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

Given: Two numbers 520 and 468

We need to find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

Solution

Let us first obtain the LCM of the given numbers:

We can do this by prime factorization method

On prime factorizing 520 we get

520=2 × 2 × 2 × 5 × 13

On prime factorizing 468 we get

468 = 2 × 2 × 3 × 3 × 13

So taking LCM from these factors we get

2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680

As per given condition, the number increased by 17 is exactly divisible by both 520 and 468

Hence 4680 – 17 = 4663

The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is 4663

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class