Complete the first 10 problems

.gitignore

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+a.out

problem1.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+int
+main(void)
+{
+	int sum = 0;
+	for (int a = 1; a < 1000; a++)
+	{
+		if (a % 3 == 0 || a % 5 == 0)
+			sum += a;
+	}
+	printf("The sum of all multiples of 3 or 5 below 1000 is: %d\n", sum);
+}

problem10.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+static int
+isPrime(int number)
+{
+	if (number <= 1)
+		return 0;
+	for (int i = 2; i * i <= number; i++)
+		if (number % i == 0)
+			return 0;
+	return 1;
+}
+
+int
+main(void)
+{
+	long sum = 0;
+	for(int i = 0; i < 2000000; i++)
+	{
+		if (isPrime(i))
+		{
+			sum += i;
+		}
+	}
+	printf("The sum of all the primes below 2,000,000 is %ld\n", sum);
+}

problem2.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+int
+main(void)
+{
+	int num = 1;
+	int prevnum = 1;
+	int sum = 0;
+	while (num <= 4000000)
+	{
+		if (num % 2 == 0)
+			sum += num;
+		int store = num;
+		num = num + prevnum;
+		prevnum = store;
+	}
+	printf("The sum of the even-valued terms of the Fibonacci sequence whose values do not exceed four million is: %d\n", sum);
+}

problem3.c

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+#include <stdio.h>
+#include <math.h>
+
+static int
+isPrime(long number)
+{
+	if (number <= 1)
+		return 0;
+	for (long i = 2; i * i <= number; i++)
+		if (number % i == 0)
+			return 0;
+	return 1;
+}
+
+int
+main(void)
+{
+	const long number = 600851475143;
+	long remainder = number;
+	long largestPrimeFactor = 1;
+	float percentComplete = 0.00;
+
+	while (remainder % 2 == 0)
+		remainder /= 2;
+	while (remainder % 3 == 0)
+		remainder /= 3;
+	while (remainder % 5 == 0)
+		remainder /= 5;
+
+	// I love types
+	long sqrtOfRemainder = (long)sqrt((double)remainder);
+
+	for (long i = 7; i < sqrtOfRemainder; i++)
+	{
+		if (number % i == 0 && isPrime(i))
+		{
+			printf("Prime factor found: %ld\n", i);
+			largestPrimeFactor = i;
+		}
+	}
+
+	printf("The largest prime factor of %ld is %ld\n", number, largestPrimeFactor);
+}

problem4.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+// This program is specific to the given problem where only 3-digit numbers are
+// considered. I _could_ have made it generalizable... but I feel lazy.
+
+int
+main(void)
+{
+	int largestPalindrome = 0;
+	char buffer[6]; // 999x999 gives a 6-digit number
+	char revbuffer[3];
+
+	for (int a = 100; a < 1000; a++)
+	{
+		for (int b = 100; b < 1000; b++)
+		{
+			int product = a * b;
+			sprintf(buffer, "%d", product);
+			revbuffer[0] = buffer[5];
+			revbuffer[1] = buffer[4];
+			revbuffer[2] = buffer[3];
+			if (strncmp(buffer, revbuffer, 3) == 0)
+			{
+				printf("%d * %d = %d is palindromic\n", a, b, product);
+				if (product > largestPalindrome)
+					largestPalindrome = product;
+			}
+		}
+	}
+	printf("The largest palindrome made from the product of two 3-digit numbers is %d\n", largestPalindrome);
+}

problem5.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+int
+main(void)
+{
+	for (int i = 1; i < 2147483647; i++)
+	{
+		for (int j = 1; j < 21; j++)
+		{
+			if (i % j != 0)
+				break;
+			if (j == 20)
+			{
+				printf("The sallest positive number that is evenly divisible by all of the numbers from 1 to 20 is: %d\n", i);
+				exit(EXIT_SUCCESS);
+			}
+		}
+	}
+}

problem6.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+int
+main(void)
+{
+	int difference = 0;
+	int squareOfSum = 0;
+	int sumOfSquares = 0;
+	for (int i = 1; i < 101; i++)
+	{
+		squareOfSum += i;
+		sumOfSquares += i*i;
+	}
+	squareOfSum *= squareOfSum;
+	difference = squareOfSum - sumOfSquares;
+	printf("The difference between the sum of the squares of the first one hundred natural numbers and the square of the sum is: %d\n", difference);
+}

problem7.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <limits.h>
+
+static int
+isPrime(long number)
+{
+	if (number <= 1)
+		return 0;
+	for (long i = 2; i * i <= number; i++)
+		if (number % i == 0)
+			return 0;
+	return 1;
+}
+
+int
+main(void)
+{
+	int primeCount = 0;
+	for(long i = 0; i < LONG_MAX; i++)
+	{
+		if (isPrime(i))
+		{
+			primeCount++;
+			if (primeCount == 10001)
+			{
+				printf("The 10,001st prime is %ld\n", i);
+				exit(EXIT_SUCCESS);
+			}
+		}
+	}
+}

problem8.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int
+main(void)
+{
+	char* thousandDigitNumber = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
+	long largestProduct = 0;
+
+	for (int i = 0; i < 988; i++)
+	{
+		long product = 1;
+		for (int j = i; j < i+13; j++) {
+			product *= (thousandDigitNumber[j]-'0');
+		}
+		if (product > largestProduct)
+			largestProduct = product;
+	}
+
+	printf("The value of the largest product of thirteen adjacent digits in the thousand digit number is: %ld\n", largestProduct);
+}

problem9.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+// This is definitely a silly and innefficient way to do this
+// But... everything is fast for small n :^)
+
+int
+main(void)
+{
+	for (int a = 1; a < 1000; a++)
+	{
+		for (int b = 1; b < 1000; b++)
+		{
+			for (int c = 1; c < 1000; c++)
+			{
+				if (a + b + c != 1000)
+					continue;
+				if (a*a + b*b != c*c)
+					continue;
+				printf("Solution found:\n");
+				printf("a = %d\n", a);
+				printf("b = %d\n", b);
+				printf("c = %d\n", c);
+				printf("Product of a, b, and c: %d\n", a*b*c);
+				exit(EXIT_SUCCESS);
+			}
+		}
+	}
+	printf("Solution not found :(\n");
+	exit(EXIT_FAILURE);
+}