Highest Power of 3 in 1!x2!x3!x............x99!x100!
Here's the solution
The above series can be written as,
1100 * 299 * 398 * 497 *.... * 1001
Note that for each number 'n' the power will be 101 - n
First we find the multiples of 3...
ie, 398, 695, 992, 1289, ............ , 992
So the no. of 3's in the above numbers is
98+95+92+................+2 = 1650...--------------------->(1)
Now, we need to add multiples of 3 contained in multiples of 9...
ie, 992, 1883, 2774, .............., 992
i.e 92+83+74+........+2 = 517 (see, the power of 9 will be 992 = (3*3)92... similarly for 18,27,...,99)-----------------(2)
Now, add multiples of 3 in 27...
ie, 74 + 47 + 20 = 141 --------------->(3)
Finally, for 81 we have 20--------------->(4)
Add (1) (2) (3) and (4) we have, 2328...
I hope this is lucid.
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