No. of strings Arnold had ⇢3
least number of equal-sized lengths each color string could have
To find the least number of equal-sized lengths of each color, we'll find the least common factor of 3,5&15
prime factor of 3 ⇢1 x 3
prime factor of 5⇢ 1 x 5
prime factor of 15⇢ 1 x 3 x 5
so,
LCM of 3,5,15 ⇢ 1 x 3 x 5 = 15