ashiima wrote:

If 1/(2^11 * 5^17) is expressed as a terminating decimal, how many non-zero digits will the decimal have?

A) 1

B) 2

C) 4

D) 6

E) 11

The question is based on very simple concepts but the application is a little tricky which actually makes it a good question.

First realize that \(2^{11} * 5^{17} = 2^{11}*5^{11}*5^6 = 10^{11}*5^6\)

So \(\frac{1}{10^{11}*5^6}\) is just \(\frac{0.00...001}{5^6}\)

Now what do you get when you divide .01 by 5? You get .002

You write 0s till you get 10 and then you get a non-zero digit.

Now what do you get when you divide .01 by 125? You get .00008

Do you notice something? The non 0 term is 8 = 2^3

The reason is this: You will only get 1 followed by as many 0s as you want in the dividend. 125 = 5^3 so you will need 2^3 i.e. you will need 10^3 as the dividend and then 125 will be able to divide it completely (i.e. the decimal will terminate)

Now, using the same logic, what will be the non zero digits if you are dividing .00001 by 625?

625 = 5^4. You will need 2^4 = 16 to get 10^4 and that will end the terminating decimal. So you will have two non 0 digits: 16

What will you get when you divide by 5^6? Your non zero digits will be 2^6 = 64 i.e. you will have 2 non-zero digits.

Try doing some calculations to better understand the concept used.

\(\frac{1}{2^{11}*5^{17}}=\frac{1}{(2^{11}*5^{11})*5^6}=\frac{1}{10^{11}*5^6}\). Multiply both nominator and denominator by \(\frac{2^6}{2^6}\) so that to have only power of 10 in denominator: \(\frac{1}{10^{11}*5^6}*\frac{2^6}{2^6}=\frac{2^6}{10^{17}}=\frac{64}{10^{17}}\), so the decimal will have two non-zero digits - 64.