//-------------------------------------------------------------------
// enigma1358.c (c) 2005 by Charles Petzold (www.charlespetzold.com)
//
// "Five Fives" by Adrian Somerfield, 
//			New Scientist, 17 September 2005, page 50.
//
// Five numbers in five different number bases. All five numbers are
// 11101, and the first four total to the fifth. What's the total?
//
// We are also told that all five numbers are even. That means that
// all the number bases are odd. (Any number in an even number base
// ending in 0 is even. Add 1 and it's odd.)
//
// We are also told that the total is less than 100,000. That restricts
// the bases to 17 or lower. 
//
// We're really only dealing with 8 possible numbers, as shown in the 
// following chart:
//
//	Base	11101 Converted to Decimal
//	----	--------------------------
//	  3 		   118
//	  5 		   776
//	  7 		 2,794
//	  9 		 7,372
//	 11 		16,094
//	 13 		30,928
//	 15 		54,226
//	 17 		88,724
//
// With these possibilities, it's not too hard to determine that the 
// solution is 776 + 2,794 + 30,928 + 54,226 = 88,724.
//
// Let's code it up anyway.
//----------------------------------------------------------------------

#include <stdio.h>

// The function assumes the first argument is a number in a number 
// base given as the second argument. The number is converted to decimal.
// 
// This function obviously cannot work for number bases greater than 10
// if the number has digits greater than 9, but that's not a problem 
// here.

int ConvertToDecimal(int num, int base)
{
	int result = 0;
	int multiplier = 1;

	while (num > 0)
	{
		result += multiplier * (num % 10);
		num /= 10;
		multiplier *= base;
	}

	return result;
}

int main(void)
{
	int base1, base2, base3, base4, base5;
	int num1, num2, num3, num4, num5;

	// Five loops for the five numbers. Let's arrange them so
	// the first number is the lowest. The number base can be
	// 3, 5, 7, or 9.

	for (base1 = 3; base1 < 11; base1 += 2)
	{
		num1 = ConvertToDecimal(11101, base1);

		// The second number base is greater than the 
		// first but can go up to 11.		

		for (base2 = base1 + 2; base2 < 13; base2 += 2)
		{
			num2 = ConvertToDecimal(11101, base2);

			// And so forth.

			for (base3 = base2 + 2; base3 < 15; base3 += 2)
			{
				num3 = ConvertToDecimal(11101, base3);

				for (base4 = base3 + 2; base4 < 17; base4 += 2)
				{
					num4 = ConvertToDecimal(11101, base4);

					for (base5 = base4 + 2; base5 < 19; base5 += 2)
					{
						num5 = ConvertToDecimal(11101, base5);

						// If the 5th number is the sum of the first four,
						// the problem is solved.

						if (num5 == num1 + num2 + num3 + num4)
						{
							printf("%i\n", num5);
						}
					}
				}
			}
		}
	}
	return 0;
}