Question

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

Answer 1

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. Smallest number which when increased by 17 is exactly divisible by both 520 and 468=4680 – 17 = 4663. So, least number divisible by both 520 and 468 is 4680.

Answer 2

See this attached picture.

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