problem_153 = undefined

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
      a <- readArray arr x
      b <- readArray arr n
      ab <- readArray arr (x*n)
      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) .
               tail . radix p
 
-- 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)