The least number that should be added to 10000 so that it is exactly divisible by 327 is:
327 is a multiple of 3.
So, a number should be added to 10000 which will make it divisible by 3.
And 327 is also a multiple of 109.
So, To make 10000 completely divisible by 327, a number should be added to that number which would be divisible by 109 also.
So, $$(10000+137)=10137$$ is divisible by 3 as well as 109.
So,Option C is correct.
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