# Difference between revisions of "Euler problems/171 to 180"

## Problem 171

Finding numbers for which the sum of the squares of the digits is a square.

Solution:

```#include <stdio.h>

static int result = 0;

#define digits 20
static long long fact[digits+1];
static const long long precision      = 1000000000;
static const long long precision_mult =  111111111;

#define maxsquare 64 /* must be a power of 2 > digits * 9^2 */

static inline int issquare( int n )
{
for( int step = maxsquare/2, i = step;;)
{
if( i*i == n )        return i;
if( !( step >>= 1 ) ) return -1;
if( i*i > n )         i -= step;
else                  i += step;
}
}

static inline void dodigit( int d, int nr, int sum, long long c, int s )
{
if( d )
for( int n = 0; n <= nr; c *= ++n, s += d, sum += d*d )
dodigit( d-1, nr - n, sum, c, s );
else if( issquare( sum ) > 0 )
result = ( s * ( fact[digits] / ( c * fact[nr] ) )
/ digits % precision * precision_mult
+ result ) % precision;
}

int main( void )
{
fact[0] = 1;
for( int i = 1; i < digits+1; i++ ) fact[i] = fact[i-1]*i;
dodigit( 9, digits, 0, 1, 0 );
printf( "%d\n", result );
return 0;
}
problem_171 = main
```

## Problem 172

Investigating numbers with few repeated digits.

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

-- how many numbers can we get having d digits and p positions
p172 0 _ = 0
p172 d p
| p < 4 = d^p
| otherwise =
(p172' p) +  p*(p172' (p-1)) + (choose p 2)*(p172' (p-2)) + (choose p 3)*(p172' (p-3))
where
p172' = p172 (d-1)

problem_172= (p172 10 18) * 9 `div` 10
```

## Problem 173

Using up to one million tiles how many different "hollow" square laminae can be formed? Solution:

```problem_173=
let c=div (10^6) 4
xm=floor\$sqrt \$fromIntegral c
k=[div c x|x<-[1..xm]]
in  sum k-(div (xm*(xm+1)) 2)
```

## Problem 174

Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements.

Solution:

```#include <stdio.h>

int main()
{
int a=0;
int c=1000000/4;
int L[1000000];
int i,x,b;
for(i=0;i<=1000000;i++)
{
L[i]=0;
}

for(x=1;x<=500;x++)
{
int q=x*x;
for(b=1;b<=250000;b++)
{
int y=q+b*x;
if(y<=c){L[y*4]++;}
}
}

for(i=0;i<=1000000;i++)
{if(L[i]<=10&&L[i]>=1)
{a++;}
}

printf( "%d\n", a);
return 1;
}
problem_174 = main
```

## Problem 175

Fractions involving the number of different ways a number can be expressed as a sum of powers of 2. Solution:

```sternTree x 0=[]
sternTree x y=
m:sternTree y n
where
(m,n)=divMod x y
findRat x y
|odd l=take (l-1) k++[last k-1,1]
|otherwise=k
where
k=sternTree x y
l=length k
p175 x y=
init\$foldl (++) "" [a++","|
a<-map show \$reverse \$filter (/=0)\$findRat x y]
problems_175=p175 123456789 987654321
test=p175 13 17
```

## Problem 176

Rectangular triangles that share a cathetus. Solution:

```--k=47547
--2*k+1=95095 = 5*7*11*13*19
lst=[5,7,11,13,19]
primes=[2,3,5,7,11]
problem_176 =
product[a^b|(a,b)<-zip primes (reverse n)]
where
la=div (last lst+1) 2
m=map (\x->div x 2)\$init lst
n=m++[la]
```

## Problem 177

Solution:

```problem_177 = undefined
```

## Problem 178

Step Numbers Solution:

```problem_178 = undefined
```

## Problem 179

Consecutive positive divisors. Solution:

```#include <stdio.h>

#define SZ 10000000
int n[SZ + 1];

int main ()
{
int i, j;
for (i = 0; i <= SZ; i++)
n[i] = 1;

for (i = 2; i <= SZ; i++)
for (j = i; j <= SZ; j += i)
n[j] += 1;

j = 0;
for (i = 1; i < SZ; i++)
if (n[i] == n[i + 1])
j++;

printf ("%d\n", j);
return 0;
}
problem_179 = main
```

## Problem 180

Rational zeros of a function of three variables. Solution:

```problem_180 = undefined
```