Question

The highest power of 2 by which the product of first 100 counting numbers can be divided without any remainder is

Answer 1

Dear student ,
This problem is solved using a direct formula. Please remember it
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the formula for the highest power of a number x contained in n!

is= [n!/x] + [n!/x^2] + [n!/x^3] + .....till it becomes 0.

where [ ] denotes graetest integer function.

here in this problem, (which means [1.2]=1, that is the greatest integer less than or equal to 1.2 here its is one)

n!=100!

x=2

so total power=[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64]

=50+25+12+6+3+1=97