500 New SAT Math Problems
Just 19.99 on Amazon

Hi everyone! The latest edition of 500 New SAT Math Problems is now available in paperback from Amazon. This edition just has been modified from the previous edition to account for the changes on the Digital SAT.

The paperback is now on sale on Amazon for only $19.99. Note that once the sale ends (by the end of today), the price of this book will go up to $42.99. 

The promotion has ended. Thanks to everyone who participated. The book is now available at its regular price here: 500 New SAT Math Problems

If you have any questions, feel free to contact me at steve@SATPrepGet800.com 

Thank you all for your continued support!

A Trick For Free Two Day Shipping

I would like to finish this post with a little trick you can use to get free 2 day shipping on any of the books you decide to purchase without making any additional purchases. If you have never used Amazon Prime you can sign up for a free month using the following link.

Sign Up For Amazon Prime For Free

If you have already had a free trial of Amazon Prime you can simply open up a new Amazon account to get a new free trial. It just takes a few minutes! You will need to use a different email address than the one you usually use.

This next part is very important! After you finish your transaction, go to your Account, select “Manage my prime membership,” and turn off the recurring billing. This way in a month’s time Amazon will not start charging you for the service.

After shutting off the recurring billing you will still continue to receive the benefit of free 2 day shipping for one month. This means that as long as you use this new Amazon account for your purchases you can do all of your shopping on Amazon for the next month without having to worry about placing minimum orders to get free shipping.

Just be aware that certain products from outside sellers do not always qualify for free shipping, so please always check over your bill carefully before you check out.

Well I hope you decide to take advantage of this very special offer, or at the very least I hope you will benefit from my Amazon “free 2 day shipping trick.” Here is the link one more time:

Sign Up For Amazon Prime For Free

If you think your friends might be interested in this special offer, please share it with them on Facebook:

Thank you all for your continued support!

Permutations On The SAT

Today I would like to talk about some more advanced methods for solving Counting Problems. The content presented here and in next week’s post is meant for more advanced students whose current SAT math score is at least a 600.

Last week we discussed some more basic methods which can be viewed by clicking the following link: Listing and the Counting Principle.

Please quickly review last week’s post before going on to the content below. Five minutes of review will really solidify the concepts from that post and will make the information in this post easier to learn.

Factorials

First let’s discuss factorials. The factorial of a positive integer n, written n!, is the product of all positive integers less than or equal to n.

n! = 1·2·3…n

0! is defined to be 1, so that n! is defined for all nonnegative integers n.

The definition for 0! may seem strange to many of you. Mathematicians have made this definition so that the formula n! = n(n – 1)! will be true for all positive integers n. Note that with this definition 1! = 1·0!

Also, I could make jokes about being really excited about factorials because we use an exclamation point in the definition – but let’s leave these for another day…

Permutations

Now a permutation is just an arrangement of elements from a set. The number of permutations of n things taken r at a time is

For example, the number of permutations of {1, 2, 3} taken 2 at a time is

These permutations are 12, 21, 13, 31, 23, and 32.

The good news is that on the SAT you do not need to know the permutation formula. You can do this computation very quickly on your graphing calculator. To compute ₃P₂, type 3 into your calculator, then in the Math menu scroll over to Prb and select nPr (or press 2). Then type 2 and press Enter. You will get an answer of 6.

As always, let’s internalize these principles by putting them into practice. Here is a simple SAT problem that can be solved using permutations.

Examples

1. Four different books are to be stacked in a pile. In how many different orders can the books be placed on the stack?

Okay, now you could imagine stacking the books in different ways. Let’s try this by listing the different permutations. To distinguish the books let’s make one red, one blue, one yellow, and one green. Let’s abbreviate each book’s color by using its first letter.

rbyg rbgy rybg rygb rgby rgyb
bryg brgy byrg bygr bgry bgyr
yrbg yrgb ybrg ybgr ygrb ygbr
grby gryb gbry gbyr gyrb gybr

We now see that there are 24 arrangements.

The thing is that on the actual SAT I would rather not take the time to write out each permutation. There is a chance, under the time pressure of the test, that an error could be made or too much time could be spent checking that I have not duplicated any of the permutations.

Here are two more solutions that are much more efficient.

First we can use the counting principle that we reviewed in last week’s post: There are 4 possible books for the bottom of the stack. After placing the first book, there are 3 possible books that can go on top of the bottom book, then 2 books for the next position, and then 1 book for the top of the stack. Using the counting principle we get a solution of (4)(3)(2)(1) = 24 arrangements.

Or we can solve this problem even faster by using our new knowledge of permutations: There are 4 books, and we are arranging all 4 of them. Therefore there are ₄P₄ = 4! = (1)(2)(3)(4) = 24 arrangements.

I think you would agree that the last solution is much faster than the original solution by listing.

Let’s do one more SAT problem:

2. Three light bulbs are placed into three different lamps. How many different arrangements are possible for three light bulbs of different colors – one red, one green, and one yellow?

Again, we can list all the possibilities, abbreviating each color by using the first letter.

rgy ryg gry gyr yrg ygr

We can easily see that there are 6 arrangements.

And again, we can also use the counting principle: There are 3 possible lamps to place the red bulb in. After placing the red bulb, there are 2 lamps to place the green bulb in. Finally, there is 1 lamp left to place the yellow bulb in. By the counting principle we get (3)(2)(1) = 6 arrangements.

And finally we can solve this problem very quickly using our knowledge of permutations: There are 3 light bulbs, and we are arranging all 3 of them. So the number of arrangements is ₃P₃ = 3! = 1·2·3 = 6.

There are many more of these types of problems in my 28 SAT Math Lessons Series. Click on the picture below for more information about these books.

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Click the following link to learn even more advanced counting techniques: Combinations

Counting Problems

This week I would like to start a thread on ‘counting.” Although counting sounds like it should be an easy concept, students tend to find SAT math problems that involve counting among the most difficult.

Writing A List

Okay let’s start with the most basic strategy, namely ‘writing a list’… Sometimes the easiest way to count the number of possibilities is to simply list them all. When doing this it is important that you have a systematic way of forming your list. This will reduce the likelihood of missing something, or listing something twice.

With that said, let’s look at a simple SAT counting problem:

1. A menu lists 2 meals and 3 drinks. How many different meal drink combinations are possible from this menu?

Some of you may already be able to solve this problem very quickly, but let’s first solve this by writing a list..

Let’s make up 2 meals and 3 drinks, say chicken and fish for the meals, and water, soda, and juice for the drinks. Now let’s list all meal-drink combinations. To save time we will abbreviate each meal and drink by using its first letter.

cw  cs  cj  fw  fs  fj

We see that there are 6 meal-drink combinations.

Now, as many of you probably already realize, we can actually solve this problem very quickly by using the counting principle.

(2)(3) = 6

The Counting Principle

The counting principle says that if one event is followed by a second independent event, the number of possibilities is multiplied. The 2 events in the above problem are “choosing a meal,” and “choosing a drink.”

Now let’s get a little more advanced and look at the counting principle in full generality.

More generally, the counting principle says that if E₁, E₂, …, E_n are n independent events with m₁, m₂, …, m_n possibilities, respectively, then event E₁ followed by event E₂, followed by event E₃, …, followed by event E_n has  m₁∙m₂ ∙∙∙ m_n possibilities.

My head is spinning after looking at all these E’s and m’s, so let’s try a problem to see this principle in action…

2. How many integers between 9 and 300 have the tens digit equal to 2, 3, or 4 and the units digit (ones digit) equal to 5 or 6?

Let’s first solve this using the more general version of the counting principle:

There are 2 possibilities for the ones digit (5 or 6). There are 3 possibilities for the tens digit (2, 3, or 4). There are 3 possibilities for the hundreds digit (0, 1, or 2). The counting principle says that we multiply the possibilities to get (2)(3)(3) = 18.

That’s not so bad, is it?

Note that we could also solve this problem by writing a list:

Let’s simply list all the numbers in increasing order:

25  26  35  36  45  46  125  126  135  136  145  146  225  226  235  236  245  246

And that’s it. Once again we see that the answer is 18.

A Difficult SAT Math Counting Problem

Okay, let’s do one more SAT counting problem. This one is Level 5:

3. How many integers between 3000 and 4000 have digits that are all different and that increase from left to right?

The best way to solve this is by writing a list:

As you can see, there are 20 integers in this list. It would be good to take note that we only wrote down the necessary information when forming our list. For example, the second entry was just written “7” instead of “3457.” This will save a substantial amount of time.

Also a clear and definite pattern was used in forming this list. In this case the list was written in increasing order. This will minimize the risk of duplicating or leaving out entries.

Follow the advice given here with  similar problems you see on the SAT and you will be able to answer them quickly and efficiently.

For more advanced counting techniques take a look at my articles on Permutations and Combinations.

Where To Find More Practice Problems

You can find lots of SAT math problems to practice with in my 28 SAT Math Lessons Series. Each of these books contains hundreds of problems, most with several different solutions. Click on the picture below for more information about these books.

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Speak to you soon!

Before I get into the bulk of this message I would like to first mention  that this offer is available for only 48 hours. So you really have to act now if you want to take advantage of this offer.

Now let me explain how this offer would be good for your SAT prep…

If you are using the College Board’s Blue Book, then you are probably aware that the College Board’s answers to their own SAT math problems are not very efficient.

The issue that teachers, tutors and students have with the College Board’s solutions is that they are very tedious in their approach to answering the question – they use long methods that you would expect a fresh-out-of-college school math teacher to use to solve problems. These methods are not time friendly. On the SAT you would be eating up valuable time using these methods.

Of course, it would be wiser to use more effective and efficient methods to solve SAT problems. Answering problems quickly will give you time to recheck your answers (preferably by using a different method) to ensure that you have gotten the correct answer.

There is now a simple solution – my latest book release: SAT Prep Official Study Guide Math Companion. The good news is that Amazon currently has this book on sale for only $9.49. My other books normally retail for $29.99 on Amazon – so this is a great savings.

This book has all the solutions to all the SAT math tests in the Blue Book – that’s tests 1 through 10, and the DVD test.

Please note that this sale is only guaranteed for the next 48 hours. The price will be going back up to $29.99 after that!

There are several reasons to use my solutions instead of the College Board’s. First, my solutions are extremely efficient. They will save you an enormous amount of time during the exam. You will be able to fly through the test and have enough time to spare to re-check your answers. Second, by reviewing and internalizing my solution methods you will become proficient in all of the most important SAT math strategies for raising your SAT score in a quick, efficient manner. Note that all relevant mathematical concepts and definitions are naturally reviewed as they come up in the questions and solutions of actual SAT math problems. And third, I do not include just one method of solution for each question – problems often have multiple efficient solutions to the problems for you to solve in different ways.

The benefit of having multiple solutions is that you can pick and choose which method is best for you to solve SAT problems on test day. Also, learning multiple solutions for SAT math problems will allow you to “re-check” answers you have gotten on test day by solving each problem with a different method. This will ensure that you are indeed answering the problems correctly.

So take a look at, what I think is, the best SAT prep book available now. For only $9.49 from Amazon, you can’t go wrong!

You only have 48 hours left, so act now!

Hey all, how are you today?

If you remember, last week we discussed the triangle rule. I provided a few SAT problems where the triangle rule was extremely useful, and it was not hard to see how this simple rule should be applied.

Recall that the triangle rule states that that the length of the third side of a triangle is between the sum and difference of the lengths of the other two sides.

This can be written symbolically as “difference < x < sum” where x is the third side of the triangle.

Today let us look at another type of problem that the triangle rule should be used to solve.

Problems involving distances between three points can often be solved using the triangle rule. After all, when you plot three points and look at the distances between each pair of points, you are looking at the lengths of the sides of a triangle.

Try plotting three points on a piece of paper and you you will immediately see what I mean.

There is a small catch however – sometimes when you plot three points, the points can be collinear: that is they all lie on the same line.

This means that the symbol “<” in the triangle rule should be replaced by the symbol “≤.”

In other words, this time we write “difference ≤ x ≤ sum.”

Let’s look at an example that could appear on the SAT.

Points Q, R and S lie in a plane. If the distance between Q and R is 18 and the distance between R and S is 11, which of the following could be the distance between Q and S?

 

Okay, to solve this problem, of course we are going to use the triangle rule… In this case, if Q, R and S form a triangle, then the length of QS is between 18 – 11 = 7 and also 18 + 11 = 29. The extreme cases 7 and 29 form straight lines. In this problem that is fine, so the distance between Q and S is between 7 and 29, inclusive. Thus, the answer is choice (E).

You can see how quickly this problem can be solved when the triangle rule is used.

If you want more basic info on the triangle rules, them please see my previous post. There are lots of problems with solutions that you an sink your teeth into.

Until next time…

This week, let’s get back to some mathematical content. Today I would like to discuss the triangle rule.

I consider using the triangle rule to be an advanced SAT prep strategy. It comes up on level 4 and 5 problems on the SAT and the only reason so many students get these problems wrong is because they have never been taught the rule in school.

The triangle rule states that that the length of the third side of a triangle is between the sum and difference of the lengths of the other two sides.

So, for example, if we knew that a triangle had two sides of lengths 5 and 7, respectively, then we could find the possible lengths of the third side by using the triangle rule. Simply note that the sum is 7 + 5 = 12 and the difference is 7 – 5 = 2. Therefore the the length of third side lies between 2 and 12.

Let’s get our heads around this with another short example: If a triangle has sides of lengths 2, 5, and x, then we have that 5 – 2 < x < 5 + 2. That is, 3 < x < 7.

The triangle rule is a very easy concept to understand. Again, the only reason why these are considered tough problems is because the rule is not emphasized in school, and sometimes not taught at all.

So let’s really get this rule into practice by solving some problems that might actually appear on the SAT.

The lengths of the sides of a triangle are x, 8, and 15, where x is the shortest side. If the triangle is not isosceles, what is a possible value of x?

The triangle rule tells us that 15 – 8 < x < 15 + 8. That is, 7 < x < 23. Since x is the shortest side, x < 8. So we must choose a number between 7 and 8. If this were a grid in problem, we could grid in 7.1 or any other decimal or improper fraction between 7 and 8. But inputting 7 or 8 as the answer would be incorrect as these integers are not between 7 and 8!

Okay, one more. I have included a few more examples in this post than I normally do. The solutions to these problems are so quick and easy that there is no point padding them out with unnecessary fluff – I would much rather provide examples of problems that require the rule to be used. This one is a multiple choice question.

If x is an integer greater than 5, how many different triangles are there with sides of length 3, 5 and x?

 

The triangle rule tells us that 5 – 3 < x < 5 + 3. That is, 2 < x < 8. Since x is an integer greater than 5, x can be 6 or 7. So there are two possibilities, choice (B).

My students find these problems very straight forward. I hope you do too. But it can be easy to forget the rule you need to use if you do not practice enough with it. If you want more practice, then please see the triangle rule SAT math problems in my books. These have more examples of problems that need to use this rule for efficient and correct answering.

Last week I gave you a diagram that guides you through the 7 steps of efficient SAT math prepping.

This diagram referenced my material within the various steps. However, some students that read this blog mentioned to me that they would like to use alternative prep materials with the 7 steps. I understand that not everyone wants to use some or all of my books to prep – for example, you may already have alternative quality resources – so I have created a slightly different diagram for you to follow.

This diagram below does not reference my books so you can use any other material you have to efficiently prepare for the SAT. Simply follow the stages and you will get to your target score within 4 months of studying for just 10 – 20 minutes per day. Of course, you will need to set aside more time on days that you are taking a timed test from the College Board’s Blue Book (approximately once per month).

Note that everyone preparing for the SAT should have the Blue Book regardless of what other preparation is undertaken.

Make sure that you continually retry questions that you have gotten wrong until you can get each question correct on your own. This includes any questions you have gotten wrong while taking practice tests from the Blue Book. Always wait at least a few days between attempts.

Other than that, just follow the diagram and you should be on the path to your SAT math target score.

A few weeks ago I wrote a blog post on how my various Get 800 SAT prep books should be used in conjunction with the College Board’s Official SAT Study Guide.

I have had a few questions about the process and I thought that it might be better to explain the 7 steps in a visual format. So I created a flow chart that takes you through the preparation process that I recommend for SAT math.

The diagram below does not tell you which step is which. Instead it takes you through each stage while asking questions that may restart the process to an earlier stage. This chart is easier to follow than the text on my website or on the previous blog post.

Also, although this flow chart specifically names books to use for your math prep, you could, in fact, replace the names of my books with alternate SAT prep material.

So for example, Instead of using the 28 SAT Math Lessons Beginner, Intermediate or Advanced books, you could essentially replace these books with level 1 – 3, 2 – 4 or 3 – 5 SAT math problems respectively.

Instead of using my 320 SAT Math Problems book to focus on areas you are not strong in, you could instead find the kinds of problems you need help with from other sources.

At the end of the flow chart is an instruction that says learn the strategies from The 32 Most Effective SAT Math Strategies. Again this can be replaced by other sources at your disposal that explain math strategies.

Finally, after taking a College Board Blue Book test, I advise that you use my Companion to compare solutions, as I believe the College Board’s solutions are not the best to use on test day and mine are more time efficient. Again, if you have alternate solutions from other sources feel free to use these instead of mine.

I may make another flow chart that is generic in terms of what products to use but the process will be the same no matter who is preparing for SAT math. I simply name my own products in this diagram because they fit perfectly within the efficient SAT math prep process that I recommend.

By now, you may have heard that College Board is going to redesign the SAT.

Before everyone begins to panic, you should be aware that the new test will not be administered in the near future. So if you are preparing to take the SAT within the next 4 months (which is the amount of time I generally recommend for SAT preparation) then this change will not affect you. So there is no need to worry.

In February, 2013  The College Board announced that the SAT will be changed. The new test will “strongly focus on the core knowledge and skills that evidence shows are the most important to prepare students for the rigors of college and career” as quoted by Peter Kaufman, the Vice President of the College Board.

Interestingly, the last time the College Board announced a redesign in 2004, the actual test was not introduced to students until 2005. In fact, it was the PSAT that changed first. This gave students an indication of the changes that would be introduced to the SAT.

So even though the College Board has not stated this (or any details for that matter) it seems to me that we should look out for a new style PSAT in 6 months to a year – early to mid 2014 – and expect a new SAT to come out about 6 months after that time.

So why the change now? Well the SAT has never been a test that remained static. It has been developed and changed with the input of Colleges, Universities and High Schools in order to better measure the test takers’ ability. For example, in 2005 there was a change to assess writing skills, and in fact, before that, the math and verbal sections used to be just one section.

So for now, there is no need to worry. But I will keep you updated as more news is released on the proposed SAT redesign.

It now seems that I have many SAT math prep books, even to me. So I would like to make clear why and when you would use each of my books.

I would first like to state that there is no duplicate material in any of my books. Each of my books have different SAT problems, and each title is designed for a specific reason. So with this in mind, let’s see how effective test preparation can be achieved using my books together with The College Board’s Blue Book.

So how should you first begin your prep? Well, you should gauge your score by taking a practice test from the College Board’s Official SAT Study Guide (the Blue Book). You should take this practice SAT under timed conditions. Use the Blue Book to tell you your current SAT score. It does not matter how many questions you got wrong, and you do not need to worry about reviewing these questions at this stage – we just want to know your score without any preparation. If you have recently taken an SAT or PSAT, you can use the score from that test in place of taking a test from the Blue Book.

Now it is time to begin your actual SAT math prep. You have a score and now you can use the appropriate level of “28 SAT Math Lessons to Improve Your Score in One Month.” These books will easily guide you to a higher score. All you have to do is complete one lesson per day, and after 28 days you should see a significant jump in your score level. My students like these books because they take out all the guess work out of what to study. These books will hand hold you to a higher SAT score.

I have three books in this series that can be used as standalone guides. So if your score is below 500 use the Beginner Course. If your score is between 500 and 600 use the Intermediate Course. If your score is over 600, use the Advanced Course unless you see that you are getting several Level 1, 2, or 3 questions wrong, Then you should begin with the Intermediate Course.

After completing the appropriate 28 Lessons book it is time to take another SAT from the College Board’s Official Study Guide under timed conditions to get your new math score.

Again, you will probably get several problems incorrect, but this time you should go over your mistakes carefully. “The Complete Official SAT Study Guide Companion“ can be used to compare your solutions to mine. You should certainly review all of your incorrect solutions, but feel free to look at solutions to problems you got correct just to compare your solutions to mine. This book provides many different ways to solve each test problem in the College Board’s Blue Book.

You may be tempted to review the problem you got wrong in the Blue Book by looking at the College Board’s solutions. This is a mistake. Although all of their solutions are correct, they are mostly tedious and inefficient. The College Board has no interest in providing students with time-saving strategies and foolproof tactics for ensuring correct answers. They only provide straightforward solutions that could actually be quite harmful to your SAT math preparation. This is why you would use “The Complete Official SAT Study Guide Companion”.

If you are happy with your new SAT math score, then you can go ahead and take your SAT. If you still want to score higher, it’s time to wash, rise and repeat:

If your score has gone to the next score range, then you can begin using the next “28 SAT Math Lessons” book in the series. For example, if your previous score was a 450, and your new score is a 550, you can now move up from the Beginner Course to the Intermediate course.

But if your score is still in the same range (score has increased since using 28 SAT Math Lessons, but it is still within the same ‘level’), you should use “320 SAT Math Problems Arranged by Topic and Difficulty Level“ to get past your sticking points. For example, if your previous score was a 300, and your new score is a 400, then you probably still need practice with Level 1 and 2 problems. Practice these problems from the 320 Problems book for a few more weeks, then take another SAT from the College Board’s Official Study Guide under timed conditions to get your new math score. Again use “The Complete Official Study Guide Companion” to compare your solutions to mine.

The nice thing about “320 SAT Math Problems Arranged by Topic and Difficulty Level” is that you can flick to the page with the types of problems that are giving you trouble. So if you are having trouble with Level 2 geometry, you can work on those problem types alone and master that particular level of that subject.

If you have completed the Advanced 28 Lessons book and would like additional practice, use “320 SAT Math Problems Arranged by Topic and Difficulty Level”  and focus on the Level 4 and 5 problems from this book.

Finally, for real SAT math mastery – especially if you are aiming for a perfect or near perfect score of 800, read “The 32 Most Effective SAT Math Strategies“ to review each of my strategies one by one and practice applying them to a wide range of SAT Math Problems. My strategies consist of quick and efficient ways to solve seemingly difficult SAT math problems with real ease.

So take another practice test and hopefully you will be at a score level you can be very happy with  – and the colleges you are applying to will be happy with your score as well – it’s time to take the SAT for real…

Hi everyone. How are we doing today?

This week I would like to help you solve SAT math problems that involve finding the greatest common divisor and least common multiple of a set of positive integers. But before I even get to this week’s post, I would like to go back to last week’s post on Prime Factorizations where I asked you to draw a factor tree for the number 6137. You can find that post here: Integers, Prime Numbers, and Prime Factorizations

Solution to Last Week’s Question

Let’s compare our drawings.

As I mentioned last week, it’s not artistic merit I am looking for, but rather that the numbers in your ‘tree’ are correct.

Let’s start by taking the square root of 6137. Punching that into your calculator (TI-84 or equivalent) we see that we get about 78.3. So using the tip that I gave to you last week, we will divide 6137 by the prime numbers less than 78.3.

The primes less than 78.3 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73.

The good news is that we do not need to divide by all of these numbers to get to find our factors – let me explain: Dividing by 2, 3, 5, 7, 11, and 13 do not produce integers, but 6137/17 = 361.

Now, the square root of 361 is exactly 19 (this is easily checked in your calculator). So 361 = (19)(19) = 19².

So 6137 = 17 · 19².

Here is my factor tree:

How does your tree look compared to mine? Get in touch if you have any questions regarding this…

GCD and LCM

Now for the main part of the post… The greatest common divisor (gcd) of a set of positive integers is the largest positive integer that each integer in the set is divisible by. The least common multiple (lcm) of a set of positive integers is the smallest positive integer that is divisible by each integer in the set.

Example 1:

Let’s find the gcd and lcm of 9 and 15.

There are a few ways we can do this.

First method: The factors of 9 are 1, 3 and 9. The factors of 15 are 1, 3, 5 and 15. So the common factors of 9 and 15 are 1 and 3. Therefore we see that gcd (9,15) = 3.

The multiples of 9 are 9, 18, 27, 36, 45, 54, 63,… and the multiples of 15 are 15, 30, 45,.. We can stop at 45 because 45 is also a multiple of 9. Therefore we see that lcm (9,15) = 45.

Second method: We first find the prime factorizations of 9 and 15. We see that 9 = 3² and 15 = 3 · 5. To find the gcd we multiply together the smallest powers of each prime from both factorizations, and for the lcm we multiply the highest powers of each prime. So we have gcd (9,15) = 3 and we have lcm (9,15) = 3² · 5 = 45.

Now take note – If you have trouble seeing where the gcd and lcm are coming from here, it may be helpful to insert the “missing” primes. In this case, 5 is missing from the factorization of 9. So it might help you if you write 9 = 3² · 5º. Now we can think of the gcd as 3¹ · 5º = 3.

Third method: Your calculator can actually do this for you! On your TI-84 calculator press MATH, scroll right to NUM. For the gcd press 9, type 9,15 and press ENTER. You will see an output of 3. For the lcm press 8, type 9,15 and press ENTER for an output of 45.

Example 2:

Let’s try another example: Find the gcd and lcm of 100 and 270.

The prime factorizations of 100 and 270 are 100 = 2² · 5² and 270 = 2 · 3³ · 5. So gcd (100,270) = 2 · 5 = 10 and lcm (100,270) = 2² · 3³ · 5² = 2700.

If we were to insert the ‘missing’ primes in the prime factorization of 100 we would get 100 = 2² · 3º · 5². So we can think of the gcd as 2¹ · 3º · 5¹ = 10.

Example 3:

Okay, now let’s try an SAT math question…

What is the least positive integer divisible by the integers 3, 7 and 14?

Calculator method: The question is actually asking for the least common multiple of 3, 7 and 14. Your calculator can only do two at a time. So first compute lcm (3,7) = 21, and then compute lcm (21,14) = 42, choice (D).

Simple!

Solution by Starting with choice (E): Begin by looking at choice (E) since it is the smallest. 28/3 comes to approximately 9.33 in our calculator. Since this is not an integer, 28 is not divisible by 3. We can therefore eliminate choice (E). We next try choice (D).

42/3 = 14, 42/7 = 6, 42/14 = 3

Since these are all integers, the answer is choice (D).

For more information on this strategy see the following blog post: Starting With Choice (C) – A Basic SAT Math Strategy.

Example 4:

Now here is a much more difficult example for us to try together:

The integer k is equal to m² for some integer m. If k is divisible by 6 and 40, what is the smallest possible positive value of k?

Did you realize that we are looking for the smallest perfect square that is divisible by the least common multiple of 6 and 40? Let’s find some prime factorizations: 6 = 2 · 3, and 40 = 2³ ·5. So lcm (6,40) = 2³ · 3 · 5. The least perfect square divisible by this number is 2⁴ · 3² · 5² = 3600.

It’s not too hard when you know how to compute the lcm.

This was quite a long blog post, but I hope you have received some good value from it.

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