Find the smallest number which when increased by 17 is exactly divisble by both 520 and 468.
This question can be solved using the concept of LCM.
Answer: The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is 4663.
Let's find the LCM of both the numbers and thereby solve the problem
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.