Exponents Power of k Problems
Posted on
27
Sep 2022

Exponents: Power of k Problems

Welcome to the penultimate article in our series on exponents on the GMAT/Executive Assessment (EA). Today we explore a problem category in which exponents are used to notating the highest power of an integer by which a larger integer is divisible. GMAT/EA problems typically use the variable k in the place of this exponent. These problems involve exponents but are also in the realm of number properties. Exponents are useful for notating the numbers of prime factors of a given integer. Here’s an example:

4680

468 * 10

234 * 2 * 2 * 5

117 * 2 * 2 * 2 * 5

13 * 9 * 2 * 2 * 2 * 5

13 * 3 * 3 * 2 *2 * 2 * 5

23 * 32 * 5 * 13

Since exponents notate successive multiplications by the same number, we can use them to consolidate lists of prime factors and immediately see how many of each prime factor exists within a number like 4680.

13 * 3 * 3 * 2 *2 * 2 * 5

23 * 32 * 5 * 13

In this example, 23 is a factor of 4680, but 24 is not a factor of 4680 because 4680 has only three prime factors of 2. 32 is a factor of 4680, but 33 is not a factor of 4680 because 4680 has only two prime factors of 3. A GMAT/EA problem might ask something like this: “What is the greatest integer k for which 2k is a factor of 4680?” The answer to this problem would be 3.

Let’s take a look at an official problem:

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?

(A) 10

(B) 12

(C) 14

(D) 16

(E) 18

This question effectively asks, “How many prime factors of 3 are in the number p?” The number p is “the product of the integers from 1 to 30 inclusive.” This is the verbalization of a piece of notation called the factorial. Factorials notate the multiplication of a positive integer by each positive integer less than itself. Here are some examples:

4! = 4 * 3 * 2 * 1

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1

9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

To generalize the formula:

n! = n * (n – 1) * (n – 2) *  . . . * 1

Because the 1 at the end of the multiplication does not affect the value, factorials are often “spelled out” without the 1 at the end of the list.

Obviously, 30! Is too large of a number for us to evaluate. As a general rule, finding the actual value of any factorial larger than 7! or so is a waste of time. We are almost always more interested in the factors contained within a factorial.

30! = 30 * 29 * 28 * . . . * 1

So how many prime factors of 3 are contained within (30!)? Well, numbers like 29 and 28 don’t contain any prime factors of 3, because these numbers are not multiples of 3. The only factors of 30! that contain prime factors of 3 are the multiples of 3.

30 27 24 21 18 15 12 9 6 3

Some of these factors themselves contain more than one prime factor of 3. 27 is 33, so it has 3 threes. 9 is 32, so it has 2 threes. And a sneaky one is 18, which is not a power of 3 but is 2 * 9, or 2 * 32. So 18 also has 2 threes.

We can tally up all our prime factors of 3 like this:

30 27 24 21 18 15 12 9 6 3

1 3 1 1 2 1 1 2 1 1

If we add up the values in the second row, we obtain an answer of 14 prime factors of 3 in the number (30!). The correct answer is C.

This approach works, but there is actually a more straightforward way involving division. Consider first that performing the division a/b is effectively a way of counting how many multiples of b are less than or equal to a. 80 / 10 = 8, so 8 multiples of 10 are less than or equal to 80. 70 / 14 = 5, so 5 multiples of 14 are less than or equal to 70. When a is not perfectly divisible by b, the quotient still gives the number of multiples of b that are less than or equal to a, and the remainder is irrelevant. 56 / 9 = 6 remainder 2, so there are 6 multiples of 9 less than or equal to 56).

How does this help us identify prime factors of 3 in a number like (30!)? Well, if we take the quotient of 30/3, we can see how many of the factors of 30! are multiples of 3. In this case, there are 10. So there are at least 10 prime factors of 3 in the number (30!). If we take the quotient of 30/9 (or 30/(32), we see that 3 of these 10 multiples of 3 are also multiples of 9. Each of these numbers (9, 18, and 27) contains another prime factor of 3 that we did not “count” when we performed the division 30/3. Finally, we need to take 30/27 (or 30/33) to see that one of our 3 multiples of 9 is also a multiple of 27. The number, of course, is 27 itself, which contains a third prime factor of 3 that we didn’t “count” when we performed the divisions 30/3 and 30/9. 

quotient of 30 / 3 = 10

quotient of 30 / 9 = 3

quotient of 30 / 27 = 1

10 + 3 + 1 = 14 prime factors of 3 within 30!

To find the highest integer k for which nk is a factor of t!, add up the quotients of t/n, t/(n2), t/(n3) . . . for every power of n that is less than or equal to t.

 

Here’s one that requires some extra steps of algebra:

If n = 9! – 64, which of the following is the greatest integer k such that 3k is a factor of n?

(A) 1

(B) 3

(C) 4

(D) 6

(E) 8

This question effectively asks, “How many prime factors of 3 are in the number 9! – 64?

Let’s start by viewing 9! and 64 in factored forms:

9! – 64

(9 * 8 * 7 * 6 * 5 * 4 * 3 * 2) – (6 * 6 * 6 * 6)

We can go one step further and break all the non-primes down to primes:

(9 * 8 * 7 * 6 * 5 * 4 * 3 * 2) – (6 * 6 * 6 * 6)

(3 * 3) * 23 * 7 * (2 * 3) * 5 * 22 * 3 * 2 – (24 * 34)

Since each factor of 6 in 64 is equal to (3 * 2), we can represent the group of 4 sixes as (24 * 34). Now to consolidate the rest:

(3 * 3) * 23 * 7 * (2 * 3) * 5 * 22 * 3 * 2 – (24 * 34)

(27 * 34 * 5 * 7) – (24 * 34)

The subtraction is preventing us from seeing how many threes are in the integer version of this number. We will have to extract a factor of (24 * 34), like this:

(27 * 34 * 5 * 7) – (24 * 34)

(24 * 34)[(23 * 5 * 7) – 1]

Now we can evaluate the bracketed expression in order to prime factorize again:

(24 * 34)[(23 * 5 * 7) – 1]

(24 * 34)[(8 * 5 * 7) – 1]

(24 * 34)[280 – 1]

(24 * 34)(279)

And we’ll break 279 down into prime factors:

(24 * 34)(279)

(24 * 34)(9 * 31)

(24 * 34)(32 * 31)

24 * 36 * 31

Finally we can see that there are 6 prime factors of 3 in the number 9! – 64, and the correct answer is D.

Here’s a final problem for today:

For the positive integers a, b, and k, ak|  |b means that ak is a divisor of b but a(k+1) is not a divisor of b. If k is a positive integer and 2k|  |72, then k is equal to

(A) 2

(B) 3

(C) 4

(E) 8

(D) 18

This problem uses a random symbol – two vertical lines – and then defines it for you. This is common practice on GMAT/EA problems and is simply a way to “disguise” the type of problem you’re looking at. Think about the way they’ve defined this symbol, and you’ll see that this problem is asking the same thing as our other examples: what is the greatest integer k for which 2k is a factor of 72? Or to put it even more simply: How many prime factors of 2 does 72 have?

72 = 8 * 9 = 23 * 32

There are 3 prime factors of 2 in 72, so the correct answer is B.

This concludes our brief study of the “power of k” problems on the GMAT/EA. Join us next time for the final article in our series on exponents to learn how to handle problems with impossibly large numbers like 287459.

If you are in the middle of studying for the GMAT/EA and are looking for a private GMAT tutor, our elite tutors have all scored over 770 on the GMAT and have years of professional experience with tutoring. You can meet with us for a 30-minute complimentary consultation call.

Contributor: Elijah Mize (Apex GMAT Instructor)

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Zeros and Nonzeros
Posted on
21
Sep 2022

Zeros and Nonzeros

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/(29+53) 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/(23+57) 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 / (29 * 53)

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 / (29 * 53)

1 / 26 * (23 * 53)

1 / 26 * 103

Three of our twos went over and joined the fives to make three tens, or 103.

Now we have 1 / (26 * 103), 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 / 26 * 103

1 / 64 * 103

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/(23+57) 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 / (23 * 57)

1 / (54 * 103)

Again, knowing your powers helps:

1 / (54 * 103)

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)

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“Almost an Integer” Problems
Posted on
20
Sep 2022

“Almost an Integer” Problems

Unless you do the math as a career or a hobby, you probably prefer integers to non-integers. Whole numbers are easier for us to conceptualize. But a certain class of GMAT/Executive Assessment (EA) problems involves numbers that are almost integers. Generally, this nearest integer is 1, so the nearby numbers look like this:

0.99999

0.9995

1.001

1.000006

Whenever you see a number like one of these on a GMAT/EA problem, you should use powers of 10 to notate the difference between the value in question and the nearest integer.

Let’s start by expressing our first example number, 0.99999, as a difference from 1 without powers of 10. Then we’ll convert that difference to a power of 10.

0.99999 = 1 – 0.00001

When the number you’re working with is of the form 0.999 . . . , the difference from 1 is all zeros with a 1 at the end, and the number of digits after the decimal remains consistent. Here there are five nines after the decimal, so our difference from 1 has four zeros and then a 1 at the end for a total of five digits after the decimal.

Now to convert our difference from 1 to scientific notation. The number in question, 0.00001, is small, so the power of 10 will be negative. The rule is to simply use the negative version of the number of digits after the decimal. According to this rule, 0.00001 = 10-5. Therefore we have this:

0.99999 = 1 – 0.00001 = 1 – 10-5

Let’s try the next example from above: 0.9995. This time there are only four digits after the decimal, but the last one is a 5 instead of a 9. Again, let’s express this value as a difference from 1 without scientific notation first:

0.9995 = 1 – 0.0005

As before, the number of digits after the decimal must remain consistent. But instead of using all zeros and then a single 1, we use all zeros and then whatever digit sums to 10 with the final digit of the original number. When the final digit of the original number is a 9, as in the first example, we use a 1 (since 9 + 1 = 10). In this case, the final digit of the original number is 5, so we need to use another 5 (5 + 5 = 10) to finish off our difference from 1.

0.9995 = 1 – 0.0005 = 1 – 5*10-4

There are four digits after the decimal, so the exponent of the 10 is -4. The coefficient of 5 is applied to the 10-4 term because 0.0005 = 5 * 0.0001 = 5*10-4.

We can solidify this with a general rule for finding decimal differences from 1. The number of digits after the decimal must remain consistent, and the digits in each place must sum to 9, except for the final digits which sum to 10. Here’s an example:

0.8653 = 1 – 0.1347

Here are the sums of the tenths, hundredths, thousandths, and ten-thousandths digits:

8 + 1 = 9

6 + 3 = 9

5 + 4 = 9

3 + 7 = 10

Here are some numbers you can use for practice. Their differences from 1 are at the end of the article.

0.23468

0.9834

0.31479

0.34098

0.999357

0.00042

0.000257

This covers numbers slightly less than 1. Numbers slightly greater than 1, like the examples from before of 1.001 and 1.000006, are easier to work with because you can convert everything after the decimal directly to a power of 10 without having to find a difference from 1.

1.001 = 1 + 10-3

1.000006 = 1 + 6*10-6

With these skills in place, you’re ready to tackle some official problems.

(1.00001)(0.99999) – (1.00002)(0.99998) =

(A) 0

(B) 10-10

(C) 3(10-10)

(D) 10-5

(E) 3(10-5)

Let’s convert each of the four numbers in the problem:

1.00001 = 1 + 10-5

0.99999 = 1 – 10-5

1.00002 = 1 + 2*10-5

0.99998 = 1 – 2*10-5

And we have a lovely pattern emerging.

(1 + 10-5)(1 – 10-5) – (1 + 2*10-5)(1 – 2*10-5)

Now all that remains is to “foil” the expressions and then simplify:

(1 + 10-5)(1 – 10-5) – (1 + 2*10-5)(1 – 2*10-5)

(1 – 10-5 + 10-5 – 10-10) – (1 – 2*10-5 + 2*10-5 – 4*10-10)

Now all the 10-5 terms cancel:

(1 – 10-10) – (1 – 4*10-10)

Now the 1s cancel as well:

-(10-10) + 4(10-10)

3(10-10)

And the correct answer is C.

Here’s another:

1 – 0.00001 = 

(A) (1.01)(0.99)

(B) (1.11)(0.99)

(C) (1.001)(0.999)

(D) (1.111)(0.999)

(E) (1.0101)(0.0909)

This one is different because the first step has already been done for us. Instead of starting with 0.999999, the problem starts in “difference from 1” form. All we have to do is convert the difference to a power of 10:

0.000001 = 10-6

1 – 0.000001 = 1 – 10-6

Now what to make of the answer choices? After a quick scan, the only ones that look very friendly to a “1 +/- 10-x” form are A and C. Choice A can’t be right because the factors 1.01 and 0.99 contain a total of only four digits after their decimals, and the product we are looking for, 1 – 10-6 or 0.999999, has six digits after the decimal. In multiplication, the number of digits after the decimal in the product always matches the total number of digits after the decimals in the factors.

So answer choice C looks like the best candidate. Let’s convert it to “power of 10” form:

(1.001)(0.999) = (1 + 10-3)(1 – 10-3)

Now we can “foil” the expression and simplify:

(1 + 10-3)(1 – 10-3) = 1 + 103 – 103 – 10-6 = 1 – 10-6

And as we suspected, answer choice C turned out to be correct.

Now you’re ready to handle “almost integers” on GMAT/EA problems. Next time we’ll use powers of 10 to address problems that ask about zeros or nonzero digits.

If you are looking for professional help to boost your GMAT/EA performance, head to our official website and book your 30 minutes complimentary assessment session now with our top tutors.

Solutions to drills:

0.23468 = 1 – 0.76532

0.9834 = 1 – 0.0166

0.31479 = 1 – 0.68521

0.34098 = 1 – 65912

0.999357 = 1 – 0.000643

0.00042 = 1 – 0.99958

0.000257 = 1 – 0.999743

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Posted on
16
Aug 2022

Understanding Exponents

A key piece of algebraic notation on GMAT and Executive Assessment (EA)  quant problems is the exponent. Exponents appear on many kinds of quantitative problems, so fluency with exponents (and radicals) is an indispensable skill for achieving a competitive quant score. Odds are, you already have some idea of what exponents “mean” in algebraic language, but let’s clarify your definition by exploring how exponents relate to the more fundamental operations of addition and multiplication.

The foundations of all arithmetic are the operations of addition and subtraction. We could even say just addition, since subtraction can be notated as the addition of a negative value. Why have I left out multiplication and division? Well, because multiplication is nothing but an efficient way to notate a special case of addition, and division is nothing but multiplication in reverse. The special case of addition is this: when you want to add up a large number of groups that are all the same size. Let’s say you want to know how many eggs are in stock at your local grocery store. You won’t count the eggs one by one; you’ll count the cartons, since you know that each carton contains 10 or 12 eggs, depending if you’re in America or Europe. You could “show your work” for counting the eggs like this:

12 + 12 + 12 + 12 + 12 + 12 . . .

But this would get out of hand. Multiplication was created for just such a job. Instead of writing out the addition of 217 dozens of eggs, you can write this:

(# of eggs) = 217 * 12

This is many times better than stringing together 217 twelves with plus signs, but the outcome is the same.

The relevance to exponents is this: just as multiplication efficiently notates successive additions of the same value, exponents efficiently notate successive multiplications by the same value.

To stay in the realm of our “eggs at the grocery store” scenario, let’s imagine that a local farm starts out with 5 hens and wants to double its egg-laying workforce every year for the next 7 years. We could notate the target number of hens at the end of the seventh year like this:

5 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 640

But this is a bit impractical. In scenarios where the population of a bacteria doubles hundreds of times, notating with multiplication simply won’t do. We need a better tool, and the tool is exponents. Returning to our hen population example, exponents work like this:

5 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 640

5 * 2 = 640

Rather than stringing together seven 2s with multiplication signs, we can place a 7 as an exponent of the 2 to notate the same thing. When we do this, the 2 is called the base of the exponential expression. The value represented by x is called the nth power of x or “x to the nth power. In the latter option, “power” is often left tacit, so in our hen scenario, we would verbalize the value of the final population as “five times two to the seventh (power)” or, for a slight simplification, “five times two to the seven” The terms squared and cubed are used for exponents of 2 and 3, respectively. 6² is “six squared”; 5³ is “5 cubed.”

This proper understanding of exponents as shorthand for multiplication makes sense of their properties. Many an algebra student has been tripped up by expressions like this:

x * x

Seeing the multiplication sign, a novice might incorrectly infer that x * x = x²⁸ and be confused by the correction that the exponents should be added, not multiplied, yielding x¹¹. Breaking down the exponential expressions x and x to their “original” multiplicative forms should add clarity.

x⁷= x * x * x * x * x * x * x

x⁴ = x * x * x * x

x⁷ * x⁴ = (x * x * x * x * x * x * x) * (x * x * x * x) = x¹¹

Now we can see that the multiplication of the exponential expressions x and x is nothing but a chain of multiplications of the variable x: 11 of them, to be exact. And the best way to notate a string of eleven “x’s” in multiplication is with an exponent of 11.

The product of equal bases with different exponents is the base raised to the sum of the exponents. x * x = xᵃ⁺ᵇ

With this rule in place, it follows that it can be reversed by “splitting” an exponential expression into two groups.

x¹¹ = x⁷ * x⁴ = x⁶ * x⁵ = x¹⁰ * x

As shown, the way you “split” your expression is flexible. Different algebraic scenarios can benefit from different “splits.”

When the operation is division instead of multiplication, the resulting exponent is calculated via subtraction instead of addition.

x⁷ / x⁴ = (x * x * x * x * x * x * x) / (x * x * x * x)

(x * x * x * x * x * x * x) / (x * x * x * x) = x³

The quotient of equal bases with different exponents is the base raised to the difference of the exponents. x / x = x⁽ᵃ⁻ᵇ⁾

This covers multiplication and division of equal bases with different exponents. Simple rules also exist for multiplication and division of different bases with equal exponents.

x³ * y³ = (xy)³

Again, breaking down the exponential expressions to their “original” multiplicative forms shows why this works:

x³ * y³ = (x * x * x) * (y * y * y)

Everything here is in multiplication, so we can reorder and regroup the factors any way we like.

(x * x * x) * (y * y * y) = xy * xy * xy = (xy)³

Don’t forget the parentheses around your base. Note that we need the parentheses to group the xy as a unit, as opposed to xy³ = x * y*y*y.

Of course, this rule works in reverse as well.

(xy)³ = x³ * y³

Again, different algebraic scenarios call for different algebraic solutions. Both the combination and the “splitting” of algebraic expressions are useful tools in different contexts.

xᵃ * yᵃ = (xy)ᵃ

(xy)ᵃ = xᵃ* yᵃ

As you probably guessed, the same rule applies for division.

x³/ y³ = (x/y)³

And the “proof:”

x³ / y³ = (x * x * x) / (y * y * y) = (x/y) * (x/y) * (x/y) = (x/y)³

And finally, the generalized form of the rule, accompanied by the reversal:

xᵃ / yᵃ = (x/y)ᵃ

(x/y)ᵃ = xᵃ/ yᵃ

One more rule remains to be covered in this introduction. To preview it, let’s return to our idea of “splitting” an exponential expression into pieces:

x¹¹ = x⁷ * x⁴ = x⁶ * x⁵ = x¹⁰ * x

No one said that we have to limit ourselves to two “pieces.” We can keep “splitting” as many times as we want.

x¹¹ = x⁸ * x³ = x⁴ * x⁴ * x³ = x² * x² * x² * x² * x³

Here we see an x³ term multiplied by a string of four “x²” terms. But isn’t there a more efficient way to notate such a string of multiplications? Yes, with exponents! An exponential expression itself can become the base of another exponent.

x³ * x² * x² * x² * x² = x³ * (x²)⁴

Remember that our (x²) term started out as x. This reveals the rule for simplifying “nested” exponential expressions, or what we call a “power to a power”:

(xᵃ)ᵇ = xᵃ * ᵇ

xᵃ * ᵇ = (xᵃ)ᵇ

This rule makes sense when you know that exponential expressions are “made of” successive multiplications. Four groups of two “x’s” in multiplication is the same thing as 8 “x’s” in multiplication. And you know the drill: the reversal of the rule – where a single exponent is factored to create a “nested” expression – is just as useful as the “original” version.

Now to assemble all of our rules:

xᵃ * xᵇ = x⁽ᵃ⁺ᵇ⁾

x⁽ᵃ⁺ᵇ⁾ = xᵃ + xᵇ

xᵃ / xᵇ = x⁽ᵃ⁻ᵇ⁾

x⁽ᵃ⁻ᵇ⁾ = xᵃ/ xᵇ

xᵃ * yᵃ = (xy)ᵃ

(xy)ᵃ = xᵃ * yᵃ

xᵃ / yᵃ = (x/y)ᵃ

(x/y)ᵃ = xᵃ/ yᵃ

(xᵃ)ᵇ = x⁽ᵃᵇ⁾

x⁽ᵃᵇ⁾ = (xᵃ)ᵇ

And before we try a few official GMAT problems, let’s take a look at some powers of integers you should know:


The main reason for knowing these powers is for something I call “backwards recognition.” If you don’t memorize these and you need to evaluate 54 or 27 in order to solve a problem, you can probably multiply your way through the powers easily enough. But it’s another thing to see 625 or 128 in a problem and immediately know “that’s 54” or “that’s 27.” Such backwards recognition can help you make sense of problems that may look confusing at first.

As a final reminder of the “power to a power” rule, powers of 4 are left out of this list because they are contained within the powers of 2. Every even power of 2 is also a power of 4. For example, 16 = 24 = 42, 64 = 26 = 43, so on and so forth.

Let’s try some official GMAT problems involving exponents. Here’s a simple one to get you started:

216 is 

  1. 2 more than 2¹⁵
  2. 16 more than 2¹⁵
  3. ½ of 2³²
  4. 2 times 2
  5. 2 times 2¹⁵

If you’ve gotten your head around exponents, this is a 15-second problem. Notice that answer choices C and D both imply the same incorrect rule: that doubling the exponent on the 2 doubles the overall value. Even if you don’t know your exponent rules, you could eliminate these answer choices on logic alone because they are indistinguishable. Answer B doesn’t make much sense, and answer A confuses the rules of exponents with the rules of multiplication. (2 * 16) is 2 more than (2 * 15), but the difference between 216 and 215 is much greater. Answer E gets it right. Increasing the value of the exponent by 1 means to multiply by the base one more time.

Here’s another:

At the start of an experiment, a certain population consisted of 3 animals. At the end of each month after the start of the experiment, the population was double its size at the beginning of that month. Which of the following represents the population size at the end of the 10 months?

  1. 2³
  2. 3²
  3. 2 * 3¹⁰
  4. 3 * 2¹⁰
  5. 3 * 10²

If you understood our example with the hens, this should be another easy one. We need to start with 3 and then double our value (multiply by 2) 10 times. Exponents enable us to notate this series of multiplications as 3 * 2¹⁰, answer choice D.

Now for some data sufficiency:

What is the value of  6ʸ ?

  1. 2⁽ˣ⁺ʸ⁾ = 32
  2. 3⁽ˣ⁺ʸ⁾ = 243

If you know your exponent rules, you should immediately note that 6x6y can be alternatively written as 6ˣ⁺ʸ. Since this is DS, you don’t necessarily need to know which power of 2 equals 32 or which power of 3 equals 243; it’s enough to recognize that each statement on its own locks in the value of the exponent “x + y” and therefore the value of the 6ˣ⁺ʸ expression that we were asked about. (But ideally you will study your powers enough to know right away that 32 = 2 and 243 = 3.) The correct answer choice is D.

Here’s our final problem for this article:

What is the smallest integer n for which 25ⁿ > 5¹²?

  1. 6
  2. 7
  3. 8
  4. 9
  5. 10

This is a case where backwards recognition of powers makes all the difference. 25 should be immediately recognizable as 52. Therefore the given inequality can be rewritten in either of the following ways:

(5²) > 5¹²

25 > 5⁽²×⁶⁾ . . . 25 > (5²) . . . 25 > 25

The second way gets you to the correct answer more quickly, but the first is rather more intuitive. The “power to a power” rule states that the exponents should be multiplied:

(5²) > 5¹²

> 5¹²

Plugging in a 6 makes the exponents equal (2 * 6 = 12), and since the bases are now equal, the expressions would be equal as well. Therefore the smallest integer that works is 7, answer choice B.

I hope you’ve enjoyed this intro to (or review of) exponent properties. As you will definitely come across exponents on the GMAT test, we think that this article is something you need to read carefully! Next time we’ll learn how to “undo” exponents with an inverse operation.

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