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

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5437

4663

6545

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answer is B.

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Detailed Solution

The smallest numbers are 520 and 468
The smallest numbers that are exactly divisible by both 520 and 468 when multiplied by 17 are obtained by subtracting 17 from the LCM of 520 and 468.

LCM of 520 and 468

Smallest number which when increased by 17 is exactly divisible by both 520 and 468

Hence, option 2 is correct.

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Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE