Hi there, can anyone help me solve this question the gregmat-way? I remembered about how 2’s and 5’s create zero’s but in the OG it’s solved way more complex than I think prepswift/gregmat would do. Any help? Thanks a lot!!

2^{11} \times 5^{11} = 10^{11} and you’ll have 5^{6} remaining as well(Here, we distributed 5^{17})

thus, your current denominator is (10^{11} \times 5^6) → Now , we need 2^6 to make another 10’s with 5^6 thus, you’ll need to divide and multiply by 2^6 which will give you 10^{17} in the denominator and 2^6 in the numerator and your equation will look like this \dfrac{2^6}{10^{17}}

Or what you can do is :

Perform this step:

2^{11} \times 5^{11} = 10^{11} and you’ll have 5^{6} remaining as well(Here, we distributed 5^{17})

Ignore the 10 raise to the power and you’re left with 64 which is two-digit thus, your ans is 2

It’s still very difficult to me… Can I rephrase and you correct me with my simplification of what I understand?

So we 2^11 and 5^17 in the denominator

As with 2’s and 5’s we make tens, I basically understand that we can ignore the couples of 2’s and 5’s. Because they will only result in zero’s, right? So taking out 2^11 and 5^11, which means 5^6 remains.

The question can be simplified to "how many nonzero digits are there in 1/15625, is that correct? Or will I lose nonzero numbers by ignore the pairs of 2’s and 5’s (11 of them)?

Then in my calculator when I do 1/15625 I get scientific notation but I’m assuming if I put in GRE calculator I will only see the 6 and the 4?

Would you identify this as a hard question?

Thanks again

It’s the same calculator that you see in GregMat full quant section quiz https://www.gregmat.com/course/gre-mini-exams

great so “my” logic works, right?