Terminating Decimals | Number Properties

If you’ve been studying for the GMAT for any length of time, you are most likely pretty solid on the basic concepts of decimals and rational numbers.

Decimals can be either rational or irrational.

Decimals which terminate and those which are non-terminating but repeating are rational. These decimals can be written as fractions in the form $\frac{a}{b}$ . As an example, $.375 = \frac{3}{8}$

Decimals which are non-terminating and non-repeating are irrational, like roots such as root 2, root 3 etc. The digits after the decimal point will continue on forever.

The problem comes when we get a question based on these rather straightforward concepts. That’s the point at which we realize that our basic knowledge and understanding are not as strong or as sufficient as we assumed them to be. As an example, let’s take a look at the following question:

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

(A) ${\frac{10}{189}}$

(B) ${\frac{15}{196}}$

(D) ${\frac{25}{144}}$

(E) ${\frac{39}{128}}$

If your first thought is that the best strategy is to simply divide the numerator by the denominator in each answer choice and thus figure out which one terminates and which ones do not, congratulations! You have a ton of computation work ahead of you, and it will be extremely time-consuming and tedious, and committing to doing that many calculations dramatically increases your chances of making a careless mistake somewhere. Assuming that the process goes well, you would eventually arrive at the right answer, but we don’t have that kind of time to burn, and we know that the GMAT does not really care if we can crank out a lot of calculations in order to brute force our way to the correct answer. Because we know that the exam is trying to evaluate our ability to reason our way to the correct answer, there simply has to be another, more logical (and significantly less exhausting and tedious!) approach to this problem. Well, let’s walk through it:

A fraction, in its most reduced form, can be expressed as a terminating decimal if and only if the denominator has powers of only 2 and/or 5. Take a look at a simple example to try to understand why that is the case:

Let’s say that the variables $a$ and $b$ represent two integers. We would also know that the following relationship is true:

$\frac{a}{b} = a \times \frac{1}{b}$

For $ab$ to be a terminating decimal, $1b$ must be a terminating decimal. Well, 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.

Taken that way, it’s a bit more obvious than it seemed, isn’t it?

$\frac{1}{3} = .333333333333333333…$

$\frac{1}{7} = .142857142857142857…$

$\frac{1}{11} = .09090909090909090…$

Now that we understand how to identify if a fraction will yield a terminating decimal, let’s revisit the question we posed above. With this new understanding in mind, it should be a little more straightforward now.

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

(A) ${\frac{10}{189}}$

(B) ${\frac{15}{196}}$

(D) ${\frac{25}{144}}$

(E) ${\frac{39}{128}}$

Instead of having to crunch through each answer choice by hand, we can simply look at the denominators of each fraction to determine which one will give us a termination decimal. Let’s take a look at the prime factors of each denominator:

$189 = 3^2 \times 7$ (will not terminate)

$196 = 2^2 \times 7^2$ (will not terminate)

$225 = 3^2 \times 5^2$ (will not terminate)

$144 = 2^4 \times 3^2$ (will not terminate)

$128 = 2^7$ (bingo!)

Because 128 is a power of 2, our fraction will terminate, and (E) is thus the correct answer. All of the denominators in the other four answer choices contain other prime numbers as well and as a result will not terminate.

Using the same concepts as the previous question, let’s look at the following problem:

Question 2: If ${1211\times517}$ is expressed as a terminating decimal, how many non-zero digits will the decimal have?

(A) 1

(B) 2

(D) 6

(E) 11

Solution:

A good first step is to realize that we can combine 2s and 5s to make 10s, which will make our job of figuring out the number of decimal places much simpler:

$2^{11} \times 5^{17} = 2^{11} \times 5^{11} \times 5^6 = 10^{11} \times 5^6$

As a result, we know that $\frac{1}{(10^{11} \times 5^6)}$ is really just $\frac{.00000…001}{5^6}$ .

From this point, we should 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 interesting there? The non zero term is 8, or $2^3$ .

The reason is this: You have 1 followed by as many 0s as you require in the dividend. 125 is $5^3$ , so you will need $2^3$ i.e. you will need $10^3$ as the dividend and then 125 will be able to divide out completely (in other words, the decimal will terminate).

Now, trying to apply the same logic, what will be the nonzero digits if you divide .00001 by 625?

Look at the exponents: $625 = 5^4$ , which tells us that, to get to a power of 10, we will need $2^4$ , or 16, to get $10^4$ , thus allowing the decimal to terminate. As a result, we know that the result will exactly have two nonzero digits: the 16 from the $2^4$ .

Going back to where we left off with the original problem, what will you get when you divide .000…0001 by $5^6$ ? Applying the same reasoning as before, we know the nonzero digits will be 64 (since we’re dividing by $5^6$ , we need $2^6$ , which is 64, to get to $10^6$ and terminate the decimal).

As usual with GMAT questions, there is another way to approach it using similar logic but a different process. If we can manipulate the fraction to give us only a power of 10 in the denominator, we can then look at the numerator to know exactly what nonzero digits will appear in the resulting terminating decimal (the power of 10 in the denominator will simply tell us how many zeroes appear between the decimal place and the nonzero digits the numerator yields). Here, we need to turn the 56 in the denominator into 10s, so let’s multiply the entire fraction by $\frac{2^6}{2^6}$ :

$\frac{1}{(10^{11} \times 5^6)} \times \frac{2^6}{2^6} = \frac{2^6}{10^{11} \times 10^6)} = \frac{2^6}{10^{17}}$

The numerator turns into 64, and since the denominator is a power of 10, we can look at the exponent to know how many zeroes will appear between the decimal point and the nonzero digits. Because the denominator has an exponent of 17, the 64 from the numerator will be shifted 17 places to the right of the decimal place. The nonzero digits are the 64 that the numerator gives us, so we end up with two nonzero digits. The correct answer is thus (B).

Let’s conclude our discussion of identifying terminating decimals on GMAT problems by taking a look at one more example, this time in the form of Data Sufficiency, everyone’s favorite kind of quant problem 😉

Question 3: Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, 0.82, and 5.096 are three terminating decimals. If ${r}$ and ${s}$ are positive integers and the ratio ${\frac{r}{s}}$ is expressed as a decimal, is ${\frac{r}{s}}$ a terminating decimal?

Statement 1: ${90 < r < 100}$

Statement 2: ${s = 4}$

Right off the bat, notice how much easier this question is if we rephrase it as “if $rs$ is in its most simplified form, does the prime factorization of the denominator consist entirely of 2’s or 5’s?”

Now that we are experts on the properties and identification of terminating decimals, Statement 1 clearly can’t be sufficient on its own because it simply tells us nothing at all about the denominator. $\frac{91}{2}$ and $\frac{91}{10}$ are terminating decimals, for example, but $\frac{91}{3}$ and $\frac{91}{12}$ are not.

Statement 2 tells us that the denominator is 4, or $2^2$ . If we’ve mastered the rule for identifying terminating decimals, we should see right away that this must be sufficient because we know anything divided by 4 will result in a terminating decimal. The answer is B, Statement 2 alone is sufficient to answer the question. Beware of the C trap here: some GMAT takers may feel uncomfortable choosing an answer that involves no knowledge of the numerator. As demonstrated on the previous examples, however, we know that the numerator of the fraction is irrelevant in the process of identifying a terminating decimal. Only the prime factors of the denominator (in its most reduced form) matter.

Our discussion of terminating decimals here also has a broader takeaway for GMAT preparation in general: many problems that appear to come down to tedious and/or time-consuming calculations are built around a concept, rule, or logical principle that will not be apparent to the unprepared GMAT student. Make sure your study plan includes time for understanding how concepts and rules work, not simply what they mean, and you will find yourself solving conceptually more often and using brute force less often, leading to great improvements in your time efficiency on the quant section!

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