Quarter Wit, Quarter Wisdom: The 3-Step Method to Solving Complex GMAT Algebra Problems

Quarter Wit, Quarter WisdomIf you have been practicing GMAT questions for a while, you will realize that not every question can be solved using pure algebra, especially at higher levels. There will be questions that will require logic and quite a bit of thinking on your part.  These questions tend to throw test-takers off – students often complain, “Where do I start from? Thinking through the question takes too much time!” Unfortunately, there is no getting away from such questions.

Today, let’s see how to handle such questions step-by-step by looking at an example problem:

N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?

(A) 29
(B) 49
(C) 58
(D) 113
(E) 131

This is not a simple algebra question, where we are asked to make equations and solve them.

We are given 6 digits: 1, 2, 3, 6, 7, 8. Each digit needs to be used to form two 3-digit numbers. This means that we will use each of the digits only once and in only one of the numbers.

We also need to minimize the difference between the two numbers so they are as close as possible to each other. Since the numbers cannot share any digits, they obviously cannot be equal, and hence, the smaller number needs to be as large as possible and the greater number needs to be as small as possible for the numbers to be close to each other.

Think of the numbers  of a number line. You need to reduce the difference between them. Then, under the given constraints, push the smaller number to the right on the number line and the greater number to the left to bring them as close as possible to each other.

STEP 1:
The first digit (hundreds digit) of both numbers should be consecutive integers – i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3** (the difference between the latter will certainly be more than 100).

We get lots of options for hundreds digits: (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**). All of these options could satisfy our purpose.

STEP 2:
Now let’s think about what the next digit (the tens digit) should be. To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1, if possible) and the tens digit of the smaller number should be as large as possible (8, if possible). So let’s not use 1 or 8 in the hundreds places and reserve them for the tens places instead, since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?

Let’s try to make a pair of numbers in the form of 2** and 3**. We need to make the 2** number as large as possible and make the 3** number as small as possible. As discussed above, the tens digit of the smaller number should be 8 and the tens digit of the greater number should be 1. We now have 28* and 31*.

STEP 3:
Now let’s use the same logic for the units digit – make the units digit of the smaller number as large as possible and the units digit of the greater number as small as possible. We have only two digits left over – 6 and 7.

The two numbers could be 287 and 316 – the difference between them is 29.

Let’s try the same logic on another pair of hundreds digits, and make the pair of numbers in the form of 6** and 7**. We need the 6** number to be as large as possible and the 7** number to be as small as possible. Using the same logic as above, we’ll get 683 and 712. The difference between these two is also 29.

The smallest of the given answer choices is 29, so we need to think no more. The answer must be A.

Note that even if you try to express the numbers algebraically as:

N = 100a + 10b + c
M = 100d + 10e + f

a lot of thought will still be needed to find the answer, and there is no real process that can be followed.

Assuming N is the greater number, we need to minimize N – M.

N – M = 100 (a – d) + 10( b – e) + (c – f)

Since a and d cannot be the same, the minimum value a – d can take is 1. (a – d) also cannot be negative because we have assumed that N is greater than M. With this in mind, a and d must be consecutive (2 and 1, or 3 and 2, or 7 and 6, etc). This is another way of completing STEP 1 above.

Next, we need to minimize the value of (b – e). From the available digits, 1 and 8 are the farthest from each other and can give us a difference of -7. So b = 1 and e = 8. This leaves the consecutive pairs of 2, 3 and 6, 7 for hundreds digits. This takes care of our STEP 2 above.

(c – f) should also have a minimum value. We have only one pair of digits left over and they are consecutive, so the minimum value of (c – f) is -1. If the hundreds digits are 3 and 2, then c = 6 and f = 7. This is our STEP 3.

So, the pair of numbers could be 316 and 287 – the difference between them is 29. The pair of numbers could also be 712 and 683 – the difference between them is also 29.

In either case, note that you do not have a process-oriented approach to solving this problem. A bit of higher-order thinking is needed to find the correct answer.

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!

The Patterns to Solve GMAT Questions with Reversed-Digit Numbers – Part II

SAT/ACTIn an earlier post, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.

The biggest takeaways from that post were:

  1. Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.
  2. Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.

Take this challenging Data Sufficiency question, for instance:

When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

(1) The integer (M –N) has 12 unique factors.

(2) The integer (M –N) is a multiple of 9.

The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.

First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.

If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.

In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.

Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:

  1. 2^3 * 3^2 (3+1)(2+1) = 12
  2. 2^5 * 3^1 (5+1)(1+1) = 12
  3. 2^2 * 3^3 (2+1)(3+1) = 12

Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.

The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.

Algebraically, if M = 10x + y, then N = 10y + x.

M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).

If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.

Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:

  • Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
  • If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.
  • For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

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 written by him here.

How to Simplify Complicated Combination and Permutation Questions on the GMAT

GMATWhen test-takers first learn how to tackle combination and permutation questions, there’s typically a moment of euphoria when the proper approach really clicks.

If, for example, there are 10 people in a class, and you wish to find the number of ways you can form a cabinet consisting of a president, a vice president, and a treasurer, all you need to do is recognize that if you have 10 options for the president, you’ll have 9 left for the vice president, and 8 remaining for the treasurer, and the answer is 10*9*8. Easy, right?

But on the GMAT, as in life, anything that seems too good to be true probably is. An easy question can be tackled with the type of mechanical thinking illustrated above. A harder question will require a more sophisticated approach in which we consider disparate scenarios and perform calculations for each.

Take this question, for example:

Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

A) 84
B) 91
C) 100
D) 105
E) 243

It’s natural to see this problem and think, “All I have to do is reason out how many options I have for each digit. So for the hundreds digit, I have 3 options (7, 8, or 9); the tens digit has to be different from the hundreds digit, and it must be non-zero, so I’ll have 8 options here; then the last digit has to be odd, so…”

Here’s where the trouble starts. The number of eligible numbers in the 700’s will not be the same as the number of eligible numbers in the 800’s -if the digits must all be different, then a number in the 700’s can’t end in 7, but a number in the 800’s could. So, we need to break this problem into separate cases:

First Case: Numbers in the 700’s  
If we’re dealing with numbers in the 700’s, then we’re calculating how many ways we can select a tens digit and a units digit. 7___ ___.

Let’s start with the units digit. Well, we know that this number needs to be odd. And we know that it must be different from the hundreds and the tens digits. This leaves us the following options, as we’ve already used 7 for the hundreds digit: 1, 3, 5, 9. So there are 4 options remaining for the units digit.

Now the tens digit must be a non-zero number that’s different from the hundreds and units digit. There are 9 non-zero digits. We’re using one of those for the hundreds place and one of those for the units place, leaving us 7 options remaining for the tens digit. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, there are 4*7 = 28 options in the 700’s.

Second Case: Numbers in the 800’s
Same logic: 8 ___ ___. Again, this number must be odd, but now we have 5 options for the units digit, as every odd number will obviously be different from the hundreds digit, which is even (1, 3, 5, 7, or 9). The tens digit logic is the same – 9 non-zero digits total, but it must be different from the hundreds and the units digit, leaving us 7 options remaining. If there are 5 ways we can select the units digit and 7 ways we can select the tens digit, there are 5*7 = 35 options in the 800’s.

Third Case: Numbers in the 900’s
This calculation will be identical to the 700’s scenario: 9___ ___. For the units digit, we want an odd number that is different from the hundreds digit, giving us (1, 3, 5, 7), or 4 options. We’ll have 7 options again for the tens digit, for the same reasons that we’ll have 7 options for the tens digit in our other cases. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, then there are 4*7 = 28 options in the 900’s.

To summarize, there are 28 options in the 700’s, 35 options in the 800’s, and 28 options in the 900’s. 28 + 35 + 28 = 91. Therefore, B is the correct answer.

Takeaway: for a simpler permutation question, it’s fine to simply set up your slots and multiply. For a more complicated problem, we’ll need to work case-by-case, bearing in mind that each individual case is, on its own, actually not nearly as hard as it looks, sort of like the GMAT itself.

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 written by him here.

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!

How to Use Units Digits to Avoid Doing Painful Calculations on the GMAT

StudentDuring the first session of each new class I teach, we do a quick primer on the utility of units digits. Imagine I want to solve 130,467 * 367,569. Without a calculator, we are surely entering a world of hurt. But we can see almost instantaneously what the units digit of this product would be.

The units digit of 130,467 * 367,569 would be the same as the units digit of 7*9, as only the units digits of the larger numbers are relevant in such a calculation. 7*9 = 63, so the units digit of 130,467 * 367,569 is 3. This is one of those concepts that is so simple and elegant that it seems too good to be true.

And yet, this simple, elegant rule comes into play on the GMAT with surprising frequency.

Take this question for example:

If n is a positive integer, how many of the ten digits from 0 through 9 could be the units digit of n^3?

A) three
B) four
C) six
D) nine
E) ten

Surely, you think, the solution to this question can’t be as simple as cubing the easiest possible numbers to see how many different units digits result. And yet that’s exactly what we’d do here.

1^3 = 1

2^3 = 8

3^3 = 27 à units 7

4^3 = 64 à units 4

5^3 = ends in 5 (Fun fact: 5 raised to any positive integer will end in 5.)

6^3 = ends in 6 (Fun fact: 6 raised to any positive integer will end in 6.)

7^3 = ends in 3 (Well 7*7 = 49. 49*7 isn’t that hard to calculate, but only the units digit matters, and 9*7 is 63, so 7^3 will end in 3.)

8^3 = ends in 2 (Well, 8*8 = 64, and 4*8 = 32, so 8^3 will end in 2.)

9^3 = ends in 9 (9*9 = 81 and 1 * 9 = 9, so 9^3 will end in 9.)

10^3 = ends in 0

Amazingly, when I cube all the integers from 1 to 10 inclusive, I get 10 different units digits. Pretty neat. The answer is E.

Of course, this question specifically invoked the term “units digit.” What are the odds of that happening? Maybe not terribly high, but any time there’s a painful calculation, you’d want to consider thinking about the units digits.

Take this question, for example:

A certain stock exchange designates each stock with a one, two or three letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be replaced and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes? 

A) 2,951
B) 8,125
C) 15,600
D) 16,302
E) 18,278 

Conceptually, this one doesn’t seem that bad.

If I wanted to make a one-letter code, there’d be 26 ways I could do so.

If I wanted to make a two-letter code, there’d be 26*26 or 26^2 ways I could do so.

If I wanted to make a three-letter code, there’d be 26*26*26, or 26^3 ways I could so.

So the total number of codes I could make, given the conditions of the problem, would be 26 + 26^2 + 26^3. Hopefully, at this point, you notice two things. First, this arithmetic will be deeply unpleasant to do.  Second, all of the answer choices have different units digits!

Now remember that 6 raised to any positive integer will always end in 6. So the units digit of 26 is 6, and the units digit of 26^2 is 6 and the units digit of 26^3 is also 6. Therefore, the units digit of 26 + 26^2 + 26^3 will be the same as the units digit of 6 + 6 + 6. Because 6 + 6 + 6 = 18, our answer will end in an 8. The only possibility here is E. Pretty nifty.

Takeaway: Painful arithmetic can always be avoided on the GMAT. When calculating large numbers, note that we can quickly find the units digit with minimal effort. If all the answer choices have different units digits, the question writer is blatantly telegraphing how to approach this problem.

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 written by him here.