First, we need to check the values of x for which the equation can be zero.
At x= 2,4,6,...20 the equation will be equal to 0 (as one of the multiple will be equal to 0)
so no of integer values of X for which eqn is 0 is 10

Next for eqn to be <0, Odd number of multiples with -ve values makes the entire equation -ve (i.e.,< 0)
This happens when X takes the values of 3,7,11,15,19 i.e., a total of 5 values.

So total number of integer values of x that satisfy the eqn are 15.

so is the answer option C or did I miss any other conditions??
What is the OA?

And btw what is the source of this question? Seems a bit of high standard for GMAT

Regards,
Chetan

Ah! The OA is indeed C

Remember doing it in some Kaplan material and it was in the advanced section, so should be 700+ level I guess

Cool problem!
IMO, Answer is C

We could start by defining the limits to this problem.
There are 10 terms in the LHS of the equation.
a) x cannot be less than 2.
For all values of x less than 2, every term in the equation will be negative.
Example - If x = 0, then equation becomes -2*-4*...-20 (10 terms). This is a product of 10 negative numbers which is a positive quantity. But equation states that LHS is <= 0. Therefore, x >= 2

b) Same way, x can't be greater than 20.
For all values of x > 20, equation becomes {positive*positive...*positive} (10 times) = positive
Not allowed.

c) Therefore, 2 <= x <= 20

d) In this range, there are 10 values where product becomes equal to 0.
x = {2,4,6,....,20)

e) Now,
when x = 3, product becomes {positive*negative*negative...9 times} = negative
when x = 5, product becomes {positive*positive*negative*negative...8 times} = positive
when x = 7, product becomes {positive*positive*positive*negative*negative...7 times} = negative
Therefore, product is negative for every other odd number between 2 and 20. Therefore, acceptable values for x = {3,7,11,15,19}

f) Combining d) and e),
Total number of possible values of x = [spoiler]10+5 = 15[/spoiler]

Cheers!