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Permutations and Combinations Part II: Formulas

Friday, March 19th, 2010

Once you’ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator.

As I mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items.  Let’s dive straight into the formulas then go through a few examples.

The number of ways of permuting r objects out of n objects is given by

n P r =  n!/(n-r)! where n! = n * (n-1) * (n-2) * … * (2) * (1)

The number of ways of combining r objects out of n objects is given by

n C r = n!/r!(n-r)!

Note that 0! is defined as 1.

Example 1

How many ways are there of arranging 4 letters out of the following F R I E N D?

FRIEND has 6 different letters, and since the order of letters matter, I know I have to use the permutation formula.  Applying the formula directly, the answer is given by

6 P 4 = 6!/(6-4)! -= 6!/2! =360

You can double-check this by using the fundamental counting principle that we covered in Part I.  Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger.

Example 2

If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entrée and 2 desserts?  Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so we know to use the combination principle.  Because we are choosing entrees and desserts separately, we have to apply the combination formula twice.

First, to choose 1 entrée out of 3, we apply the formula 3 C 1 = 3!/1!(3-1)! = 3!/1!2! = 3

Next, to choose 2 desserts out of 5, we apply the formula 5 C 2 = 5!/2!(5-2)! = 5!/2!3! = 10

Then because for each entrée, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entrée, to get the total number of possibilities, we take 3*10 = 30.

Example 3

What if the question throws you a curveball and asks you to permute something that has a repeated item.  For example, how many ways are there of arranging the letters A G H A S T?

The ‘A’ is repeated twice so first we pretend that the letters are distinct and find the number of possibilities.  Then divide that value by 2! because ‘A’ is repeated twice.

AGHAST has 6 letters, and if we permute all the letters, we get 6!

Because A is repeated, we divide 6! by 2! to get the answer 360

Supposing I asked you to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 P 4 as per normal, and divide that answer by 2! since A is repeated.  The answer should be 180.

Check out Grockit for more GMAT permutations and combinations practice!

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GMAT Solving Linear Equations with 1 Unknown

Wednesday, March 17th, 2010

A linear equation is any equation where the highest power of the unknown, which I shall call x, is 1.  To illustrate more clearly with a few examples:

x+1 = 4; 10x = 3; x = 18 – 4x are three examples of linear equations

x2 + 2 = 2x and x3 = 8 are not linear equations because there are x’s that are raised to a higher power than 1.

A linear equation with 1 variable is the simplest type to solve.  There is 1 equation and 1 unknown, which means that the unknown can always be determined.  To solve such an equation, you need to rearrange the equation to have like terms on either side of the equal sign.  Put another way, you are trying to isolate x (or whatever the variable is called) on one side of the equation.

For example, if 2x = 234, to isolate x, we have to divide the entire equation by 2.  Doing this, we get x = 117.

If there are x’s and numbers on either side of the equal sign, we add and subtract values to isolate x on one side.  Suppose 2x – 17 = 18 – 3x

The first thing we could do is to add 17 to both sides to get: 2x – 17 + 17 = 18 – 3x + 17

This reduces to 2x = 35 – 3x

Now, we need to have all the x’s on one side so we add 3x to both sides to get: 2x + 3x = 35 – 3x + 3x

This reduces to 5x = 35

Dividing by 5 on both sides, we get x = 7

What I just went through was a fairly simple algebraic equation.  The questions on the GMAT will look more complicated but you are essentially doing the same thing: manipulating both sides of the equation in the same way to isolate x.  Let’s try a practice problem from Grockit.

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To tackle this question, we multiply both sides by 2+3/x to get 3 = 2+3/x. .  (This is also known as cross multiplying where in general if a/b-c/d,

then ad = bc

To simplify 3 = 2+3/x,

we multiply the entire equation by x to get 3x = 2x + 3.  This leaves you with a much simpler equation that you already know how to solve.

What’s a little trickier than manipulating algebraic equations is translating a word problem into an algebraic equation.  Here’s another practice problem:

Jack and his brother are sharing a monster piece of licorice that is 28 inches long. Since Jack is older, his share is 8 inches longer than his brother’s. How long, in inches, is Jack’s brother’s piece?

The way to solve this problem is to let something be x.  Here’s what happens if we let Jack’s piece be x inches.

Jack’s piece = x inches

Jack’s brother’s piece = x – 8 inches

Total length of licorice = Jack’s piece + Jack’s brother’s piece = 28 = x + (x-8)

This means that x = 18 inches.  But remember that the question wants the length of Jack’s brother’s piece, which we have defined as x – 8.  So the correct answer is 10 inches.

Here’s what happens if we let Jack’s brother’s piece be x inches.

Jack’s brother’s piece = x inches

Jack’s piece = x + 8 inches

Total length of licorice = 28 = x + (x+8) and we determine that x = 10.  In this case, since we have already defined Jack’s brother’s piece to be x, there is no further step we need to take.

In general, here are a few things to keep in mind.

  • if there is only one unknown, you only need one equation to determine the value of the unknown
  • in dealing with algebraic equations, remember that anything you do to one side (be it adding, subtracting, multiplying or dividing) you need to do to this other side too.
  • in dealing with word problems, define something to be x and see if you can define other things in terms of x only.  (For example, in the question about licorice, you would not want to let Jack’s piece be x inches and his brother’s be y inches)  Don’t introduce unnecessary variables if it can be expressed in terms of an existing variable.

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Permutations and Combinations Part I: Counting

Friday, March 12th, 2010

When dealing with permutations and combinations, you are essentially trying to find the number of different outcomes given a set of items and a number of restrictions.  The difference between permutation and combination merely depends on whether the order matters.  Let me illustrate with an example.  Suppose you have three food items, apples, bananas and carrots.  If you were asked to only pick two of them, how many possibilities are there?  If you got 3, that’s the right answer.  You have apples & bananas, apples & carrots and bananas and carrots.  This is combination.  If I told you however, that the order in which you eat the food matters, then you have more possibilities, because instead of just apples & bananas, you have to consider bananas & apples too.  The latter is permutation.

Listing

What I did earlier, when I listed out the 3 choices is called, not surprisingly, listing.  But there is a method to listing to ensure that you don’t leave any possibilities out.  Everyone tends to have his or her own method, let me share mine.  If I have four items – a, b, c and d – and I’m supposed to choose 2:

Step 1: I take the first item a and combine it with b. Then I go down the list and combine it with c and d to get a total of three possibilities: ab, ac, ad.

Step 2: I take the second item b and combine it with the rest of the list.  I don’t combine it with a because I already did that in step 1.  So I get: bc, bd

Step 3: Keep doing the same thing for the next thing on the list, which is c.  For c, there’s only one thing behind it on the list – d.  So there’s only 1 possibility here: cd

Step 4: Keep going until you get to the second last item.  In this case, we already reached it in step 3.  Tally up your choices: ab, ac, ad, bc, bd, cd

Tree Diagram

If you noticed, listing is a good method for figuring out combinations.  The Tree Diagram is a graphical method that helps you with permutation. Using the previous example of 4 items, here’s the tree diagram that illustrates the number of ways of permuting items.

3

If you look at the last line of the tree diagram and count the number of boxes, you will see that there are 12 possible ways to permute 2 items out of 4.  Each item can be combined with 3 other things.  Another way of seeing this is that you have two spaces you need to fill ___ and ___.  Looking at the first space, you have 4 possible items you could place there.  Looking at the second space, you have 3 possible items you could place there, since you have already put 1 item in the first space.  So there are 4 x 3 = 12 ordered possibilities.

Fundamental Counting Principle

The spaces-and-number-of-options-to-fill-that-space method of thinking is essentially the fundamental counting principle.  The principle states that

if task A can be done in m ways and after Task A is complete, Task B can be done in n ways, then there are m*n ways of completing Task A then Task B.

Let’s see if you understand the principle with a quick example.  How many different ways are there are arranging these letters: T A N G O

Did you get 120?

Imagine five spaces ___   ___   ___   ___   ___

You have 5 possible letters to put in the first space, and when that’s done, you have 4 possible letters to put in the second space, 3 letters for the third space and so on.  So the answer is 5*4*3*2*1 or 5!

If you’re comfortable using these methods to solve permutation and combination problems, stay tuned for the second installment that uses direct formulas to calculate the number of possibilities. In the meantime, check out Grockit for GMAT quantitative practice.

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Interest and Compound Interest Problems

Wednesday, March 10th, 2010

Interest and Cmoneyompound Interest

There are two types of interest problems on the GMAT, and they include simple interest and compound interest. Simple interest is the most basic and is a function of P, the principle amount of money invested, the interest rate earned on the principle, i, and the amount of time the money is invested, t (this is usually stated in periods, such as years or months). The resulting equation is:       I = iPt

In basic terms, the above equation tells us the amount of interest that would be earned on a principle amount invested (P), for a given time (t) at a given interest rate (i).

Example:

If you invested $1,000 (P = your principle) for one year (t = one year) at 6% simple interest (i = given interest rate), you would get $60 in interest at the end of the year and would have a total of $1,060.

For compound interest, you would earn slightly more.

Let’s look at similar type problem, though this one involves compound interest.

Mr. Riley deposits $500 into an account that pays 10% interest, compounded semiannually. How much money will be in Mr. Riley’s account at the end of one year?

For compound interest, first you need to divide the interest rate by how many compound periods there are. So for in the above question, because we are compounding semiannually, we need to divide 10% by 2 (because of 2 compounding periods), and if we were compounding quarterly, we would need to divide 10% by 4.

In the above question, Mr. Riley deposited $500 into his account at a rate of 10% compounded semiannually and the bank will divide his interest into two equal parts. They will pay 5% interest (10%/2) at the end of six months, and then will pay another 5% at the end of the year. Compound interest can essentially be translated into “interest paid on interest”, meaning that after one period, you are paid interest on the interest that was paid in prior periods, hence the phrase “compounding”.

So at the end of the six months, Mr. Riley has $525 because the bank paid $25 in interest ($500*5%) into his account. For the second half of the year, Mr. Riley is then paid 5% on the $525 balance that was in his account at the end of the first six months. This interest is equal to $525*5% = $26.25. Therefore, at the end of the year, Mr. Riley has $551.25, which is equal to his balance of $500, plus the $25 interest paid at the end of 6 months, plus $26.25 paid at the end of the year. Mr. Riley earns $1.25 more with this compound interest than he would have been paid if he were paid only 10% simple interest (would have been only $550). The lesson? Compound interest always pays more!

Let’s look at another similar type of problem that involves interest.

Money invested at x%, compounded annually, triples in value in approximately every 112/x years. If $2500 is invested at a rate of 8%, compounded annually, what will be its approximate worth in 28 years?

A. $3,750

B. $5,600

C. $8,100

D. $15,000

E. $22,500

At first glance, this one seems pretty tricky because you are given x% as the interest rate and it asks you about compounding and it might seem difficult where to find a starting point for this. For this one, it might be a bit easier to think about this without the use of compound interest, which might unnecessarily confuse you. Here, we are given x% as 8%, so all we need to do is take 112/8 = 14. Thus, we know that the money triples in value every 14 years. Further, we know that the money will triple exactly twice in 28 years, once in 14 years and one more time at the 28th year. So first we need to multiply the original $2500 invested by 3 to get the balance at the end of year 14 (because it triples), to get $7,500 (or $2,500*3). Now, we know that this balance of $7,500 will triple again, so the final balance at the end of the next 14 year period will be $22,500 (or $7,500*3). The correct answer choice is E.

Overall, the three types of interest problems you will most likely encounter come test day will be simple interest, compound interest, and word problems involving the mention of interest, but that can be solved without the application of interest or compound interest methods. The key to deciphering between compound interest and simple interest is to see how many periods the interest is paid….interest paid in one period is simple interest and interest “paid on interest” in multiple periods is compound interest. Finally, remember that some questions can be solved intuitively.

Check out Grockit for more GMAT quantitative practice!

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Rectangular Solids and Cylinders

Monday, March 1st, 2010

Questions involving rectangular solids, particularly data sufficiency questions, test whether you understand the concept of volume and surface area.   You essentially need to remember that you need three different values to find volume and surface area (the length, the width and the height).  If the prompt and statements 1 and 2 are lacking some these values or some way to find them, neither of the statements will be sufficient.

A rectangular solid is formed by 3 pairs of similar rectangular faces.  In other words, 6 rectangular faces in total.

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The formulas you need to remember for a rectangular solid are

Volume = Length (l) x Width (w) x Height (h)

Surface Area = (2 x Length x Width) + (2 x Length x Height) + (2 x Width x Height)

If length = width = height, that means that the rectangular solid is, in fact, a cube.

Other vocabulary that might be important is the terms vertex and edge. A vertex is a mathematical way of referring to the corner of any figure.  The rectangular solid above has 8 vertices (plural of vertex), can you identify them?  The edge is simply the lines you see in the diagram above: the line where two surfaces meet.

Questions involving cylinders are similar and perhaps easier because there are only two values you need to know to solve cylinder-problems – the radius (r) and the height (h).

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If you don’t know the radius, anything that enables you to determine the radius, such as the diameter (radius = diameter / 2) or the circumference (radius = circumference / 2pi) will suffice.

Regarding cylinders, the formulas you need to know are

Volume = area of the base circle x height = pi x (radius)2 x height

Surface Area = (2 x pi x (radius)2 )+ (pi x (diameter) x height)

Let’s try a problem: A cylindrical water tank has a stripe painted around its circumference, as shown in the figure provided. What is the surface area of this stripe?
(1) y = 0.7
(2) The height of the tank is 2 meters.

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To find the surface area of the stripe, you need to know the circumference of the cylinder, but there is not data in the question that gives you the radius or diameter to let you find the circumference.  Hence the answer should be that neither statement together is sufficient.

Posted in Data Sufficiency, GMAT, Quantitative | No Comments »

Weighted Averages On The GMAT

Friday, February 26th, 2010

This post will introduce weighted average questions you’ll see on the GMAT.  There is one main formula you need to solve simple GMAT Average questions:6

  • Average = SUM / # of observations

Note that this can be rearranged to read:

  • SUM = Average x (# of obs)
  • # obs = SUM / Average

So, if you are given ANY 2 of the 3 values, you should be able to find the 3rd. For example:

John drinks an average of 1.5 cups of water/day. After how many days has he drank 3 gallons of water? (1 gallon = 16 cups.)

In this case, we are looking for the number of days (or observations) such that we total 48 cups (3 gallons) of water.

# = SUM / Average

  • # days = 48 cups / 1.5 cups/day
  • # days = 32 days

NEVER AVERAGE AVERAGES!

Class A  has 15 students and an average height of 60”. Class B has 20 students. What is class B’s average height if the average height of both classes is 65”.

One might say:  (A + B) / 2 = 65”; A = 60”; so B must be 70”. However, keep in mind:

TOTAL AVERAGE = TOTAL SUM / TOTAL OBS

CLASS A + CLASS B = BOTH

  • 15 students + 20 students = 35 students
  • 60” average + 68.75” average = 65” average
  • 900” total in A + 1375” total in B = 2275” total in Both

The given information is in black. The necessary intermediate steps are in blue, and the red is your answer. Note that the average of the averages ≠ total average. We must calculate each average separately, and to do this we need the SUM and # of observations for each category. This brings us to the idea of WEIGHTED AVERAGES.

A WEIGHTED AVERAGE is needed when you are taking average of a large group in which there are subgroups with a different number of observations in each. Take a look at this generalized formula, assuming there are 3 groups A, B and C.

(Average of A x Obs in A) + (Average of B x Obs in B) + (Average of C x Obs in C)

(Obs in A) + (Obs in B) + (Obs in C)

Think of weighted averages like a tug of war between numbers. The “stronger” one side (dog) is, the more that weighted average (tennis ball) will be “pulled” in that direction.

In the previous question, we had:

CLASS A + CLASS B = BOTH

  • 15 students + 20 students = 35 students
  • 60” average + 68.75” average = 65” weighted average

Note that the weighted average is CLOSER to B’s average than it is to A’s. This is because there are 20 students in Class B compared to only 15 students in Class A.

Two More Examples

At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2

B. 5

C. 9

D. 15

E. 30

WITHOUT using the formula, we can see that today the restaurant served 30 customers above the average. The total amount ABOVE the average must equal total amount BELOW the average. This additional 30 customers must offset the “deficit” below the average of 90 created on the x days the restaurant served only 75 customers per day.

30/15 = 2 days. Choice (A).

WITH the formula, we can set up the following:

  • 90 = (75x + 120)/(x + 1)
  • 90x + 90 = 75x + 120
  • 15x = 30

x = 2  Answer Choice (A)

Use whichever makes more sense to you!

Anita spent a total of $780 on 52 bottles of wine for her wedding. She then decided to buy 8 bottles of sparkling wine for the toasts, as well. Was the average (arithmetic mean) price per bottle of wine less than $20?

(1) Each bottle of sparkling wine cost more than $15.

(2) Each bottle of sparkling wine cost less than $40.

Take another look at what exactly the question is calling for: the TOTAL average price of all the wine at the wedding. We should look at the suggested average ($20) and use that as our threshold amount.

  • 60 bottles * $20/bottle = $1200 total
  • $1200 total – $780 (given) = $420 (left for sparkling wine)
  • $420 / 8 bottles = $52.50/bottle of sparkling wine (for the total average to equal $20)

Which of the answer choices are conclusively above or below $52.50/bottle of sparkling wine? Only (2). With (1), we can be below OR above the threshold, so (1) is not sufficient.

Answer Choice (B)

Now, you’re score will be above average! Please visit the Grockit forum or leave a comment here if you have more questions on weighted averages.

Good luck!

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Estimation Strategies For GMAT Problem Solving

Tuesday, January 19th, 2010

The GMAT is not testing who is the fastest at long division. It is a test that seeks to measure problem solving skills that are not necessarily the “textbook” ways to discover solutions. Let’s discuss some estimation strategies, which are not used as often as they should be.

1. Round Up AND Round Down When Multiplying

Be aware of the direction in which you are altering the result. If you want to estimate a product of two “ugly” numbers, you can move one up and one down, which is an attempt to minimize the error in your estimation. For example:

658*436 = 286,888

If we round 658 UP to 700 and 436 DOWN to 400, we can approximate using:

700*400 = 280,000

2. Round In The Same Direction When Dividing

When you want to approximate a fraction, you can either adjust only the numerator (or denominator) or move both in the same direction. For example:

8/19 = .4210526…

8/20 = 0.4 (Note that increasing the denominator, will decrease the fraction.)

9/20 = 0.45 (Note that increasing both top and bottom will increase the fraction.)

Your estimate is somewhere between .40 and .45.

3. Remember These Other Helpful Tips

  • Peek at your answer choices: If your answer choices are relatively far apart, this could be hint that approximation is helpful. If the answers are very tight together, you may still estimate, but you have to be more careful and do due diligence.
  • Geometry shortcut 1: √2 =~ 1.4 and √3 =~ 1.7. Try to commit these to memory, as they are very common.
  • Geometry shortcut 2: Be careful when using pi = 3. Recognize that you are using a smaller number, so your result will be smaller too. Test makers love to give tempting answer choices that assume pi = 3. It’s not.
  • Geometry shortcut 3: Even though you cannot assume charts are drawn to scale, they can still be a resource. Obtuse/acute angles are typically shown as much, and angles can be approximated in many circumstances. That’s not to say “if it looks like a right angle, it must be 90.” But you can use the drawing as a guide to your estimation.
  • Use the extremes: If you are given a range, it helps to plug in those extremes to see between which values your answer falls. This will focus your attention on the cases that are above (or below) those endpoints.

Two Examples

If a square has a perimeter of 80 inches, what is the approximate length of its diagonal, in inches?

A. 20

B. 28

C. 40

D. 56

E. 112

This question uses the word “approximate,” so that should be a very big hint that you will need to find a number “close enough.” If P = 80, then s = 20. The diagonal is essentially a hypotenuse of a 45-45-90 triangle, so d = 20√2.

Two strategies:

1) 20√1 = 20 and 20√4 = 40. Therefore 20 < 20√2 < 40. (B) 28 is the only option.

2) Since we remember that √2 =~1.4, we can simply multiply 20*1.4 = 28. (B).

Addison High School’s senior class has 160 boys and 200 girls. If 75% of the boys and 84% of the girls plan to attend college, what percentage of the total class plan to attend college?

A. 75

B. 79.5

C. 80

D. 83.5

E. 84

84 is an obscure number. When you see obscure numbers, that is another sign that you may want to look for an approximating shortcut.

Firstly, we should eliminate the overtly incorrect choices. This will be (A) 75 (since that’s the low extreme) and (E) 84 and (D) 83.5 (since they are both essentially equal to the high extreme).

Secondly, find the average of the given percents. Since there are more girls than boys, we know that the weighted average will be closer to the girls’ percent than the boys’ percent. By finding 79.5% as the mean of 75% and 84%, we are given the low extreme. Again, we recognize the weight placed on 84%, making the answer higher than 79.5. (C) 80 it is!

(For similar questions in the future where we actually need to calculate, we could drop the extra “0” from 160 and 200. The ratio of 16:20 is the same (4:5), and the calculation is much easier.)

Any other estimation tricks? Just post in the comment field or check out Grockit’s forums for more strategies on GMAT math.

Posted in GMAT, Problem Solving, Quantitative | No Comments »

Sets: Matrix or Venn?

Tuesday, January 5th, 2010

Sets, Matrices and Venn Diagrams: They are all the Same

It’s really a matter of preference; some students like to use Venn diagrams, and others make matrices. Personally, I prefer a matrix format, but there is no “better” way.

On your GMAT, you will encounter 1-3 questions that contain overlapping groups with specific characteristics. You will almost never see more than two characteristics (since you can’t draw 3D on your scratch paper). For illustration, let’s take a look at the following Data Sufficiency example:

Of the 70 children who visited a certain doctor last week, how many had neither a cold nor a cough?

(1) 40 of the 70 children had a cold but not a cough.

(2) 20 of the 70 children had both a cold and a cough.

There are two characteristics (cough and cold) and two categories for each (yes and no), so there are four total categories, as indicated by this matrix:

1

I’ve filled in the given information from both statements, and the parenthetical information is inferred. This clearly lays out the 4 combinations of options. If we sum vertically, we can infer that there are 60 total children with colds. Because there are 70 total children, this also means that 10 do NOT have colds. The bottom-right quadrant cannot be found because we do not know how those 10 children get divided between the two empty boxes. Choice E – together the statements are insufficient.

We may also visualize the question as Venn diagram, in which there are still two characteristics, represented by overlapping circles. You will notice that there are still two undefined regions, so the given information is insufficient.

2

For any Data Sufficiency or Problem Solving Set question, map out the provided information and mark the region that you need to find. Note that there may be implicitly defined regions, such as “60 children have a cold” above. Let’s look at one more example in matrix format.

Each of the dogs in a certain kennel is a single color.  Each of the dogs in the kennel either has long fur or does not.  Of the 45 dogs in the kennel, 26 have long fur, 17 are brown, and 8 are neither long-furred nor brown. How many long-furred dogs are brown?

  1. A. 26
  2. B. 19
  3. C. 11
  4. D. 8
  5. 6

3

Again, we are provided a small amount of overlapping information and our matrix can simplify our visualization. The parenthetical information is inferred. In fact, in order to have found the correct answer (6) in the top-left quadrant, we had to have derived either of the two empty quadrants.

Of all the types of Quantitative questions on the GMAT, overlapping sets are some of the most like puzzles. This should make you very, very excited. Because who doesn’t like puzzles?

Got a tough set question? Post below or email me at jakeb@grockit.com!

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Sequences On The GMAT Math Section

Monday, December 28th, 2009

There are four main categories of Sequences that appear GMAT Quantitative Section: Arithmetic, Geometric, Repeating and Sums. Think of sequences as a simple pattern, and detecting this pattern is probably the most difficult part.

Arithmetic Sequences

An Arithmetic Sequence is when subsequent terms in a sequence increase (or decrease) by a constant amount. Here’s the standard formula:

  • a_n = a_1 + (n – 1)d, where a_n is the value at term n, and d is the constant change.

Note that a_n – a_(n-1) = d

If given the following sequence, we can derive both a_1 and d, to solve for any term.

8, 11, 14, 17, 20, 23…..    (a_1 = 8 and d = 3)

So, if asked what term number 86 is, we can just plug in to the formula:

a_n = a_1 + (n – 1)d

a_n = 8 + (86 – 1)3 = 8 + 85*3 = 8 + 255 = 263

Geometric Sequences

The same principle applies to Geometric Sequences, in which each subsequent term is multiplied by a certain constant. Compound interest in an example of a geometric sequence. Here’s the standard formula.

a_n = a_1 * r^(n-1), where a_n is the value at term n, and r is multiplicative rate of increase.

Note that a_(n+1) / a_n = r

Similarly, when given the rate of increase and the value of any term, we can find any other term. For example:

John originally put $8 in his piggy bank in 2001. If his parents double the money in the piggy bank once a year (without John adding anymore himself), how much money will he have in 2009?

r = 2

a_n = a_1 * r^(n-1)

a_1 = 8

a_n = 8*2^7

n = 8

a_n = $1,024

Repeating Patterns

Many times, repeating patterns will yield a remainder question.

The first term of a sequence is  -2 and the second term is 2. Each subsequent odd term is found by adding 2 to the previous term, and each subsequent even terms is found by multiplying the previous term by -1. What is the sum of the first 669 terms?

Clearly, we are not looking to enter all 669 terms and see what the last one is. But, we can do a few and check out the pattern.

n = 1, a_n = -2

n = 2, a_n = 2

n = 3, a_n = 4

n = 4, a_n = -4

n = 5, a_n = -2

n = 6, a_n = 2

Since we can see that the pattern will repeat every 4 terms, we can solve for the remainder after dividing 669/4. Since 4 goes into 668 evenly, we know that the value will be equivalent to that of the first term, which = -2.

Sums

With sequence questions involving sums, identify the pattern. This is the most important first step toward finding the solution. Check out the following example.

A set of consecutive integers begins at -19. After which term will the sum of all the terms equal 41?

Again, we don’t want to write all these out. BUT, we can start, and try to see if we can detect the pattern. In this particular example, we will be adding integers in the negative space until 0, after which we will be adding the positive pair of the same integers all the way up to +19. The sum after +19 equals zero, equals 20 after +20 and equals 41 after +21.

That means 19 negative terms + 1 (zero) + 21 positive terms = 41 total terms.

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All About Remainders

Tuesday, December 22nd, 2009

Remainders are the NUMERATOR of a fraction from a mixed number that results from division. For example, 19/3 leaves a remainder of 1, since 19/3  = 6 1/3.

Some quick tips:

  1. Your remainder can only range from zero to the denominator of the fraction. For example, when dividing by 9, your remainder options are 0-8, since a remainder of 9 leaves you a new whole number (with a remainder of 0).
  2. Look for the closest whole number and count up or down from there. For example, when trying to find the remainder of 146/15, you can see that 15 would go into 150 evenly. You then count down four from 150 to 146, so your remainder is (15 -4) = 11. This is easier than recognizing 135/15 is a whole number and counting up.
  3. Become familiar with common trends or patterns. For example, multiples of even numbers are even, so 167/(even #) must have an ODD remainder.
  4. The remainder should NOT be reduced. 18/4 = 4 2/4. The remainder stays equal to 2, even though you can reduce 4 2/4 to  4 1/2.

I recently came across this question, which I think is a good introduction:

What is the remainder of 3^(4n+3) divided by 5, assuming n is a positive integer?

Firstly, when dividing by 5, we are looking for the remainder above a one’s digit of either 0 or 5. In this scenario, we only care about the one’s digit, so we only need to look at the one’s digit while multiplying.

We can break 3^(4n+3) into 3^4n * 3^3 by the rules of exponents.

If n = 1, 3^4n = 3^4 = 81 = one’s digit of 1.

If n = 2, 3^4n = 3^8 = 81*81 = one’s digit of 1.

We detect the pattern that regardless the value of n, we will be multiplying a term with a one’s digit of 1 with a term with a one’s digit of 7 (3³), so the result will have a one’s digit of 7. When and number with a one’s digit of 7 dividing by 5, we are left with a remainder of 2.

Pattern questions with division are many times Remainder questions at their core

The 4 members of the Jones Family rotate who takes out the trash on a daily basis. The order goes as follows: Mom, Dad, Brother, Sister. If Dad takes out the trash on January 18th, who takes out the trash on March 26th? (There are 31 days in January and 28 days in February.)

It’s clear that we don’t want to whip out our calendars and start counting. (A general rule of thumb is that if you think it’s taking too long, it probably is….)

Instead, we see how many days pass between January 18 and March 26:

January 19-31: 13 +

February 1-28: 28 +

March 1 – 26:  26  = 67 days.

67/4 leaves you will a remainder of 3, so we count 3 from Dad, leaving us with Mom on March 26th.

Fractions and Decimals are the same thing

You should be familiar with common decimals, mainly:

1/2 = .5

1/3 = .33 repeating

1/4 = .25

1/5 = .20

1/6 = .166 repeating

1/8 = .125

1/9 = .11 repeating

Note that multiplying these by constants will leave similarly instructive results, such as:

3/8 = 3*1/8 = 3*0.125 = 0.375

The more familiar with these you become, the quicker you can eliminate answer choices are clearly wrong. For example:

If x is an integer, which of the following is a possible value of (x² +2x – 7)/9?

A. 0.268

B. 4.555 repeating

C. -2.4

D. 1.166 repeating

E. 8.125

We don’t have to start plugging in. We know that when divided by 9, the remainder will be a number repeating to the right of the decimal place. Only choice (B) fits that description. ((C) is divided by a factor of 5, (D) by a factor of 6, and (E) by a factor of 8.)

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