Question

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

Answer 1

Answer:

4663

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.

Answer 2

Answer:

Smallest number divisible by 520 and 468 when it is increased by 17 is 4663

Explanation:

Solution:

The given numbers are 520 and 468.

Let us find the LCM of 520 and 468

520 = 2 × 2 × 2 × 5 × 13

468 = 2 × 2 × 3 × 3 × 13

LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 ×5 × 13

LCM of 520 and 468 = 4680

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 = LCM of 520 and 468 – 17

The smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 – 17

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