How to Solve “Unsolvable” Equations on the GMAT

Quarter Wit, Quarter WisdomThe moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.

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

Quarter Wit, Quarter Wisdom: A GMAT Quant Question That Troubles Many!

Quarter Wit, Quarter WisdomWhat determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”.  Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.

Along those same lines:

AB = 10A + B
BA = 10B + A

Going back to our original question:

(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

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

Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT!

Quarter Wit, Quarter WisdomYour first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0
(B) 1
(C) 5
(D) 8
(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

(i) 100-29
100
-29
071

(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0
–  pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

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!

Quarter Wit Quarter Wisdom: What is Your Favorite Number?

Quarter Wit, Quarter WisdomFans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

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

GMAT Tip of the Week

Think Globally, Act Locally, Succeed Exponentially

(This is one of a series of GMAT tips that we offer on our blog.)

As mentioned previously in this space, the GMAT’s quantitative section is increasingly emphasizing problem solving skills over calculation abilities, and often does so in the form of “Number Properties” questions. The authors of the exam are also quite adept at recognizing “mathematical psychology”, and creating questions that increase an examinee’s anxiety by enough to make that process of problem solving a bit more difficult. One of the major themes that arises as a result is the use of exponents, which carry with them a number of properties extremely useful to the writers of the GMAT.

Exponents:

  • Inspire fear (or at least apprehension) in test takers
  • Lead to cumbersome, time-consuming calculations involving large numbers
  • Are actually quite pattern-driven, and reward those who seek out those patterns rather than attempt to perform the extensive calculations

How can this help you on the exam?

If you embrace the pattern-driven quality of exponents, you can rest easy on exponent questions involving large numbers, knowing that you can test the pattern with small numbers, and simply extrapolate it to solve the overall question. Take, for example, a question that asks:

What is the sum of the digits of 10^25 – 37?

Listing out the numbers will be time consuming and contains the potential for error (counting to 25 when writing out the zeroes, then writing out that many digits in the difference, is a tedious process). But if you recognize that the first number will simply be “1” followed by 25 zeroes, and that the difference of the two will end in 63, preceded by a series of 9s, you can make quick work of this problem.

Try using a smaller exponent to see what the result will look like:

10^4 – 37:

10000
– 37
9963

In this case, if the exponent is 4, we end up with 4 digits in the answer: a 6, a 3, and the rest are 9s. Trying again with another exponent, we can see that the pattern holds for any exponent:

10^6 – 37:

1000000
– 37
999963

Again, we have the same number of digits in the difference as the value of the exponent, and two of those are 6 and 3, with the others 9s. So, we can conclude that 10^x – 37 will give us a solution in which we have (x-2) 9s, a 6, and a 3. Because 6+3 is 9, the sum of the digits will be (x-2) * 9 + 9, or (x-1) * 9.

In the original example, the exponent is 25, so we’ll have (25-1) * 9, or 24*9, and the answer is 216.

For more GMAT prep tips and resources, be sure to visit Veritas Prep. And, while you’re at it, follow us on Twitter!