Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

The solution can be done based on LCM.

Answer: 4663 is the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Let's explore the least common multiple and how to apply LCM to the given question.

Explanation:

Let the required number be x.

Given that, the number x when increased by 17 is exactly divisible by 520 and 468.

x + 17 = least common multiple of both 520 and 468.

x + 17 = LCM(520, 468)

x + 17 = 4680

x = 4680 - 17

x = 4663

Thus, the smallest number which when increased by 17 is exactly divisible by both 520 and 468 is 4663.