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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Exponent Properties in Data Sufficiency
Posted on
08
Sep 2022

Exponent Properties in Data Sufficiency

Exponents have properties that make them ideal pieces for data sufficiency problems on the GMAT and Executive Assessment (EA) quant sections. We’ve surveyed most of these properties in our first four articles in this series, but a couple of key ones haven’t been mentioned explicitly.

First, x0 = 1. A full mathematical explanation of this property is beyond the scope of this series of articles and is unnecessary for GMAT/EA preparation. If you know the rule, you can employ it as needed. The debated exception 00 exists but does not occur on GMAT quant.

Second, the numbers 0 and 1 are both “immune” to exponents. We mentioned this rule for 1 in our last article, but it is important to know that the rule applies to 0 as well. 0x = 0, and 1x = 1.

Another important rule for exponent DS problems has already been mentioned: even powers of negative numbers are positive, and odd powers of negative numbers are odd. To extend this, xeven is always greater than or equal to 0 (it is only equal to 0 when x = 0; in all other cases xeven is positive).

A basic rule to remember is that positive numbers have two square roots which are negative and positive versions of the same value. If b2 = 16, b may equal 4 or -4. Forgetting about the possibility of the negative square root usually leads to incorrect answers.

One more important rule to remember is that if x is positive, then xy is positive. No exponent can cause an exponential expression with a positive base to have a negative value or a value of 0. To state the same rule differently, exponential expressions with negative values have negative bases.

Let’s get into some official problems:

Is zp negative?

  1. pz4 < 0
  2. p + z4 = 14

We will need to know the signs of the variables p and z. If they are both positive or both negative, then the product zp is positive. If z or p is negative and the other positive, then the product zp is negative.

Statement 1 tells us that the product p * z4 is negative. Therefore either of p or z4 is negative and the other is positive. Since the exponent 4 is positive, z4 is always positive (or 0, but the statement rules out that possibility). This means that p must be negative. This isn’t enough to answer whether zp is negative, but it may be useful if we have to combine statements 1 and 2.

Statement 2 tells us that the sum of p and z4 equals 14. There are simply too many possibilities for this statement on its own to be sufficient.

Now we must combine statements 1 and 2. We know from statement 1 that p is negative. Therefore z4 must be positive in order for p + z4 to have the positive sum of 14. But we already know that z4 is positive because the exponent 4 is even! A common mistake in these DS problems is to accidentally transfer information about z4 back to the variable z itself. Doing so on this problem might lead you to incorrect answer choice C. In this case, statements 1 and 2 together are still not sufficient, and the correct answer is E.

Let’s try another:

If r, s and t are nonzero integers is r5s3t4 negative?

  • rt is negative.
  • s is negative.

Here’s another even exponent of 4. This means that t4 is either positive, or 0 in the case of t = 0. If t = 0, then the product r5s3t4 is 0 and not negative.

Statement 1 tells us that the product rt is negative. Therefore either of r or t is negative and the other is positive. We still know nothing about the variable s, so this can’t be sufficient on its own.

Statement 2 tells us that s, and therefore s3, is negative, but it says nothing about r or t. This can’t be sufficient on its own.

Combining the statements, we can approach statement 1 with two pieces of info: that t4 is positive (or 0) and that s3 is negative (since s is negative). Therefore the product r5s3t4 is (?) * (-) * (+), and the sign of the variable r makes all the difference.

Returning to statement 1, we consider again that the product rt is negative. This means that either r or t is negative, but unfortunately, we don’t know which one. Again, if we mistakenly jump from the fact “t4 is positive” to the unfounded conclusion “t is positive,” we will select incorrect answer choice C. Again, the statements together are insufficient, and the correct answer is E.

Here’s another:

If a and b are integers is a5< 4b?

  1. a3= -27
  2. b2= 16

The upshot of statement 1 is that a is negative. Therefore a5 is also negative, since 5 is another negative exponent. To get specific (which is probably unnecessary for the problem), a = -3 and a5 = -243.

We don’t know anything about b, so it’s tempting to conclude that statement 1 alone is insufficient. But we are asked to compare a5 against 4b, not against b itself. If you remember the rule that exponential expressions with positive bases have positive values, you’ll see that statement 1 alone is sufficient. a5 is negative, and 4b is positive.

Statement 2 tells us that b = 4 or -4. Therefore 4b = 256 or 1/256. Either way, without statement 1, we know nothing about a5. Statement 2 on its own is insufficient, and the correct answer is A.

Here’s a final DS exponents problem:

If x and y are integers, is x > y?

  1. x + y > 0
  2. yx < 0

Statement 1 is certainly insufficient by itself, but we should still think about what it tells us. Either x and y are both positive, or one is negative and the other is positive, with the positive number having the greater absolute value.

Statement 2 tells us that yx is negative. The rule we need is that exponential expressions with negative values have negative bases. This means y is negative.  On its own, this statement is still insufficient. But combined with statement 1, we know that x must be greater than y, because x must be positive in order to produce a positive sum with a negative number y. The statements together are sufficient, and the correct answer is C. 

We are halfway through our series on exponents and have covered all the basics. The remaining five articles will cover specific problem types involving exponents.

If you are in the middle of studying for the GMAT 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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Posted on
07
Sep 2022

Bases Between -1 and 1

Many GMAT and Executive Assessment (EA) exponent problems – especially data sufficiency ones – require you to consider fractional bases. By this I mean proper fractions with values between -1 and 1, not improper fractions whose numerators exceed their denominators.

There are four “kinds” of bases separated by the three “boundary points” of -1, 0, and 1. Numbers in each of the four “zones” separated by these values behave similarly as bases of exponents. On many DS problems, we need to consider numbers less than -1, negative fractions, positive fractions, and numbers greater than 1, as well as the boundary points of -1, 0, and 1. This sounds like a lot of work, but practice will build your “spidey sense” for when a certain kind of base leads to an exception and an insufficient statement.

Let’s handle the “boundary points” first. As mentioned in a prior article, the boundary point of 1 is simple because it is “immune” to exponents: 1anything = 1. Likewise, 0positive = 0. For -1, (-1)even = 1 and (-1)odd = -1. Since (-1)anything equals either 1 or -1, it is worth noting that the absolute value of -1anything is 1. If you are unfamiliar with absolute value, don’t worry, it’s a simple concept: absolute value is a number’s distance from 0 on a number line. The absolute value of a positive number is . . . itself. The absolute value of a negative number is simply the positive version of the number. For positive numbers, “normal” value and absolute value are the same and trend together. For negative numbers, absolute value increases as “normal” value decreases. Absolute value is notated with vertical lines on either side of a value, variable, or expression. |(-1)anything| = 1.

Now for the four ranges of numbers. Numbers greater than 1 are the simplest. For these numbers, the higher the exponent, the higher the overall value. For numbers greater than 1, higher powers have higher values.  Numbers less than -1 are only slightly more complex. For numbers less than -1, higher powers have higher absolute values, but odd powers are negative and even powers are positive.

Now for the positive and negative fractions: the more times you multiply a fraction by itself, the closer the resulting value gets to 0. (¾)2 or ¾ * ¾, which we can read as “¾ of ¾,” is less than ¾. For positive fractions, higher powers have lower values. For negative fractions, higher powers have lower absolute values, but odd powers are negative and even powers are positive.

Let’s demonstrate all of our rules with the examples of 2, -2, ½, and -½.

2 < 22 < 23 . . .

|2| < |22| < |23| . . .

 

(-2)5 < (-2)3 < (-2) < 0 < 22 < 24 < 26 . . .

|-2| < |(-2)2| < |(-2)3| . . .

 

½ > (½)2 > (½)3 . . .

|½| > |(½)2| > |(½)3| . . .

 

-½ < (-½)3 < (-½)5 < 0 < (-½)6 < (-½)4 < (-½)2

|-½| > |(-½)2| > |(-½)3| . . .

The patterns for the negatives can take some getting used to, so study these rules frequently and, more importantly, build your fluency with practice problems! Here’s a straightforward one:

Is x2 greater than?

  1. x2 is greater than 1.
  2. x is greater than -1.

For statement 1, it helps to remember that only numbers greater than 1 or less than -1 can have powers greater than 1. Powers of fractions are always fractions. So if x2 is greater than 1, x is either greater than 1 or less than -1. If x is less than -1, then x2, which according to the statement is greater than 1, is greater than x. And if x is greater than 1, it still gets larger when it is squared, so x2 is always greater than x, and statement 1 is sufficient.

Statement 2 tells us that x is greater than -1. If we remember our boundary points, we can solve this one without having to think about fractions. X could be 0 or 1, and in either case, x2 is equal to, not greater than, x. But for any number greater than 1, x2 is greater than x. So statement 2 on its own is insufficient, and the correct answer is A.

Let’s try another DS problem:

Is xy > x2y2?

  1. 0 < x2 < 1/4
  2. 0 <  y2 < 1/9

To verbalize the question: is the absolute value of the product xy greater than the square of the product xy?

Statement 1 tells us that x2 is a positive fraction, which means that x itself is a fraction with a greater absolute value, but we don’t know whether it is positive or negative. Without knowing anything about y, this isn’t enough. Statement 1 alone is insufficient. Statement 2 is similar and also insufficient. Taking the statements together, we know that the absolute value of x is greater than x2, and the absolute value of y is greater than y2. It follows that the absolute value of the product xy is greater than x2y2. We don’t know whether x and y are positive or negative, but we’re talking absolute value so it doesn’t matter. The statements together are sufficient, and the correct answer is C.

Finally, let’s see what happens when fractional bases meet negative exponents:

(1/2)-3(1/4)-2(1/16)-1=

A. (1/2)(-48)

B. (1/2)(-11)

C. (1/2)(-6)

D. (1/8)(-11)

E. (1/8)(-6)

This problem benefits from “translating” out of the negative-exponent form. (½)-3 = 23, (¼)-2 = 42, and (1/16)-1 = 16. Ideally, you recognize that everything in the expression can be converted to powers of 2. 42 and 16 both equal 24, so the full expression is 23 * 24 * 24, or, remembering your exponent rules, 211. If you noticed how the fractional base and the negative exponent “canceled” each other, you should recognize 211 as (½)-11, answer choice B.

If you are looking for extra help in preparing for the GMAT, we offer extensive one-on-one GMAT tutoring for high-achieving students. You can schedule a complimentary, 30-minute consultation call with one of our tutors to learn more! 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Negative Exponents and Negative Bases_
Posted on
30
Aug 2022

Negative Exponents and Negative Bases

Welcome back to our series on exponents. Today we will see what happens when we throw negatives into our exponential expressions. We will explore both negative bases and negative exponents.

First, the bases. The rule to remember for negative bases is that odd powers of negative bases are negative and even powers of negative bases are positive. This rule makes sense when you remember that exponents simply notate a number of multiplications by the base (and remember your rules about multiplication with negative factors). 

Multiplication with an odd number of negative factors yields a negative product: 

(-x)3 = -x * -x * -x

(-x)3 < 0

Multiplication with an even number of negative factors yields a positive product:

(-x)4 = -x * -x * -x * -x

(-x)4 > 0

A note on notation: parentheses should always be used around a negative value as the base of an exponent. If they are not, then the order of operations dictates that the exponent be applied before the negative sign. To avoid confusion, whenever the negative is meant to be left out of the exponential operation, parentheses are used like this: -(x)4 to make the order clear. Please note that -(x)4 is less than 0, and (-x)4 is greater than 0.

Now for negative exponents. Here is a simple rule is the best way to explain it:

x-n = 1/xn

A negative exponent indicates a value reciprocal to the value with a positive exponent. It’s good practice to “translate” any exponential expressions with negative exponents to their reciprocal positive forms. Seeing it both ways can help you make sense of problems.

5-4 = 1 / (54) = 1 / 625

17-2 = 1 / (172) = 1 / 289

(9 / 16)-2 = (16 / 9)2 = 256 / 81

Integer bases with negative exponents go under a numerator of 1; fractional bases with negative exponents simply flip. Let’s look at some “double negative” exponential expressions.

(-6)-3 = 1 / (-6)3 = 1 / -216

(-2)-10 = 1 / (210) = 1 / 1024

(-4 / 3)-4 = (-¾)4 = 81 / 256

Let’s get into some official GMAT problems. Be careful with this first one!

From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

  1. (-10)20
  2. (-10)10
  3. 0
  4. -(10)19
  5. -(10)20

One incorrect answer is chosen far more often than any other on this problem: answer choice D. Trying to minimize the product, many people consider taking the maximum number (20) of the lowest value (-10). The common mistake is then thinking that -(10)20 (answer choice E) is actually a positive value since it involves an even number (20) of negative factors. Many people then take “the next best thing” in answer choice D, which shifts the exponent to the next odd number down from 20.

In fact, -(10)20 does not involve an even number of negative factors, since the negative sign is excluded from the exponential expression by the parentheses. Answer E means “take 1020 and make it negative.” It is true that taking 20 negative tens and multiplying them all together produces a large positive value (the opposite of what we are aiming for on this problem), but this misguided idea is notated by answer choice A – not answer choice E. Remember that the answer choices notate the product of the 20 factors, not necessarily a condensed list of the 20 factors. 

It is possible to choose 20 integers from -10 to 10 inclusive that, when multiplied, yield a product of -(10)20 (answer choice E). The “least possible value” is obtained by finding the greatest absolute value (distance from 0) in negative form. So we want all 20 of our factors to be either 10 or -10 since this will maximize the absolute value (distance from 0) of the product. To ensure that the product is also negative, we simply need an odd number of negative tens. We can use nineteen negative 10s and 1 positive 10, 1 negative 10 and 19 positive 10s, or any odd combination in between. Any of these options will yield a product of -(10)20. Read the notation carefully!

Let’s try another:

The value of 2(-14)+2(-15)+2(-16)+2(-17)/5 is how many time the value of 2(-17)?

  1. 3/2
  2. 5/2
  3. 3
  4. 4
  5. 5

This problem benefits from the skill of noticing patterns and “checking” them. You should see the pattern in the numerator and generalize it by saying “the negative exponent on the 2 keeps decreasing by 1.” Then you can see how this pattern “works” by checking a single case. 

2-17 = 1 / 217

2-16 = 1 / 216

Since 217 = 2 * 216, (1 / 217) is half the value of (1 / 216). Or, to say it a more useful way, 2-16 = 2 * 2-17 This pattern will continue through the numerator. Since we are looking for how many copies of 2-17 we have in this expression, we can replace 2-17 with 1 and follow the pattern.

(214 + 2-15 + 2-16 + 2-17) / 3

(8 + 4 + 2 + 1) / 3

15 / 3 = 3

And the correct answer is C.

Here’s a final problem for today:

a is a nonzero integer. Is 

a2greater than 1?

  • a < -1
  • a is even.

To evaluate statement 1, simply start by checking a = -2. (-2)-2 = 1 / (-2)2  = ¼. Moving on to a = -3, (-3)-3 = 1 / (-3)3 = 1 / -27. This time the value is negative, but the positive value from a = -2 is still less than 1. If you imagine continuing with a = -4, a = -5, etc., you will just keep making smaller and smaller fractions. Statement 1 alone is sufficient.

Statement 2 on its own is easy to check since we already know from checking statement 1 that some even values for the variable a yield an aa with a value less than 1. And it shouldn’t be hard to imagine an even value for variable a where aa is greater than 1. For example, 22 = 4 and 44 = 256. So statement 2 alone is insufficient, and the correct answer is A.

If you went with answer choice C, here’s what might have happened. Noticing that statement 1 tells you the base is negative, you might have seen next that statement 2 tells you the exponent is even. You might have thought that this “evenness” of the exponent makes the difference since it determines the positivity or negativity of the expression. Very often in DS problems with negative bases, the even/odd identity of the exponent really matters. But in this case, it’s a trap, because we were asked whether aa is greater than 1 (not 0), and the fact that the exponent is also negative means that it’s even/odd identity is irrelevant – the value is always less than 1.

The rules governing negative exponents and negative bases are simple, but the GMAT and EA problems that employ these rules can catch you if you aren’t careful. Next time we will look at another tricky exponential scenario: when the base is between -1 and 1.

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Contributor: Elijah Mize (Apex GMAT Instructor)

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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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