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+// Autor: Ahmed Gamal
+
+// the idea for this solution is so simple, let's look at the following observations:
+// 1- each bulb will be on after the number of its factors is odd
+// 2- each number has even number of factors except the perfect squares (because each factor has a pair in the other side of the number, except the perfect squares)
+// 3- so the number of bulbs that will be on is the number of perfect squares less than n
+
+// since the perfect squares are growing so fast, we will loop until the square of the number is less than or equal n (at most sqrt(n))
+
+// or we can use the following formula to calculate the number of perfect squares less than n:
+// Math.floor(Math.sqrt(n)), why?
+// because the perfect squares are the squares of the numbers from 1 to Math.floor(Math.sqrt(n)), so every integer from 1 to Math.floor(Math.sqrt(n)) will have a perfect square less than or equal n, so the number of perfect squares less than n is Math.floor(Math.sqrt(n))
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var bulbSwitch = function(n) {
+    return Math.floor(Math.sqrt(n));
+};
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