029.js

// Consider all integer combinations of ab for 2 <= a <= 5 and 2 <= b <= 5:

//   2^2=4, 2^3=8, 2^4=16, 2^5=32
//   3^2=9, 3^3=27, 3^4=81, 3^5=243
//   4^2=16, 4^3=64, 4^4=256, 4^5=1024
//   5^2=25, 5^3=125, 5^4=625, 5^5=3125

// If they are then placed in numerical order, with any repeats removed, we get
// the following sequence of 15 distinct terms:

//   4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

// How many distinct terms are in the sequence generated by a^b for 
// 2 <= a <= 100 and 2 <= b <= 100?

function power(n, power) {
	var digits = [1];
	while (power--) {
		var cr = 0;
		for (var i = 0; i < digits.length; i++) {
			var num = n * digits[i] + cr;
			digits[i] = num % 10;
			cr = Math.floor(num/10);
		}
		while (cr > 0) {
			digits.push(cr % 10);
			cr = Math.floor(cr/10);
		}
	}
	return digits.reverse().join('');
}

function distinctPowersNum(min, max) {
	var out = [];
	for (var n = 2; n <= max; n++) {
		for (var pow = 2; pow<= max; pow++) {
			out.push( power(n, pow) );
		}
	}
	return out.filter(function(val, i, self) {
		return self.indexOf(val) === i;
	}).length;
}

console.log( distinctPowersNum(2, 100) );