public static long Problem37()
{
/*
* The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
*
* Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
*
* NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
*/
long res = 0L;
List <long> primes = Formulas.PrimeSieve(1000000);
List <long> truncatablePrimes = new List <long>();
int i = 8;
int j = 0;
while (j < 11)
{
if (i > primes.Last())
{
return(-2);
}
if (Formulas.isTruncatable(i))
{
j++;
res += i;
}
i++;
}
return(res);
}
public static long Problem43()
{
/*
* The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
*
* Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
*
* d2d3d4=406 is divisible by 2
* d3d4d5=063 is divisible by 3
* d4d5d6=635 is divisible by 5
* d5d6d7=357 is divisible by 7
* d6d7d8=572 is divisible by 11
* d7d8d9=728 is divisible by 13
* d8d9d10=289 is divisible by 17
* Find the sum of all 0 to 9 pandigital numbers with this property.
*/
List <string> numbers = Formulas.PandigitalNumbers(0, 9);
List <long> primes = Formulas.PrimeSieve(20);
long result = 0L;
foreach (string str in numbers)
{
bool buf = true;
for (int i = 1; i <= 7; i++)
{
if (Int64.Parse(String.Concat(String.Concat(str[i], str[i + 1]), str[i + 2])) % primes.ElementAt(i - 1) != 0)
{
buf = false;
}
}
if (buf)
{
result += Int64.Parse(str);
}
}
return(result);
}
public static long Problem49()
{
/*
* The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
* There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
* What 12-digit number do you form by concatenating the three terms in this sequence?
*/
List <long> primes = Formulas.PrimeSieve(10000);
List <long> numbers = new List <long>();
List <string> results = new List <string>();
int limit = 10000;
foreach (long n in primes)
{
if (n > 1000)
{
numbers.Add(n);
}
}
for (int i = 0; i < numbers.Count(); i++)
{
for (int j = i + 1; j < numbers.Count; j++)
{
long k = numbers[j] + (numbers[j] - numbers[i]);
if (k < limit && numbers.Contains(k))
{
if (Formulas.IsPermutation((int)numbers[i], (int)numbers[j]) && Formulas.IsPermutation((int)numbers[i], (int)k))
{
if (numbers[i] != 1487)
{
results.Add(String.Concat(numbers[i], numbers[j], k));
}
}
}
}
}
return(Int64.Parse(results[0]));
}
public static long Problem50()
{
long result = 0L;
int resCount = 0;
long limit = 1000000;
List <long> primes = Formulas.PrimeSieve(limit);
for (int i = primes.Count - 1; i >= 0; i--)
{
/*
* The prime 41, can be written as the sum of six consecutive primes:
*
* 41 = 2 + 3 + 5 + 7 + 11 + 13
* This is the longest sum of consecutive primes that adds to a prime below one-hundred.
*
* The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
*
* Which prime, below one-million, can be written as the sum of the most consecutive primes?
*/
long l = 0L;
long j = i;
int c = 0;
while (l < limit && j >= 0)
{
l += primes.ElementAt((int)j);
j--;
c++;
if (c > resCount && Formulas.isPrime(l))
{
resCount = c;
result = l;
}
}
}
return(result);
}
public static long Problem41()
{
/*
* We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
* What is the largest n-digit pandigital prime that exists?
*/
List <long> primes = Formulas.PrimeSieve(10000000);
long result = 0L;
foreach (long p in primes)
{
if (Formulas.isPandigital(p))
{
result = p;
}
}
return(result);
}