Welcome back to our series on exponents. Last time we used powers of 10 to express “almost integer” numbers. Today we will use powers of 10 to handle problems that ask us to count zeros or nonzero digits. These problems can be baffling if you haven’t learned about them. Let’s start by comparing two official GMAT/Executive Assessment (EA) problems in this category:

If *t *= 1/(2^{9}+5^{3}) is expressed as a terminating decimal, how many zeros 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

If *d *= 1/(2^{3}+5^{7}) is expressed as a terminating decimal, how many nonzero digits will *d* have?

A. One

B. Two

C. Three

D. Seven

E. Ten

In each of these problems, we see the number 1 divided by powers of 2 and powers of 5. The first problem asks us to count the number of zeros between the decimal and the first nonzero digit. The second problem asks us to count the nonzero digits instead.

Whether counting zeros or nonzeros, the best way to start on these problems is to “extract” the powers of 10 in the denominator. Let’s take the first problem:

1 / (2^{9} * 5^{3})

Recall our fundamental fact about exponents: they notate successive multiplications by the same value. So this denominator is the product of nine twos and three fives. We can make powers of 10 by pairing two with fives (since 2 * 5 = 10).

1 / (2^{9} * 5^{3})

1 / 2^{6} * (2^{3} * 5^{3})

1 / 2^{6} * 10^{3}

Three of our twos went over and joined the fives to make three tens, or 10^{3}.

Now we have 1 / (2^{6} * 10^{3}), and we need to see how many zeros appear after the decimal but before any nonzero digits. At this point, knowing your powers of 2 comes in handy and allows you to find the full value of the denominator.

1 / 2^{6} * 10^{3}

1 / 64 * 10^{3}

So we have 1 / 64,000. We can find the number of zeros in the decimal form of this number by simply *subtracting 1 from the number of digits in the denominator*. The denominator, 64,000, has five digits, so 1 / 64,000 has four zeros between the decimal and the first nonzero digit. **The correct answer is B.**

Let’s look again at our second example problem:

If *d *= 1/(2^{3}+5^{7}) is expressed as a terminating decimal, how many nonzero digits will *d* have?

A. One

B. Two

C. Three

D. Seven

E. Ten

This problem asks us about the nonzeros instead of the zeros. Since every digit is either a nonzero or a nonzero, we can find the number of nonzero digits most easily by subtracting the number of zeros after the decimal from the total number of digits after the decimal.

# of nonzero digits after decimal = (# of digits after decimal) – (# of zeros after the decimal)

We know how to find the number of zeros after the decimal, but first, we will find the total number of digits after the decimal. On these problems, *this number is always equal to the larger exponent in the base. *Here the larger exponent is 7, so there are a total of 7 digits after the decimal.

Now, all we have to do is find the number of zeros before the first nonzero digit (like we did on the previous problem) and subtract this number from 7.

1 / (2^{3} * 5^{7})

1 / (5^{4} * 10^{3})

Again, knowing your powers helps:

1 / (5^{4} * 10^{3})

1 / (625 * 1000)

1 / (625,000)

The denominator has six digits, so there are 6 – 1 = 5 zeros after the decimal before the first nonzero digit. There are a total of seven digits after the decimal.

# of nonzero digits after decimal = (# of digits after decimal) – (# of zeros after the decimal)

# of nonzero digits after decimal = 7 – 5

# of nonzero digits after decimal = 2

And **the correct answer choice is B.**

Now you’re ready for “zero or nonzero” GMAT/EA problems. Next time we will look at a problem category that merges exponents with number properties.

**Contributor: ***Elijah Mize (Apex GMAT Instructor)*