Question

13 sbse chota multiple jisse 4 5 6 7 8 se bhaag dene pr ,remain 2 bacge​

Answer 1

Answer:

Lcm of 4,5,6,7,8= 840

Thus the smallest number required is 840+2=842. Thus the remainder is added to the lcm. This is because the lcm is exactly divisible by all these numbers. And the remainder is left only when a part is not divided. If I add that part to a perfectly divisible number, it gives us the required number.

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