The post GRE Exponents: Basics & Exponent Practice Question Set appeared first on Magoosh GRE Blog.

]]>In this post, we’ll take a look at GRE exponent basics (yes, that’s the stuff you learned—and, if you’re like most of us, probably forgot—back in high school). Then, we’ll do a whole lot of GRE exponent practice. Consider this your own private GRE tutoring session about powers! Let’s dive in.

- GRE Exponent Rules and Basics
- GRE Exponent Practice Questions
- GRE Exponent Practice Answers and Explanations

We have practice questions for each question type. If you’re already familiar with exponent rules, use this table to jump directly to a practice set.

Question Type | Questions | Answers/Explanations |
---|---|---|

Adding/Subtracting Exponents | Questions | Answers and Explanations |

Multiplying/Dividing Exponents | Questions | Answers and Explanations |

Exponents Raised to Exponents | Questions | Answers and Explanations |

Bonus Exponent Problems | Questions | Answers and Explanations |

Before any other exponent rules, you need to know about the base. You’ll see this in all GRE problems that involve powers. The base is the number underneath the exponent. In , two is the base and three is the exponent. means that you are multiplying (the base) three times: .

You’ll need to know about the base for every GRE problem involving powers. Other than that, it’s a matter of understanding which arithmetic rules to use in which scenario–after all, at the end of the day, powers are part of GRE Arithmetic, which is one of the core concept areas tested in GRE Quant!

What happens when you are adding similar bases? Can you just add the powers? For example:

Can we add the and the ? The answer is an unequivocal no. The only way you can change the above number is by factoring: . This last step is relatively advanced, and I wouldn’t worry about it too much unless you are going for a high score. Just remember, **if you are adding the bases—whether the same or different—then you cannot add the powers**.

Takeaways about adding exponents:

- You cannot add exponents when adding bases
- It doesn’t make a difference whether bases are the same or different—you can’t add them!
- To solve problems that require adding exponents, you need to factor

Everything I just said about adding powers? **It’s also true for subtracting them.** If you have a hard time remembering this, just keep in mind that subtracting is the same thing as adding a negative number.

To prove this, let’s take a look at a variation on the previous example.

The answer is NOT ! Instead, what you need to do is factor:

Takeaways about subtracting exponents:

- You cannot subtract exponents when subtracting bases
- It doesn’t make a difference whether bases are the same or different—you can’t subtract them!
- To solve problems that require subtracting exponents, you need to factor

The rule is as follows: **if the bases are the same and you are multiplying them, add the powers**. So . That is, I keep the base the same (I don’t multiply it) and then I add the powers.

What about ? Well, the bases are different so we cannot combine them as we did before. Remember: **you can only multiply exponential terms when the bases are the same**. In this case, you would just have to multiply the long way, .

However (yup, just to make it a little more complicated!), there is one exception to this different base rule. **If the exponents are the same but the base is different, you can multiply the bases**. In , the bases are different but both are to the fifth power. In this case, we keep the power the same and we multiply the bases. Therefore, .

Takeaways about multiplying exponents:

- Add powers when multiplying terms with the same bases
- You can only multiply exponential terms with the same bases OR the same power

Just like adding and subtracting exponential terms, multiplying and dividing exponential terms go hand in hand–except that when you’re dividing, you subtract the powers. So . That is, I keep the base the same (I don’t multiply it) and then I subtract the powers.

Similarly, if you’re looking at ? The bases are different—you guessed it, we cannot combine them as we did before. Remember: **you can only divide exponential terms when the bases are the same**. In this case, you would just have to divide the long way, .

BUT, when the bases are different and the powers are the same, you can divide the bases. In , the bases are different but both are to the fifth power. In this case, we keep the power the same and we divide the bases. Therefore, .

Takeaways about dividing exponents:

- Subtract powers when dividing terms with the same bases
- You can only divide exponential terms with the same bases OR the same power

So, I know you’re all wondering: what happens when you take an exponent to an exponent?

?

In this case, you **multiply the powers while keeping the base**, 4, the same.

.

Therefore, not .

Takeaways about raising an exponent to an exponent:

- Keep the base the same
- Multiply the powers

Think you got it? See if you can answer the following exponent practice questions correctly!

*We’ve provided clickable radio buttons for you to select your answer as you go through these GRE exponent practice questions. This way, you can keep track of your answers and check your work at the end. However, please note that there’s no option to submit them!*

Reduce to its simplest form.

If , what is the value of ?

Reduce to its simplest form.

If , what is the value of y?

6

12

18

36

72

*, what is the value of n?*

-2

0

1

2

4

What is the sum of the digits of integer x, where x = ?

13

11

10

8

5

If then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

2

3

5

7

9

Reduce to its simplest form.

Check the answer here!

If , where n is a non-negative integer, what is the greatest value of ?

1

5

32

64

(5/4)^{-n} < 16^{-1}

What is the least integer value of n?

Check the answer here!

The equation is true for how many unique integer values of n, where n is a prime number?

7

4

2

1

None of the above

Now that you’ve had some GRE exponent practice, check your answers to see how you did!

**Answer: **. Because you cannot mix bases, leave as is.

**Answer: C.** Do not simply add the powers together. You can only do that when you are multiplying powers, not adding them. Instead, notice how there are 4 of the same number, . Therefore, .

**Answer: E**.The best way to deal with this problem is by plugging in the answer choices. B, C, and D will all result in some funky irrational number because you are taking the square or cube root of the number. That leaves us with (A) -1 and (E) 1. Plugging in (A), you end up getting . Therefore, the answer has to be (E) 1. Remember, also, that any number taken to a negative exponent results in the reciprocal of that number, e.g. . And .

** Answer: **. Multiply the bases and leave the powers alone; .

**Answer: C**

**Answer: D**

**Answer: B**. Notice that can be rewritten as . We can now express as . The logic here is that . That is, 10 to any integer power greater than 1,will be a 1 followed by zeroes.

So now, let’s rewrite the problem again so we get . Combine and we get . That is, we get 1 followed by 13 zeroes. If you are taking the sum, it’s straightforward: 1 plus 13 zeroes is 1. We are not done yet as we have the . When you multiply this out, you get 128. 128 x 1,000 thirteen zeroes is equal to 128 followed by the thirteen zeroes. Ignore the zeroes and we get , which equals 11.

**Answer: D**

**Answer: **. First, add the powers inside the parenthetical to get . Then, combine both powers by multiplying.

**Answer: D** First, you must multiply the power in parentheses to the power outside the parentheses, giving you: . Remember, that does not add up to . If you factor from both and , you get: . Also, you can use approximation and eliminate those answers that are too small and too big. Answer choice E is far too big, and Answer choice A too small.

**Answer: B.** Don’t think big – think small. That is the smaller n becomes the greater ½^n becomes. So what is the smallest value? You may be tempted to say 1, which would give us ½. But remember n = 0, because . Therefore The “hidden zero,” as I like to call it, is a classic GRE math trick. So always keep your eyes open, especially when you see “non-negative integer,” which includes zero.

**Answer: 13.** The best place to start here is by getting rid of the unseemly negative signs and translating the equation as follows:

(4/5)^{n} < 1/16

A good little trick to learn using 4/5 taken to some power is that (4/5)^{3} = 64/125, which is slightly — but only slightly — greater than ½. Therefore, we can translate (4/5)^{3} to ½.

(1/2)^{4} = 1/16

That would make (4/5)^{12} a tad larger than 1/16. To make it less than 1/16 we would multiply by the final 4/5, giving us n = 13.

**Answer: D.** The most important piece of info is n is a prime number. So do not start by plugging in zero or one. Neither is a prime. The lowest prime is 2. When we plug in ‘2’ we get:

This is clearly true. Thus we have one instance.

As soon as we plug in other prime numbers a pattern emerges.

is always a negative number if n is odd. Because all of the primes greater than 2 are odd, the number in the middle will always be negative:

Because in each case n is a positive number we can never have the middle of the dual inequality be positive, if n is an odd prime.

Thus the only instance in which the inequality holds true is if we plug in ‘2’, the answer is (D).

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]]>The post GRE Division, Mixed Numerals, and Negatives appeared first on Magoosh GRE Blog.

]]>This is a post to clarify a potentially confusing passage in the OG. In the GRE OG 2e, on p. 134, as well as in ETS’s GRE Mathematical Conventions PDF on p. 5, point 9, we find a paragraph discussing quotients and remainders that utilizes the formula *n = qd + r*. This formula pertains to the situation where an integer *n* is divided by an integer *d*. *q* represents the quotient, and *r* represents the non-negative remainder. The paragraph then goes on to discuss the results of the following expressions: 20 ÷ 7, 21 ÷ 7, and −17 ÷ 7.

Well, using their formula, we can verify those examples:

a) 20 = (2)(7) + 6

b) 21 = (3)(7) + 0

c) –17 = (–3)(7) + 4

All true. So far so good. The problem comes when we want to use all of this to convert an improper fraction to a mixed numeral. The guiding formula is what I will call the “Division Equation”:

This equation is true 100% of the time. With example (a), if we just plug everything into this formula we magically change the improper fraction into a mixed numeral.

Voila! The correct mixed numeral magically results. With example (b), we don’t need to worry about mixed numerals, because the divisor goes into the dividend evenly, and thus the result of the division is an integer: 21/7 = 3. With example (c), things get tricky:

The Division Equation above still works, but for negatives, we can’t just dump the numbers in and magically get the correct negative mixed numeral result. This has to do with the way ETS has chosen to define remainders for negative numbers. Let’s explore this.

In the passage from p. 134 quoted above, every math book on planet Earth, and probably most written by aliens on other planets, would agree 100% with what ETS said about quotients and remainders for positive numbers. All that is absolutely undisputable. The vast majority of math books deliberately ignore the entire topic of what happens to quotients and remainders with negative numbers, because of the mathematical trade-offs that arise. Most math books have the good sense to avoid this troublesome topic, but ETS brashly charges in and firmly establishes a position. (To follow knowledge like a sinking star beyond the utmost bound? or, fools rush in where angels fear to tread?)

The *disadvantage* of defining remainders of negative numbers in this way is that makes finding negative mixed numerals more difficult: we can no longer use the Division Equation — we’ll get back to that point below. The *advantage* is: when you divide by a certain divisor, say 7, and talk about all the numbers that have a remainder of 4, those numbers are equally spaced on the number line in both the positive and negative direction. The numbers which, when divided by 7, have a remainder of 4 are:

Notice that –17 appears in that pattern, and follows the same pattern as the familiar positive numbers which, when divided by 7, have a remainder of 4. It’s very important: if ETS mentions a dividend which, when divided by 7, has a remainder of 4, don’t automatically assume the dividend is positive: it could be negative, and it would follow this sort of pattern.

What exactly is a mixed numeral? That is to say, what do we mean when we write, say:

What this really means is:

In other words, ** any mixed numeral implicitly contains addition between the integer and the fraction.** That is the

It’s relatively easy to see that, for positive numbers, the Fraction Equation will result in a whole number quotient plus a fraction less than one, which automatically fits the pattern for a mixed number.

What’s going on with a negative mixed number? Well, that’s a little different:

The negative sign in front of the whole mixed number applies to both terms and distributes, so you get subtraction instead of addition. If we are using ETS’ definition of remainder for negative numbers, then the Fraction Equation still works, but it doesn’t help us convert a negative improper fraction to a negative mixed number, because it results in *adding *the fraction instead of *subtracting* it. Again, this part is the disadvantage of the convention ETS is following on this particular topic. The Fraction Equation, so helpful for converting positive improper fractions to positive mixed numerals, is ostensibly useless in helping us analogously with negative numbers.

The question arises, then: how do we convert a negative improper fraction to a negative mixed numeral? For example, suppose we have the fraction –17/7. Again, as the OG p. 134 tells us, “when –17 is divided by 7, the quotient is –3, and the remainder is 4.” That will be consistent with the Fraction Equation, but, as we have seen above, that in turn will not lead to a correct mixed numeral. Therefore, everything OG says about quotients and remainders for negative number is useless if what we want to find is a negative mixed numeral. What do we do?

Actually, what you do is remarkably simple: pretend the negative sign isn’t there for the moment, and just convert the positive improper fraction to positive mixed numeral!

Now, we are dividing +17 by 7: of course, the quotient is 2 and the remainder is 3, so the Fraction Equation conveniently gives us

So far, so good. As a result of working with the positive numbers, we know:

But what we wanted was –17 divided by 7, not +17 divided by 7. How do we get that? We get that simply by multiplying the previous equation by a negative sign on both sides.

Again, to convert negative improper fraction to a negative mixed numeral: just pretend the improper fraction is positive, do the conversion with positive numbers, and then simply make the output negative when you are done.

1) Here, we must follow the ETS convention. According to this convention, when a negative number is divided the remainder must be positive. What is the multiple of 6 that is less than –23 and closest to –23? That would be –24, so 6 goes into –23 negative four times, with a remainder of (–23) – (–24) = +1. This checks out with the ETS’s equation: n = –23, d = 6, q = –4, and r = 1, and it’s true that

–23 = (6)( –4) + 1

So the quotient is –4 and the remainder is +1. Answer = **E**.

2) Here, we will first change it to a positive improper fraction. When +23 is divided 6, the quotient is 3 and the remainder is 5, so:

Now, just multiply both sides by a negative sign.

Answer = **B**.

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]]>The post Proportions on the GRE appeared first on Magoosh GRE Blog.

]]>Proportions are extremely common on the GRE. If you don’t have a strong grasp of them, and you are busy trying to figure out combinations/permutations or probability, stop. Focus your attention on mastering proportions before moving on to more tertiary concepts.

So let’s start basic. Proportions can be broken up into two groups: direct proportions and indirect proportions. In this post I am going to focus on direct proportions. They are more intuitive than indirect proportions and are also the more common on the GRE.

Below is an example of a direct proportion:

To solve for x we cross multiply, giving us: ; .

This is a direct proportion because as the ‘5’ becomes larger (namely it quadruples to become 20) the x also gets larger (it quadruples to become 8). That is both sides are getting larger.

I can almost guarantee that you will not see such a straightforward equation on the GRE Quant section. Instead, you will be given either a word problem or a graph and you will have to translate the information into an equation like the one above.

Let’s take a look at two problems:

In 2004, 2,400 condos sold, 15% of the total housing units sold that year. If 25% of the homes sold in 2004 were four-bedroom houses, then how many four-bedroom homes sold in 2004?

(A) 3,600

(B) 4,000

(C) 4,200

(D) 4,800

(E) 6,000

**Explanation**

Here we want to set up an equation. ; ; . (B).

Some things to note: you can take off the last two zeroes in 2,400 to get 24, a step which will make the math easier. Remember to bring the two zeroes back, which makes sense: x =40, is clearly too low and not amongst the answer choices.

Speaking of answer choices, notice that 25% is less than double of 15%. Therefore, the number of four-bedroom houses sold has to be less than twice the number of condos sold. (D) and (E) cannot be answer. (A) 3600 is only 50% greater than 2,400, so it is probably too low as well. Elimination, esp. if you are short on time, or getting tangled up in the calculation can be very effective.

Now let’s put a spin to this question. Nothing too tricky; indeed you should be able to solve this using the method above.

In 2004, 2,400 condos sold, 15% of the total housing units sold that year. How many units sold in 2004 were not condos?

(A) 16,000

(B) 13,600

(C) 12,400

(D) 11, 200

(E) 8,600

**Explanation:**

To approach this question, a good idea is to find the number of units in the total market. Then to find the number that are not condos subtract the condos from the total.

My reason for this approach is it is easier to do the math when we are working with 100 vs. 85 (which would be the percent of home that are not condos).

The solution is as follows:

; . Remember to add the two zeroes: 16,000. Now we have to subtract the total condos (2,400) to find the number of units that are not condos: 16,000 – 2,400 = 13,600.

Setting up a proportion is essential to solving a range of GRE math questions. Make sure you can confidently and quickly solve this question type before more on to more challenging – but less common – concepts.

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]]>The post Student Question on ETS Material: Graphs of Absolute Value Functions appeared first on Magoosh GRE Blog.

]]>Here’s a question that one of our students sent in– ask and you shall receive! 🙂

“Hello team Magoosh. First I’d like to say I love this site, I’ve learned so much! I’m posting a question about a certain math problem I found on the official ETS website. I tried to solve it and got it wrong.

…do you think you post a video as to how and why this is the answer?

– James”

You may also want to check out absolute value basics. Let us know if you have any questions about this (or anything else GRE-related!).

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]]>The post GRE Math: Absolute Values appeared first on Magoosh GRE Blog.

]]>On average you will see at least one question on the Revised GRE dealing with absolute values. You may even see a few. Yet, absolute value gets lost in the prep fray amongst the more popular concepts. So if you don’t want this relatively innocuous concept to surprise you test day read on.

What do -4 and 4 have in common? They are both four units from the 0 on a number line. Think of absolute value has how far from the zero a given number is on a number line.

For positive numbers finding the absolute value is easy – it is always the number between the absolute value signs, which look like this .

When we take the absolute value of a negative number, we drop the negative, and the absolute value sign.

What is the value of x in the equation ? Well, the absolute value of both and is , so or .

Now take a look at the following:

Anything seem off? Well, the absolute value of any number can never be a negative therefore there is no value for x.

To solve for a variable inside an absolute value sign, we want to remove the absolute value sign and solve the equation. However, there is a slight twist: you will want to create two separate equations. For one remove the absolute value signs and solves for x. For the other, make the side of the equation not inside the absolute value equal to a negative. Let’s try a practice problem:

?

Our two equations are:

If this seems strange, think of it this way: when you find a value for x that makes the equation equal -4, you have also solved for positive four. Remember, the absolute value of -4 is 4. Solving for x we get:

,

, .

Therefore x can equal 6 or -2.

Now let’s complicate things a little and throw an inequality in to the picture. Have a look:

We turn the inequality sign into an equal sign and solve for x the way we did above.

so .

However, when we turn 3 into -3, we have to reverse the sign. This is the one critical step.

so . You can plug in different values for x to see how this is the case.

This is a basic overview to absolute value and should help you with most of the sub-150 problems. For the harder problems, however, you will want to make sure to practice with more advanced problems.

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]]>The post GRE Math Basics: Quick Tips appeared first on Magoosh GRE Blog.

]]>**Prime numbers**

2 is the smallest prime number

2 is the only prime even number

1 is not a prime

3, 5, 7 are the only three consecutive odd integers that are each a prime.

**Digits**

In the integer 5,432, 5 is the thousands digit, 4 is the hundreds digit, 3 is the tens digit, and 2 is the units digit.

In a multiple-digit number, there are 10 possible units digits (0 – 9). In the first digit, i.e. the digit farthest to the left, there are only nine possible digits. A zero as the first digit would not be valid. For instance, 0,453 is not a four-digit number. It’s – if anything – the three-digit number 453.

**Divisibility Rules**

3: Add up the digits in any number. If they are multiple of 3, then the number is divisible by 3. For example, 258 is divisible by 3 because 2 + 5 + 8 is 15. And 15 is a multiple of 3.

4: If the tens and units digits of integer X is a multiple of 4, then integer X is divisible by 4. For example, 364 is divisible by 4, because 64 is a multiple of 4.

5: If a number ends in 5 or 0, it is divisible by 5.

6: If the rules to divisibility to 2 and 3 are fulfilled, then a number is divisible by 6. One hundred and eleven is not divisible by 6. Two hundred and forty six, because it is divisible by both 2 and 3, is divisible by 6.

9: If the sum of the digits of a number is divisible by 9, then that number is divisible by 9. For example, adding up the digits in 4,536, we get 4+5+3+6 = 18. Because 18 is divisible by 9, the number 4,536 is also divisible by 9.

**Odds (so to speak) and Ends**

Zero is an even integer

Integers can be negative numbers (-3, -6, -17, etc.)

0! = 1

x^0 = 1

The problems below can be solved by referencing the above. Good luck!

**Practice Problems**

1. What is the smallest integer that can be divided by the product of a prime number and 7 while yielding a prime number?

(A) 7

(B) 14

(C) 24

(D) 28

(E) 35

2. The sum of five consecutive even integers is 20. What is the product of the median of the series and the smallest integer in the series?

(A) 12

(B) 10

(C) 6

(D) 4

(E) 0

3. Which of the following integers is NOT divisible by 3?

(A) 624

(B) 711

(C) 819

(D) 901

(E) 969

Answers:

1. D

2. E

3. D

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]]>The post GRE Math Video Lessons: Series and Counting Basics appeared first on Magoosh GRE Blog.

]]>A relatively popular concept on the GRE math is series – when you have a list of numbers that follow a certain pattern(e.g. consecutive numbers). Usually, the question will ask you to add up all the numbers that are part of the series.

Today, we are not going to go quite so far. Instead, we are going to focus on a step that is far more basic, yet one that often eludes students. Try the following problem:

*How many integers are there in Set A, if Set A includes all the numbers 1 – 10?*

* ***Really Easy, Right?**

* *The answer seems pretty straightforward – 10. That is, there are ten possible integers: 1, 2, 3… 9, 10.

Now, you are probably thinking that this question is really obvious. Of course, there are ten numbers, if you are counting the numbers 1 – 10. I agree, but try this almost identical question:

*How many integers are there in Set A, if Set A includes all the numbers 5 – 15?*

The answer is ten, right? Actually, the answer is eleven. And that’s what makes this question so tricky. In answering ten, you probably went something like this – 15 – 5 = 10; therefore, there are ten numbers.

**The Twist**

Whenever you are counting the total numbers in a list by subtracting, you always have to count an extra number.”For instance, with the first question, if you just subtract 10 – 1 = 9, you get one fewer number in the set than there actually is.”The reason we always add an extra one is because we want to include the number we are counting. For instance, if I have read pages 1 – 3 in a book, I would want to make sure I count page 1 as one of the pages I read.

With this in mind, now try the following problem:

*Set A consists of the digits 50 – 100. How many digits are there in set A?*

* **(A) **49*

*(B) **50*

*(C) **51*

*(D) **99*

*(E) **100*

**Solution:**

The simple formula for counting the number of elements in a consecutive sequence, like the one above, is L – F + 1. 100 – 50 + 1 = 51.

**Practice Time**

The next three questions pertain to a 210-page novel:

*How many pages are there between 45 and 111, not including either of those pages?*

* **(A) **45*

*(B) **65*

*(C) **66*

*(D) **67*

*(E) **76*

*If I am at the top of page 125 of a book, then how many pages will I have to read if I want to read to the end of page 152?*

* **(A) **25*

*(B) **26*

*(C) **27*

*(D) **28*

*(E) **31*

* *

* **I am printing a book, and, for each digit I print on the page, I have to pay 5 cents (pg. 25 = 10 cents, pg. 134 = 15 cents). How much will it cost to print the page numbers, starting on page 9 and continuing for another 100 pages?*

* **(A) **9.05*

*(B) **10.15*

*(C) **10.20*

*(D) **10.55*

*(E) **10.60*

**Answers:**

1. B

2. D

3. D

To find out where series and counting (aka sequences) sit in the “big picture” of GRE Quant, and what other Quant concepts you should study, check out our post entitled:

What Kind of Math is on the GRE? Breakdown of Quant Concepts by Frequency

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]]>The post GRE Math – Essential Tips For Factoring appeared first on Magoosh GRE Blog.

]]>Let’s say you have a large number, and you need to break it down into its prime factors (prime factors are prime numbers, e.g. 2, 3, 5, 7). When dealing with small numbers, such as 24 (2 x 2 x 2 x 3), finding the prime factors isn’t too tough. But what about 324?

Many students will freeze when they see such a number. However, on the GRE, there is always a way to break down a large number.

Divide by 2

Is the number even? If so, simply divide by 2. If that results in a number that is also divisible by 2 continue dividing until the number is no longer even. 324/2 = 162. 162/2 = 81. At this point, you should be able to break down 81 into 9 x 9. Next, 9 x 9 can further be broken down into 3 x 3 x 3 x 3. Therefore, the prime factors of 324 are 3, 3, 3, 3, 2, 2.

Now, let’s try another question:

What is the sum of the unique prime factors of 207?

(A) 23

(B) 26

(C) 29

(D) 69

(E) 207

Divide by 3

Well, 207 is a very unpleasant number. Is it even divisible by any number, besides 1? Well, a good rule when dealing with odd numbers is to add up the digits, e.g. 2 + 0 + 7 = 9. If the sum is divisible by 3, then the number is divisible by 3. Therefore, 207 is divisible by 3 (9 is divisible by 3). 207/3 = 69. What about 69? Add up the digits, and you get 15, which is also divisible by 3. Therefore, 69/3 = 23. 23 is a prime number because it can’t be divided by any number, except 1.

The factors of 207 are therefore 3, 3, and 23. Notice, the question asks for the sum of the unique factors. The factor 3 appears twice, so we can discount one of the threes (because it is not unique; there is already another 3). The answer is 23 + 3 = 26 (B).

Finally, let’s try one last question.

What is the range of the prime factors of x, where x is 275?

(A) 5

(B) 6

(C) 11

(D) 30

(E) 55

Divide by 5

If a number ends in a 5, it is always divisible by 5. 275/5 = 55. 55/5 = 11. Therefore, the prime factors are 5, 5, and 11. The range of these factors is the greatest number minus the smallest number. 11 -5 = 6. Answer (B).

**Takeaway**

Factoring/breaking down numbers is something you will need to do often on the GRE. Becoming adept at factoring quickly – either on paper or with a calculator – will save you a lot of time. So, practice as much as possible, and have more time test-day to walk through the really tough problems.

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]]>The post GRE Math Basics: Distance, Rate and Time appeared first on Magoosh GRE Blog.

]]>The 4^{th} of July weekend is upon us, and many of us will be taking road trips, or even taking a plane somewhere. To commemorate this collective movement, let’s learn the most fundamental formula when dealing with movement over time. First, let’s meet Bob…

*Bob drives at an average rate of 50 mph from Berkeley to Los Angeles, a distance of 350 miles. How long does it take him to complete the trip?*

* *

*(A) **4 hrs*

*(B) **5 hrs*

*(C) **7 hrs*

*(D) **10 hrs*

*(E) **12 hrs*

When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time.

With the problem above, the distance between Berkeley and Los Angeles is 350 miles. So D = 350. Bob is traveling at 50 mph, so that is his rate. The question is how long will it take him to complete this trip. Therefore, we have to solve for T. Let’s set up the equation, plugging in the values for D and R:

350 = 50T.

Solving for T, we get 7. Because we are dealing with miles per hours, the 7 corresponds to hours. So T = 7. Answer (C).

Now let’s try another problem:

*Charlie takes 2.5 hours to fly from Los Angeles to Mexico City, a distance of 1200 miles. What is the average speed of his plane in miles per hour?*

* *

*(A) **200 mph*

*(B) **240 mph*

*(C) **410 mph*

*(D) **480 mph*

*(E) **533 mph*

Setting up the equation, we get 1200 (Distance) = 2.5 (Time) x R; 1200 = 2.5R. Before solving (assuming this is on the current GRE, in which you do not have a calculator), let’s change 2.5 to 5/2, as it is much easier to do the math with fractions than with decimals. We get 1200 = 5/2R. Solving for R, we multiply both sides by the reciprocal, 2/5. This gives us R on the right-hand side of the equation, and 1200 x 2/5 on the left-hand side. 2400/5 = 480 (D).

Note you could have solved this problem back-solving, in which you put the answer choices back into the question. Or, even better, we could just plug-in 500 mph. Of course that is not one of the answer choices, but using 500 will make the math very easy. If the number is a little too high when we plug-in 500, then the answer must be (D). If it is a little low, then the answer must be (E) 533. Again, you never try to back solve with an ugly number like 533—it will take too long.

Using 500 mph, we get 1200 (Distance) = 500 (Rate) x 2.5 (Time). You can see that multiplying these two numbers gives us 1250. Meaning, we flew too fast and missed Mexico City by 50 miles. Therefore, we have to slow down the plane a little—the answer is 480.

So, this weekend, whether you are traveling abroad or just making a trip to a friend’s BBQ, you can figure out your average distance, rate, or time. Who said GRE doesn’t relate to the real world?

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]]>The post GRE Math Tips – Getting Rid of Your x’s and y’s appeared first on Magoosh GRE Blog.

]]>The GRE Math section is full of problem solving questions, which test exactly the way you do that—solve problems. The key is that you solve the problem in the shortest amount of time possible. Unfortunately, many students have difficulty either finding a solution for a problem, or, if they do, finding that solution quickly.

Part of the reason is that many students persist in thinking in terms of and ; that is, they continue to think in terms of algebra. Algebra—that cauldron of variables that bubbles up in their mind whenever they see a difficult word problem—is often times the enemy. When solving a difficult word problem, the sooner you can let go of the ’s and s, the better.

Let’s have a look at the following problem:

*The product of two integers is 40. Which of the following CANNOT be the sum of these two integers?*

*(A) -41*

*(B) 3*

*(C) 13*

*(D) 14*

*(E) -14*

Many students would proceed as follows:

, = ??

At that point, students freeze. Sometimes for seconds, sometimes for minutes. Some may try to plug the answers into the above equation. For instance, starting with *(C) 13*, they try to set up the equation and solve via substitution:

, .

After a few tedious steps, they may get: .

You could factor this equation and find that the sum of the roots is 13. But just getting to this point is very difficult and time-consuming. And, until you get the correct answer, you would have to solve for each of the answer choices in this fashion (definitely not a good use of your 45 minutes). So, instead, let’s dispense with the and and break down the problem as follows, using a tip that I call “Getting Rid of Your x’s and y’s”:

What are the factors of ?

Each of these pairs could also be negative (remember a negative negative is a positive).

Now we just need to find the sum of each pair, and see which one is not represented above.

The only number not represented is *(B) 3*.

Again, the GRE Math section is testing the way you think. It wants logical thinkers who are able to see the quickest way to the solution, so use this tip to your advantage. Not that algebra is unhelpful, but, for the most part, the quickest way to the solution starts with getting rid of your ’s and ’s!

What other GRE Math tips would you like to see? Let us know. 🙂

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