nkaur wrote:Hi,

I have a question regarding question 108 in the OG 12:

If t=1/2^9x5^3 is expressed as a terminating decimal, how many ezros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

When I started finding a solution, I wrote t= 1 x 2^-9 x 5^-3 and then wanted to multiply the whole thing to zee how many zeros the term has. However I did not succeed. When I checked the answer in the OG i understood it but I dont think I would use this method in the test. So, can someone present another way to solve this problem? It would be great if someone could tell me if "my" approach, mentioned above, can be applied to this problem at all?

Thanks a lot

It's useful to start by looking for powers of 10 "hiding" in the product.
1/[(2^9)(5^3)] = 1/[(2^6)(2^3)(5^3)]
From here we can use the property (A^n)(B^n) = (AB)^n to rewrite (2^3)(5^3) as 10^3
So, we get 1/[(2^6)(10^3)] = 1/[(64)(1000)]
Now let's break this into the product of 2 fractions --> (1/64)(1/1000)

Let's begin with 1/64: We know that 1/10 = 0.1 and we know that 1/100 = 0.01
So, 1/64 is between 0.01 and 0.1. Let's just say that 1/64 = 0.0??? (it doesn't matter what the digits are here. All we need to know is that there is one zero to the right of the decimal point.

When we take 0.0??? and multiply by 1/1000 (or we could say we are dividing by 1000), we must move the decimal point 3 spaces to the left to get 0.0000???, which means we will have 4 zeros to the right of the decimal point before hitting the first non-zero digit.