I think the answer is A

No.. 10^1 is a factor and 10^2 is also a factor

So D can be 1 or 2...

I will post the OA later

OK. Yes. Starting from
10^1 thru 10^7 are all factors
d=1,2,..7

From (2) d>6

So the answer is C

i think answer is e since d coule have more values.

OA is C

This question can be easily done if we know the no of tens in f.
we know that f = 30!
if we want to know the no of tens in a factorial here is the method
lets say we want to know the no of tens in 30!
it will be [30/5] +[30/5^2] + [30/5^3] + ... so on till the division produces 0.
here the no of tens is 6 +1 + 0 = 7.
Now using d>6 we easily approach the answer

[email protected] wrote:[30/5] +[30/5^2] + [30/5^3] + ... so on till the division produces 0.

saurabh, can you please explain this a lil more.. how did you decide to divide by 5

if I want to know the no of tens I need to find 5*2 in a number, rite?
now if a number has 3 5s and 6 2s.
How many tens will it make .obviously it will be 3 as no of five is only 3.
So i have divided the number(factorial) by 5 to know the no of tens in a number.
I read this method in a book long ago and hence thought of applyin here too
Please let me know if you need further explanation

Coool... thanks,

I Agree the ans should be C

now the nos that we are considering here is 30!

so highest value of d would be the nos of trailing 0's in the nos 30!

now 30! is

1X2X3X4.........X29X30

now in this product a trailing 0 would be produced in two cases

1) when an even nos is multiplied by 5

2) when any nos is multiplied by 10

now in this product we have the nos

5,10,15,20,25,30

which would lead to trailing 0's
e.g. 4X5 or 5X6 will give u a trailing 0, take any ways

i.e product upto 6!= 4! X (5X6)

or think of this as product upto 5! is 3!X (4X5)

However 25 = 5X5 so we can visualise prodcut upto 25! as

23! X (24X5) X (5X26)

here 24X5 as well as 5X26 would produce trailing 0's so this would add 2 0's

hence total nos of 0's =7 (d>6)

so C is SUFF