GMAT Tip of the Week: The Song Remains the Same

Welcome back to hip hop month in the GMAT Tip of the Week space, where we’re constantly asking ourselves, “Wait, where have I heard that before?” If you listen to enough hip hop, you’ll recognize that just about every beat or lyric you hear either samples from or derives from another track that came before it (unless, of course, the artist is Ol’ Dirty Bastard, for whom, as his nickname derives, there ain’t no father to his style).

Biggie’s “Hypnotize” samples directly from “La Di Da Di” (originally by Doug E. Fresh – yep, he’s the one who inspired “The Dougie” that Cali Swag District wants to teach you – and Slick Rick). “Biggie Biggie Biggie, can’t you see, sometimes your words just hypnotize me…” was originally “Ricky, Ricky, Ricky…” And right around the same time, Snoop Dogg and 2Pac just redid the entire song just about verbatim, save for a few brand names.

The “East Coast edit” of Chris Brown’s “Loyal”? French Montana starts his verse straight quoting Jay-Z’s “I Just Wanna Love U” (“I’m a pimp by blood, not relation, I don’t chase ’em, I replace ’em…”), which (probably) borrowed the line “I don’t chase ’em I replace ’em” from a Biggie track, which probably got it from something else. And these are just songs we heard on the radio this morning driving to work…

The point? Hip hop is a constant variation on the same themes, one of the greatest recycling centers the world has ever known.

And so is the GMAT.

Good test-takers – like veteran hip hop heads – train themselves to see the familiar within what looks (or sounds) unique. A hip hop fan often says, “Wait, where I have heard that before?” and similarly, a good test-taker sees a unique, challenging problem and says, “Wait, where have I seen that before?”

And just like you might recite a lyric back and forth in your mind trying to determine where you’ve heard it before, on test day you should recite the operative parts of the problem or the rule to jog your memory and to remind yourself that you’ve seen this concept before.

Is it a remainder problem? Flip through the concepts that you’ve seen during your GMAT prep about working with remainders (“the remainder divided by the divisor gives you the decimals; when the numerator is smaller then the denominator the whole numerator is the remainder…”).

Is it a geometry problem? Think of the rules and relationships that showed up on tricky geometry problems you have studied (“I can always draw a diagonal of a rectangle and create a right triangle; I can calculate arc length from an inscribed angle on a circle by doubling the measure of that angle and treating it like a central angle…”).

Is it a problem that asks for a seemingly-incalculable number? Run through the strategies you’ve used to perform estimates or determine strange number properties on similar practice problems in the past.

The GMAT is a lot like hip hop – just when you think they’ve created something incredibly unique and innovative, you dig back into your memory bank (or click to a jazz or funk station) and realize that they’ve basically re-released the same thing a few times a decade, just under a slightly different name or with a slightly different rhythm.

The lesson?

You won’t see anything truly unique on the GMAT. So when you find yourself stumped, act like the old guy at work when you tell him to listen to a new hip hop song: “Oh I’ve heard this before…and actually when I heard it before in the ’90s, my neighbor told me that she had heard it before in the ’80s…” As you study, train yourself to see the similarities in seemingly-unique problems and see though the GMAT’s rampant plagiarism of itself.

The repetitive nature of the GMAT and of hip hop will likely mean that you’re no longer so impressed by Tyga, but you can use that recognition to be much more impressive to Fuqua.

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

By Brian Galvin.

2 Tips to Make GMAT Remainder Questions Easy

stressed-studentSeveral months ago, I wrote an article about remaindersBecause the concepts of division, quotients, and remainders show up so often on the GMAT, I thought it would be useful to revisit this frequently-tested topic. (After all, there always remains at least something unsaid about remainders, right?) When you encounter quotient/remainder problems on the GMAT, at times, it will be helpful to know the kind of division terminology we’re taught in grade school – in particular the quotient + remainder formula you’ll see detailed below – while at other times, you will simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement.

To ensure that you’re prepared for all types of quotient/remainder problems on the GMAT, let’s explore each of these division-related scenarios in a little more detail. A simple example can illustrate the important division terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + 3/4.

7, the term we’re dividing by something else, is called the dividend. 4, which is doing the dividing, is called the divisor. 1, the whole number component of the mixed fraction, is the quotient. And 3 is the remainder. This probably feels familiar even if the terminology takes a little reminding to come back to you.

In the abstract, the classic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get another helpful variant of the remainder formula: Dividend = Quotient*Divisor + Remainder.

Simply knowing this terminology and the remainder equation will be sufficient to answer the following official GMAT question:

When N is divided by T, the quotient is S and the remainder is V. Which of the following expressions is equal to N? 

A) ST
B) S + V
C) ST + V
D) T(S+V)
E) T(S – V) 

In this division problem, N – which is getting divided by something else – is our dividend, T is the divisor, S is the quotient, and V is the remainder. Plugging the variables into our remainder equation of Dividend = Quotient*Divisor + Remainder, we get N = ST + V… and we’re done! The answer is C.

(Note that if you forgot the remainder formula, you could also pick simple numbers to solve this problem. Say N = 7 and T = 3. 7/3 = 2 + 1/3.  The Quotient is 2, and the remainder is 1, so V = 1. Now, if we plug in 3 for T, 2 for S, and 1 for V, we’ll want an N of 7. Answer choice C will give us an N of 7, 2*3 + 1 = 7, so this is correct.)

When we need to generate a list of potential values to test in a data sufficiency question, often a statement will give us information about the dividend in terms of the divisor and the remainder.

Consider, for example, the following question: when x is divided by 5, the remainder is 4. Here, the dividend is x, the divisor is 5, and the remainder is 4. We don’t know the quotient, so we’ll just call it q. In equation form, it will look like this: x = 5q + 4. Now we can generate values for x by picking values for q, bearing in mind that the quotient must be a non-negative integer.

If q = 0, x = 4. If q = 1, x = 9. If q=2, x = 14. Notice the pattern that emerges with our x values: x = 4 or 9 or 14, or 19… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. Without much math at all, you could continue this cycle indefinitely: 4, 9, 14, 19, 24, 29, etc. This is a handy shortcut to use in complicated data sufficiency problems, such as the following:

If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

1) When x – y is divided by 5, the remainder is 1
2) When x + y is divided by 5, the remainder is 2

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
(C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
(D) EACH statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient

In this problem, Statement 1 gives us potential values for x – y. Remember the pattern discussed above: x – y must be 1 greater than a multiple of 5. If we begin with the remainder (1) and continually add the divisor (5), we know that x – y = 1 or 6 or 11, etc. If x – y = 1, we can say that x = 1 and y = 0. In this case, x^2 + y^2 = 1 + 0 = 1, and the remainder when 1 is divided by 5 is 1. If x – y = 6, then we can say that x = 7 and y = 1. Now x^2 + y^2 = 49 + 1 = 50, and the remainder when 50 is divided by 5 is 0. Because the remainder changes from one scenario to another, Statement 1 is not sufficient by itself.

Statement 2 gives us potential values for x + y. Once again, let’s use that pattern: x + y must be 2 greater than a multiple of 5. If we begin with the remainder (2) and continually add the divisor (5), we know that x + y = 2 or 7 or 12, etc. If x + y = 2, we can say that x = 1 and y = 1. In this case, x^2 + y^2 = 1 + 1 = 2, and the remainder when 2 is divided by 5 is 2. If x + y = 7, then we can say that x = 7 and y = 0. Now x^2 + y^2 = 49 + 0 = 49, and the remainder when 49 is divided by 5 is 4. Because the remainder changes from one scenario to another, Statement 2 is also not sufficient on its own.

Now that we find ourselves in the classic C or E scenario, let’s test them together – simply select one scenario from Statement 1 and one scenario from Statement 2 and see what happens. Say x – y = 1 and x + y = 7. Adding these equations, we get 2x = 8, or x = 4. If x = 4, y = 3. Now x^2 + y^2 = 16 + 9 = 25, and the remainder when 25 is divided by 5 is 0.

Now remember this: for us to pick a non-E answer on Data Sufficiency, we must know that the value will stay the same in any scenario allowed by the question and statements. To be safe, let’s try another scenario. Say x – y = 6 and x + y = 12. Adding the equations, we get 2x = 18, or x = 9. If x = 9, y = 3, and x^2 + y^2 = 81 + 9 = 90. The remainder when 90 is divided by 5 is, again, 0. No matter which values we select, this will be the case – we can prove definitively that the remainder is 0. Together, the statements are sufficient, so the correct answer is C.

Now let’s summarize some important takeaways regarding GMAT quotient/remainder problems and the ever-important remainder formula. You’re virtually guaranteed to see remainder questions on the GMAT, so you want to make sure you have this concept mastered. First, make sure you feel comfortable with the remainder formula: Dividend = Divisor*Quotient + Remainder. Second, if you need to select values, you can simply start with the remainder and then add the divisor over and over again. If you internalize these two ideas, remainder questions will become considerably less daunting.

Remember, also, that division is something you were once quite good at as a primary school student, so do not let the terminology intimidate you as an adult! The GMAT thrives on abstraction in these problems, so if you find yourself distracted by terminology or abstraction, simply try using small numbers to remind yourself how the operation works. The remainder equation Dividend = Divisor*Quotient + Remainder is an important one, but if you blank on it you can reconstruct it. Try, as we did at the beginning, 7 divided by 4. The result of that is 1, remainder 3. And the quotient is 1 because 4 goes into 7 one time, leaving 3 left over. So you can reconstruct the equation: to get back to 7, multiply the divisor (4) by the 1 time it went in to 7, and then add back the remaining 3: 7 = 4(1) + 3. Once you’ve stripped away the abstraction in quotient/remainder problems, the remainder is a concept you’ve been quite adept with your whole life!

Interested in more practice with GMAT division problems and the remainder equation? Check out some of our other articles on this frequently-tested GMAT topic, or try your hand at some practice questions via the Veritas Prep GMAT Question Bank or practice tests.

*GMATPrep questions courtesy of the Graduate Management Admissions Council.

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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.

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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!