# How to Identify Terminating Decimals on the GMAT We know the basics of decimals and rational numbers.

–   Decimals can be rational or irrational.

– Decimals which terminate and those which are non-terminating but repeating are rational. They can be written in the form a/b.

–  Decimals which are non-terminating and non-repeating are irrational such as ?2, ?3 etc.

The problem comes when we get a question based on these basics. That’s when we realize that our basics are not as strong as we assumed them to be. For example, look at this question:

Question: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189
(B) 15/196
(C) 16/225
(D) 25/144
(E) 39/128

If your first thought is that we will simply divide the numerator by the denominator in each case and figure out which terminates and which doesn’t, you must realize that that is a very time consuming process. There has to be another logical approach to this problem. Well, here it is:

A fraction in its lowest term can be expressed as a terminating decimal if and only if the denominator has powers of only 2 and/or 5. Let’s try to understand the logic behind it.

Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s. You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?

1/3 = .333333333333333333…

1/7 = .142857142857142857…

1/11 = .09090909090909090…

Now the question we posed above is quite simple. Let’s look at it again.

Question 1: Which of the following fractions has a decimal equivalent that is a terminating decimal?

(A) 10/189
(B) 15/196
(C) 16/225
(D) 25/144
(E) 39/128

Only option (E) has a denominator of the form 2^a*5^b.

128 = 2^7

Therefore, 39/128 will terminate. All the other denominators have other prime numbers as well and hence will not terminate.

Using the same concepts, let’s look at another question.

Question 2: 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

Solution:

First realize that 2^11 * 5^17 = 2^11 * 5^11 * 5^6 = 10^11 * 5^6
So 1/(10^11 * 5^6)  is just 0.00…001/5^6.

Now let’s try to figure out the answer intuitively:

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.

What do you get when you divide .01 by 125 (which is 5^3)? You get .00008.

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

The reason is this: You have 1 followed by as many 0s as you require 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 .000…0001 by 5^6? Your non zero digits will be 2^6 = 64 i.e. you will have 2 non-zero digits.

Another way to look at the problem is this:

1/(10^11 * 5^6)  = 2^6/(10^17) (multiply and divide by 2^6)

= 64/(10^17)

Since the denominator is a power of 10, it will just move the decimal 17 places to the left. The non-zero digits will remain 64 only i.e. 2 digits.