Special pythagorean triplet: Difference between revisions

<br>

Incidentally, even though we are '''told''' there is only one solution, it is almost as quick to verify this by observing that, since a < b < c, the maximum value of a must be such that 3a + 2 = 1000 or max(a) = 332. The following version ran in 0.015 seconds and, of course, produced the same output:

<lang wren>for (a in 3..332) {

var b = a + 1

while (true) {

var c = 1000 - a - b

if (c <= b) break

if (a*a + b*b == c*c) {

System.print("a = %(a), b = %(b), c = %(c)")

System.print("a + b + c = %(a + b + c)")

System.print("a * b * c = %(a * b * c)")

}

b = b + 1

}

}</lang>