Question

10) Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468
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Answer 1

Answer:

4663

Step-by-step explanation:

Answer: The given numbers are 520 and 468. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4663

hope it's helpful to you

Answer 2

Answer:

4646 is the answer.

To find this take LCM of 520 and 468.

LCM = 4663

Now subtract 17 from 4663.

We'll get 4646. So when 4646 is increased by 17 it becomes the smallest no. which can divide 520 and 468.

Hope it helps you.