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

This question can be solved using the concept of LCM.

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

Let's find the LCM of both the numbers and thereby solve the problem.

Explanation:

Let the required number be x

Given that the required number when increased by 17 will be exactly divisible by both 520 and 468.

That means, x + 17 is divisible by both 520 and 468 and it should be the smallest number.

As we know that LCM (520,468) is the least possible number which is exactly divisible by the given numbers.

LCM (520,468) is 4680.

So, x + 17 = 4680

⇒ x = 4680 - 17

⇒ x = 4663.

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