Solution to question 1 .....

see whenever u r being asked to count the no of zeros in products like this ....

remember one thing...that 10 comes when u multiply primes nos 2 & 5 .Now there will be many 2's in it so the restricting factor will be no. of 5's ..thats it

Now to find the no of 5's ......

100,95,90,85,...........5 will contain 5's so there are 100+95+90+.....5 (as these will be the powers of 5 also) ie. 1050 (Its an AP series with a=5,d=5,n=20)

again, we left some nos in which 5 came twice...so now consider 100,75,50,25 only ie.......250 5's

so total is 1300 a)

*brain_anand <brain_anand@...>* wrote:

question 1

r=100^100x 99^99 x
98^98x ............3^3x 2^2x 1 how many zeros will

be there at end of r

1)1300

2)1325

3)1350

4)none of these

Q 2. Vaibhav has only 25 paise&50 paise coin worth Rs.15 find the

number of 25paise coins.

1)10

2)20

3)30

4)40

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