It’s All Greek to Me: How to Use Greek Concepts to Beat the GMAT

Aero_img084The ancient Greeks were, to put it mildly, really neat. They created or helped to create the foundations of philosophy, theater, science, democracy, and mathematics – no small accomplishment for a small war-torn civilization from over two millennia ago. Many of our contemporary ideas, beliefs, and traditions are rooted in contributions made by Greek thinkers, and the GMAT is no exception.

A few months ago, I wrote about this difficult Data Sufficiency question.

When I first encountered this problem I couldn’t help but wonder what kind of mad scientist question-writer engineered it. Where would such an idea even come from? It turns out, it wasn’t a GMAC employee at all, but Archimedes, the famous Greek geometer and coiner of the phrase “Eureka!”

The question is based on his attempt to trisect an angle with only a straight edge and a compass. (Alas, Archimedes’ work, though ingenious, was not technically a correct solution to the problem, as it provides only an approximation.) The reader is hereby invited to contemplate the kind of person who encounters a proof by Archimedes and instinctively thinks, “This would make an excellent Data Sufficiency question on the GMAT!” We’d like to believe that the good folks at GMAC are just like you and me, but perhaps not.

So this got me thinking: what other interesting Greek contributions to mathematics might be helpful in analyzing GMAT questions? In Euclid’s work Elements, he offers a simple and elegant proof for why there is no largest prime number. The proof proceeds by positing a hypothetical largest prime number “p.” We can then construct a product that consists of every prime number 2*3*5*7….*p. We’ll call this product “q.”

The next consecutive number will be q + 1. Now, we know that “q” contains 2 as a factor, as “q,” supposedly, contains every prime as a factor. Therefore q +1 will not contain 2 as a factor. (The next number to contain 2 as a factor will be q + 2.) We know that “q” contains 3 as a factor. Therefore q + 1 will not contain 3 as a factor. (The next number to contain 3 as a factor will be q + 3.)

Uh oh. If “p” really is the largest prime number, we’ve got a problem, because q + 1 will not contain any of the primes between 2 and p as factors. So either q + 1 is itself prime, or there is some prime greater than p and less than q + 1 that we’ve failed to consider. Either way, we’ve proven that p can’t be the largest prime number – I told you the Greeks were neat.

One axiom that’s worth internalizing from Euclid’s proof is the notion that two consecutive numbers cannot have any factors in common aside from 1.  When q contains every prime from 2 to p as a factor, q + 1 contains none of those primes. How would this be helpful on the GMAT? Glad you asked. Check out this question:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x + 1 must be:

(A) Between 1 and 10

(B) Between 11 and 15

(C) Between 15 and 20

(D) Between 20 and 25

(E) Greater than 25

We’re given information about x, and we’re asked about x + 1. If x is the product of all even numbers from 2 to 50, we can write x = 2 * 4 * 6 …* 50. This is the same as (1*2) * (2*2) * (3*2)… (25*2), which means the product consists of all the integers from 1 to 25, inclusive, and a bunch of 2’s.

So now we know that every prime number between 2 and 25 will be a factor of x. What about x + 1? (Paging Euclid!) We know that 2 is not a factor of x + 1, as 2 is a factor of x, and so the next multiple of 2 would be x + 2. We know that 3 is also not a factor of x + 1, as 3 is a factor of x, and so the next multiple of 3 would be x + 3. And once we’ve internalized that two consecutive numbers cannot have any factors in common aside from 1, we know that if all the primes between 2 and 25 are factors of x, none of those primes can be factors of x + 1, meaning that the smallest prime of x, whatever is, will be greater than 25. The answer, therefore, is E.

Takeaway: One of the beautiful things about mathematics is that fundamental truths do not change over time. What worked for the Greeks will work for us. The same axioms that allowed ancient mathematicians to grapple with problems two millennia ago will allow us to unravel the toughest GMAT questions. Learning a few of these axioms is not only interesting – though I’d caution against bringing up Archimedes’ trisection proof at a dinner party – but also helpful on the GMAT.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTube, Google+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

GMAT Tip of the Week: Kanye, Wiz Khalifa, Twitter Beef…and GMAT Variables

GMAT Tip of the WeekThis week, the internet exploded with a massive Twitter feud between rappers Kanye West and Wiz Khalifa, with help from their significant others and exes. For days now, hashtags unpublishable for an education blog have topped the trending lists, all as a result of the epic social media confrontation. And all of THAT originated from a classic GMAT mistake from the Louis Vuitton Don – a man who so loves his hometown Kellogg School of Management that he essentially named his daughter Northwestern – himself:

Kanye didn’t consider all the possibilities when he saw variables.
A brief history of the beef: there was musical origin, as Wiz wanted a bit of credit for his young/wild/free friends for the term “Wave,” as Kanye changed his upcoming album title from Swish to Waves. But where things escalated quickly all stemmed from Wiz’s use of variables in the following tweet:

Hit this kk and become yourself.

Kanye, whose wife bears those exact initials, K.K., immediately interpreted those variables as a reference to Kim and lost his mind. But Wiz had intended those variables kk to mean something entirely different, a reference to his favorite drug of choice. And then…well let’s just say that things got out of hand.

So back to the GMAT: Kanye’s main mistake was that he didn’t consider alternate possibilities for the variables he saw in the tweet, and quickly built in some incorrect assumptions that led to disastrous results. Do not let this happen to you on the GMAT! Here’s how it could happen:

1) Forgetting about not-obvious numbers.
If a problem, for example, defines k as 10 < k < 12, you can’t just think “k = 11” because you don’t know that k has to be an integer. 11.9 or 10.1 are also possibilities. Similarly if k^2 = 121, you have to consider that k could be -11 as well as it could be 11.

Ultimately, that was Yeezy’s mistake: he saw KK and with tunnel vision saw the most obvious possibility. But why couldn’t “KK” have been Krispy Kreme or Kyle Korver or Kato Kaelin? Before you leap to conclusions on a GMAT variable, see if there’s anything else it could be.

2) Assuming that each variable must represent a different number.
This one is a bit more nuanced. Suppose you were asked:

For positive integers a and b, is the product ab > 1?

(1) a = 1

With that statement, you might start thinking, “Well if a is 1, b has to be something else…” but all the variable b really means is “a number we don’t know.” Just because a problem assigns two different variables does not mean that they represent two different numbers! B could also be 1…we just don’t know yet.

Where this manifests itself as a problem most often is on function problems. When people see the setup, for example:

The function f is defined for all values x as f(x) = x^2 – x – 1

They’ll often be confused when that’s paired with a question like, “Is f(a) > 1?” and a statement like:

(1) -2 < a < 2

“I know about f(x) but I don’t know anything about f(a),” they might say, but the way these variables work, f(x) means “the function of any number…we just don’t know which number” so when you then see f(a), a becomes that number you don’t know. You’ll do the same thing for a: f(a) = a^2 – a – 1. What goes in the parentheses is just “the number you perform the function on” – the function doesn’t just apply to the variable in the definition, but to any number, variable, or combination that is then put in the parentheses.

The real lesson here is this: variables on the GMAT are a lot like variables in Wiz Khalifa’s Twitter feed. You might think you know what they mean, but before you stake your reputation (or score) on your response to those variables, consider all the options. Hit this GMAT and become yourself.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTubeGoogle+ and Twitter!

By Brian Galvin.

GMAT Tip of the Week: Make 2016 The Year Of Number Fluency

GMAT Tip of the WeekWhether you were watching the College Football Playoffs or Ryan Seacrest; whether you were at a house party, in a nightclub, or home studying for the GMAT; however you rang in 2016, if 2016 is the year that you make your business school goals come true, hopefully you had one of the following thoughts immediately after seeing the number 2016 itself:

  • Oh, that’s divisible by 9
  • Well, obviously that’s divisible by 4
  • Huh, 20 and 16 are consecutive multiples of 4
  • 2, 0, 1, 6 – that’s three evens and an odd
  • I wonder what the prime factors of 2016 are…

Why? Because the GMAT – and its no-calculator-permitted format for the Quant Section – is a test that highly values and rewards mathematical fluency. The GMAT tests patterns in, and properties of, numbers quite a bit. Whenever you see a number flash before your eyes, you should be thinking about even vs. odd, prime vs. composite, positive vs. negative, “Is that number a square or not?” etc. And, mathematically speaking, the GMAT is a multiplication/division test more than a test of anything else, so as you process numbers you should be ready to factor and divide them at a moment’s notice.

Those who quickly see relationships between numbers are at a huge advantage: they’re not just ready to operate on them when they have to, they’re also anticipating what that operation might be so that they don’t have to start from scratch wondering how and where to get started.

With 2016, for example:

The last two digits are divisible by 4, so you know it’s divisible by 4.

The sum of the digits (2 + 0 + 1 + 6) is 9, a multiple of 9, so you know it’s divisible by 9 (and also by 3).

So without much thinking or prompting, you should already have that number broken down in your head. 16 divided by 4 is 4 and 2000 divided by 4 is 500, so you should be hoping that the number 504 (also divisible by 9) shows up somewhere in a denominator or division operation (or that 4 or 9 does).

So, for example, if you were given a problem:

In honor of the year 2016, a donor has purchased 2016 books to be distributed evenly among the elementary schools in a certain school district. If each school must receive the same number of books, and there are to be no books remaining, which of the following is NOT a number of books that each school could receive?

(A) 18

(B) 36

(C) 42

(D) 54

(E) 56

You shouldn’t have to spend any time thinking about choices A and B, because you know that 2016 is divisible by 4 and by 9, so it’s definitely divisible by 36 which means it’s also divisible by every factor of 36 (including 18). You don’t need to do long division on each answer choice – your number fluency has taken care of that for you.

From there, you should look at the other numbers and get a quick sense of their prime factors:

42 = 2 * 3 * 7 – You know that 2016 is divisible by 2 and 3, but what about 7?

54 = 2 * 3 * 3 * 3 – You know that 2016 is divisible by that 2 and that it’s divisible by 9, so you can cover two of the 3s. But is 2016 divisible by three 3s?

56 = 2 * 2 * 2 * 7 – You know that two of the 2s are covered, and it’s quick math to divide 2016 by 4 (as you saw above, it’s 504). Since 504 is still even, you know that you can cover all three 2s, but what about 7?

Here’s where good test-taking strategy can give you a quick leg up: to this point, a savvy 700-scorer shouldn’t have had to do any real “work,” but testing all three remaining answer choices could now get a bit labor intensive. Unless you recognize this: for C and E, the only real question to be asked is “Is 2016 divisible by 7?” After all, you’re already accounted for the 2 and 3 out of 42, and you’ve already accounted for the three 2s out of 56.

7 is the only one you haven’t checked for. And since there can only be one correct answer, 2016 must be divisible by 7…otherwise you’d have to say that C and E are both correct.

But even if you’re not willing to take that leap, you may still have the hunch that 7 is probably a factor of 2016, so you can start with choice D. Once you’ve divided 2016 by 9 (here you may have to go long division, or you can factor it out), you’re left with 224. And that’s not divisible by 3. Therefore, you know that 2016 cannot be divided evenly into sets of 54, so answer choice D must be correct. And more importantly, good number fluency should have allowed you to do that relatively quickly without the need for much (if any) long division.

So if you didn’t immediately think “divisible by 4 and 9!” when you saw the year 2016 pop up, make it your New Year’s resolution to start thinking that way. When you see numbers this year, start seeing them like a GMAT expert, taking note of clear factors and properties and being ready to quickly operate on that number.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+ and Twitter!

By Brian Galvin.

How to Choose the Right Number for a GMAT Variable Problem

Pi to the 36th digitWhen you begin studying for the GMAT, you will quickly discover that most of the strategies are, on the surface, fairly simple. It will not come as a terribly big surprise that selecting numbers and doing arithmetic is often an easier way of attacking a problem than attempting to perform complex algebra. There is, however, a big difference between understanding a strategy in the abstract and having honed that strategy to the point that it can be implemented effectively under pressure.

Now, you may be thinking, “How hard can it possibly be to pick numbers? I see an “x” and I decide x = 5. Not so complicated.” The art is in learning how to pick workable numbers for each question type. Different questions will require different types of numbers to create a scenario that truly is simpler than the algebra. The harder the problem, the more finesse that will be required when selecting numbers. Let’s start with a problem that doesn’t require much strategy:

If n=4p, where p is prime number greater than 2, how many different positive even divisors does n have, including n? 

(A) 2

(B) 3

(C) 4

(D) 6 

(E) 8 

Okay in this problem, “p” is a prime number greater than 2. So let’s say p = 3. If n = 4p, and 4p = 4*3 = 12. Let’s list out the factors of 12: 1, 2, 3, 4, 6, 12. The even factors here are 2, 4, 6, 12. There are 4 of them. So the answer is C. Not so bad, right? Just pick the first simple number that pops into your head and you’re off to the races. Bring on the test!

If only it were that simple for all questions. So let’s try a much harder question to illustrate the pitfalls of adhering to an approach that’s overly mechanistic:

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?

(A) x + 10

(B) 10x + 1

(C) 100(x + 10)

(D) 100 * (x+10)/(x+100)

(E) 100 * (10x + 1)/(10x+10)

You’ll notice quickly that if you simply declare that x = 10 and r =20, you may run into trouble. Say, for example, that the starting value from one week ago was 100 liters. If x = 10, a 10% increase will lead to a volume of 110 liters. If we remove 20% of that 110, we’ll be removing .20*110 = 22 liters, giving us 110-22 = 88 liters. But we’re also told that the resulting volume is 90% of the original volume! 88 is not 90% of 100, therefore our numbers aren’t valid. In instances like this, we need to pick some simple starting numbers and then calculate the numbers that will be required to fit the parameters of the question.

So again, say the volume one week ago was 100 liters. Let’s say that x = 20%, so the volume, after water is added, will be 100 + 20 = 120 liters.

We know that once water is removed, the resulting volume will be 90% of the original. If the original was 100, the volume, once water is removed, will be 100*.90 = 90 liters.

Now, rather than arbitrarily picking an “r”, we’ll calculate it based on the numbers we have. To summarize:

Start: 100 liters

After adding water: 120 liters

After removing water: 90 liters

We now need to calculate what percent of those 120 liters need to be removed to get down to 90. Using our trusty percent change formula [(Change/Original) * 100] we’ll get (30/120) * 100 = 25%.

Thus, when x = 20, r =25. Now all we have to do is substitute “x” with “20” in the answer choices until we hit our target of 25.

Remember that in these types of problems, we want to start at the bottom of the answer choice options and work our way up:

(E) 100 * (10x + 1)/(10x+10)

100 * (10*20 + 1)/(10*20+10) = 201/210. No need to simplify. There’s no way this equals 25.

(D) 100 * (x+10)/(x+100)

100 * (20+10)/(20+100) = 100 * (30/120) = 25. That’s it! We’re done. The correct answer is D.

Takeaways: Internalizing strategies is the first step in your process of preparing for the GMAT. Once you’ve learned these strategies, you need to practice them in a variety of contexts until you’ve fully absorbed how each strategy needs to be tweaked to fit the contours of the question. In some cases, you can pick a single random number. Other times, there will be multiple variables, so you’ll have to pick one or two numbers to start and then solve for the remaining numbers so that you don’t violate the conditions of the problem. Accept that you may have to make adjustments mid-stream. Your first selection may produce hairy arithmetic. There are no style point on the GMAT, so stay flexible, cultivate back-up plans, and remember that mental agility trumps rote memorization every time.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

How to Use Difference of Squares to Beat the GMAT

GMATIn Michael Lewis’ Flashboys, a book about the hazards of high-speed trading algorithms, Lewis relates an amusing anecdote about a candidate interviewing for a position at a hedge fund. During this interview, the candidate receives the following question: Is 3599 a prime number? Hopefully, your testing Spidey Senses are tingling and telling you that the answer to the question is going to incorporate some techniques that will come in handy on the GMAT. So let’s break this question down.

First, this is an interview question in which the interviewee is put on the spot, so whatever the solution entails, it can’t involve too much hairy arithmetic. Moreover, it is far easier to prove that a large number is NOT prime than to prove that it is prime, so we should be thinking about how we can demonstrate that this number possesses factors other than 1 and itself.

Whenever we’re given unpleasant numbers on the GMAT, it’s worthwhile to think about the characteristics of round numbers in the vicinity. In this case, 3599 is the same as 3600 – 1. 3600, the beautiful round number that it is, is a perfect square: 602. And 1 is also a perfect square: 12. Therefore 3600 – 1 can be written as the following difference of squares:

3600 – 1 = 602 – 12

We know that x– y= (x + y)(x – y), so if we were to designate “x” as “60” and “y” as “1”, we’ll arrive at the following:

60– 1= (60 + 1)(60 – 1) = 61 * 59

Now we know that 61 and 59 are both factors of 3599. Because 3599 has factors other than 1 and itself, we’ve proven that it is not prime, and earned ourselves a plumb job at a hedge fund. Not a bad day’s work.

But let’s not get ahead of ourselves. Let’s analyze some actual GMAT questions that incorporate this concept.

First:

999,9992 – 1 = 

A) 1010 – 2

B) (106 – 2) 2   

C) 105 (106 -2)

D) 106 (105 -2)

E) 106 (106 -2)

Notice the pattern. Anytime we have something raised to a power of 2 (or an even power) and we subtract 1, we have the difference of squares, because 1 is itself a perfect square. So we can rewrite the initial expression as 999,9992 – 12.

Using our equation for difference of squares, we get:

999,9992 – 12  = (999,999 +1)(999,999 – 1)

(999,999 + 1)(999,999 – 1) = 1,000,000* 999,998.

Take a quick glance back at the answer choices: they’re all in terms of base 10, so there’s a little work left for us to do. We know that 1,000,000 = 106  (Remember that the exponent for base 10 is determined by the number of 0’s in the figure.) And we know that 999,998 = 1,000,000 – 2 = 106 – 2, so 1,000,000* 999,998 = 106 (106 -2), and our answer is E.

Let’s try one more:

Which of the following is NOT a factor of 38 – 28?

A) 97

B) 65

C) 35

D) 13

E) 5

Okay, you’ll see quickly that 38 – 28 will involve same painful arithmetic. But thankfully, we’ve got the difference of two numbers, each of which has been raised to an even exponent, meaning that we have our trusty difference of squares! So we can rewrite 38 – 28 as (34)2 – (24)2. We know that 34 = 81 and 24 = 16, so (34)2 – (24)2 = 812 – 162. Now we’re in business.

812 – 162 = (81 + 16)(81 – 16) = 97 * 65.

Right off the bat, we can see that 97 and 65 are factors of our starting numbers, and because we’re looking for what is not a factor, A and B are immediately out. Now let’s take the prime factorization of 65. 65 = 13 * 5. So our full prime factorization is 97 * 13 * 5. Now we see that 13 and 5 are factors as well, thus eliminating D and E from contention. That leaves us with our answer C. Not so bad.

Takeaways:

  • The GMAT is not interested in your ability to do tedious arithmetic, so anytime you’re asked to find the difference of two large numbers, there is a decent chance that the number can be depicted as a difference of squares.
  • If you have the setup (Huge Number)2 – 1, you’re definitely looking at a difference of squares, because 1 is a perfect square.
  • If you’re given the difference of two numbers, both of which are raised to even exponents, this can also be depicted as a difference of squares, as all integers raised to even exponents are, by definition, perfect squares.

*Official Guide question courtesy of the Graduate Management Admissions Council.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Quarter Wit, Quarter WisdomLast week, we reviewed the concepts of cyclicity and remainders and looked at some basic questions. Today, let’s jump right into some GMAT-relevant questions on these topics:

 

 

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Statement 1: k is divisible by 10

Statement 2: k is divisible by 4

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

Statement 1: k is divisible by 10

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

Statement 2: k is divisible by 4

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

What is the remainder of (3^7^11) divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = 3

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Rounding Up Some Official GMAT Questions!

Quarter Wit, Quarter WisdomLast week we looked at some rounding rules. Today, let’s go over some official questions on rounding. They are quite simple and if we just keep the “Slip to the side and look for a 5” rule in mind, they can be easily solved.

Question 1: If n = 2.0453 and n* is the decimal obtained by rounding n to the nearest hundredth, what is the value of n* – n?

(A) -0.0053
(B) -0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153

Solution: A quick note on place value nomenclature:

Given a decimal 345.789, we know that 5 represents the units digit, 4 the tens digit and 3 the hundreds digit. Also, 7 represents the tenths digit, 8 the hundredths digit and 9 the thousandths digit and so on…

Now let’s go back to this question:

n = 2.0453

We need to round n to the nearest hundredth which means we will retain 2 digits after the decimal. The third digit after the decimal is 5 so 2.0453 rounded to the nearest hundredth is 2.05.

Thus n* – n = 2.05 – 2.0453 = 0.0047

Answer (D)

Question 2: If digit h is the hundredths digit in the decimal n = 0.2h6, what is the value of n, rounded to the nearest tenth?

Statement 1: n < 1/4

Statement 2: h < 5

Solution: Given that n = 0.2h6

We need to find the value of n rounded to the nearest tenth i.e. we need to keep only one digit after the decimal.

Statement 1: n < 1/4

In decimal form, it means n < 0.25

If h were 5 or greater, n would become 0.256 or 0.266 or higher. All these values would be more than 0.25 so h must be less than 5 such as 0.246 or 0.236 etc. In all such cases, n would be rounded to 0.2

This statement alone is sufficient.

Statement 2: h < 5

This is even simpler. Since we have been given that h is less than 5, when we round n to the tenths digit, we will get 0.2

This statement alone is also sufficient.

Answer (D)

Question 3: If d denotes a decimal number, is d >= 0.5?

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

Statement 2: When d is rounded to the nearest integer, the result is 1.

Solution: Again, a simple question!

We need to find whether d is greater than or equal to 0.5 or not.

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

This means that whatever d is, when we round it to the nearest tenth, we get 0.5. What are the possible values of d? If d is anywhere from 0.450 to 0.5499999…, it will be rounded to 0.5

Some of these numbers are less than 0.5 and others are greater than 0.5 so this statement alone is not sufficient.

Statement 2: When d is rounded to the nearest integer, the result is 1.

In this case d must be at least 0.5; only then can it be rounded to 1.

d can be anything from 0.50 to 1.499999… In any case, d will be greater than or equal to 0.5.

This statement alone is sufficient to answer the question.

Answer (B)

We hope you see that if we just remember the rules, we can solve most rounding questions very quickly and efficiently.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Rounding Rules on the GMAT: Slip to the Side and Look for a Five!

Quarter Wit, Quarter WisdomThe famous rounding song by Joe Crone is pretty much all you need to solve the trickiest of rounding questions on GMAT:

You just slip to the side, and you look for a five.

Well if the number that you see is a five or more, 

You gotta round up now, that’s for sure.

If the number that you see is a four or less, 

   You gotta round down to avoid a mess.

To put it in our own words, when we round a decimal, we drop the extra decimal places and apply certain rules:

–          If the first dropped digit is 5 or greater, we round up the last digit that we keep.

–          If the first dropped digit is 4 or smaller, we keep the last digit that we keep, the same.

For Example, we need to round the following decimals to two digits after decimal:

(a) 3.857

We drop 7. Since 7 is ‘5 or greater’, we are left with 3.86

(b) 12.983

We drop 3. Since 3 is ‘4 or smaller’, we are left with 12.98

(c) 26.75463

We drop 463. Since 4 is ‘4 or smaller’, we are left with 26.75

(d) 8.9675

We drop 75. Since 7 is ‘5 or greater’, we are left with 8.97

Note example (c) carefully:

When we round 26.75463 to two decimal places, we do not start rounding from the rightmost digit i.e. this is incorrect: 26.75463 becomes 26.7546 which becomes 26.755 which further becomes 26.76 – this is not correct. .00463 is less than .005 and hence should be ignored. You only need to worry about the digit right next to the digit you are keeping. Just slip to the side, and look for a five!

A logical question arises: what happens when we have, say, 2.5 and we need to round it to the nearest integer? 2.5 is midway between 2 and 3. In that case, why do we round the number up, as the rule suggests? Note that a 2.5 is a tie and we have many tie breaking rules that can be used. They are ‘Round half to odd’, ‘Round half to even’, ‘Round up’, ‘Round down’, ‘Round towards 0’, ‘Round away from 0’ etc. We don’t need to worry about all these since GMAT uses only Round up i.e. 2.5 will be rounded up to 3.

Let’s take a look at a question now which uses these fundamentals.

Question: The exact cost price to make each unit of a widget is $7.6xy7, where x and y represent single digits. What is the value of y?

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65.

Solution: The question is based on rounding. We need to figure out the value of y given some rounding scenarios. Let’s look at them one by one.

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

When rounded to the nearest cent, the cost becomes 7 dollars and 65 cents. 6xy7 cents got rounded to 65 cents. When will .6xy7 get rounded to .65? When .6xy7 lies anywhere in the range .6457 to .6547. Note that in all these cases, when you round the number to 2 digits, it will become .65.

Say price is 7.6468. We need to drop 68 but since 6 is ‘5 or greater’, 4 gets rounded up to 5.

Similarly, say the price is 7.6543. We need to drop 43. Since 4 is ‘4 or smaller’, 5 stays as it is.

So x and y can take various different values. This statement alone is not sufficient.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65

Now the cost is rounded to the tenth of a cent which means 3 places after the decimal. But the cost is given to us as $7.65. Since we need 3 places, the cost must be $7.650 (which will be written as $7.65)

When will 7.6xy7 get rounded to 7.650? Now this is the tricky part of the question – from 7.6xy7, you need to drop the 7 and round up y. When you do that, you get 7.650. This means 7.6xy7 must have been 7.6497. Only in this case, when we drop the 7, we round up the 9 to make 10, carry the 1 over to 4 and make it 5. This is the only way to get 7.650 on rounding 7.6xy7 to the tenth of a cent. Hence x must be 4 and y must be 9. This statement alone is sufficient to answer the question.

Answer (B)

Hope you see that a few simple rules can make rounding questions quite easy.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!