Math, asked by toppoaman287, 1 month ago

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

Answers

Answered by elenstudent365

Answer:

4663.

The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LC M of 520 and 468. Hence, the smallest number which when increased by 17 is exactly divisible by both 520 and 468 is 4680−17 = 4663.

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