Hey guys,

I've asked this in the GMAT Club forums, but didn't get an answer, but wanted to see what you guys thought.

They finally unblocked GMAT Club here in Afghanistan (initially had it marked as a "spam" site for some reason) so I registered for the paid tests.

I'm taking them with no time limit right now, and I am still getting ravaged, battling under .500!

This is actually starting to not just frustrate me, but undermine my confidence a little bit. I'm trying to practice with the toughest quant possible, because I need a 49 or 50 Q to get the score I want, but I'm starting to think that my skills are eroding because I'm getting killed on these.

Anyone else try them? I don't think they are reflective of the actual GMAT's difficulty. They seem way tougher, but I'd like to hear everyone else's opinions.

Edit to add:

Example:

How many zeros does 100! end with?
20
24
25
30
32

Explanation:

Trailing zeros in 100!: 1005+10052=20+4=24

THEORY:

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

For example 125,000 has 3 trailing zeros;

The number of trailing zeros n!, the factorial of a non-negative integer n, can be determined with this formula:

n5+n52+n53+...+n5k, where k must be chosen such that 5(k+1)>n

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

325+3252=6+1=7. Notice that the last denominator (52) must be less than 32. Also notice that we take into account only the quotient of the division, that is 325=6.

So there are 7 zeros in the end of 32!.

Another example, how many trailing zeros does 125! have?

1255+12552+12553=25+5+1=31,

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.



Are you kidding me? trailing zero theory? what in the world is that????