Posts tagged with: Plug In

It almost never fails when I’m working with a new student that the first time a question involving remainders rears its ugly head, we need to spend some time talking about what remainders are and how to find them. This happens fairly early in the process with me, intentionally. I always start with new students by working through the plug in and backsolve chapters in my math book, and the plug in chapter prominently features a remainder question. That’s what’s discussed in the video above.

A remainder, just so we’re clear, is what’s left over when one positive integer is divided by another. When 17 is divided by 5, for example, the remainder is 2: 5 goes into 12 three times—making 15—but there’s still 2 left over to get to 17. Remember long division?

long division

Remainders are always whole numbers, never decimals. However, there’s a handy shortcut that’ll allow you to convert decimals to remainders:

  1. Do the division on your calculator.
  2. If there’s no decimal, then there’s a remainder of 0. If there is a decimal, then there’s a remainder.
  3. Convert the decimal to a remainder by subtracting the part before the decimal point from the quotient you have on your calculator.
  4. You’ll be left with only a decimal. Multiply that decimal by the original divisor.
  5. Boom! You’ve got a remainder.

 
Here, I’ll show you. Let’s do 17/5 again. When you put 17/5 into your calculator, you get 3.4. Subtract 3 from that, and you’re left with 0.4. Multiply that by 5, and you’re left with 2—that’s your remainder! Note that if there’s a repeating decimal, you shouldn’t round it or you won’t get an integer remainder.

Here’s another example with the exact keystrokes I enter into my TI-83, and what the calculator displays.

What is the remainder when 52,343 is divided by 92?

52343/92 [ENTER]
                     568.9456522
Ans–568 [ENTER]
                     .9456521739
Ans*92 [ENTER]
                              87

 
Therefore, 52,343/92 = 568 R 87.

Cool right? Any remainder operations you’ll be doing on the SAT will be far less involved and easily done with long division, so you don’t need to memorize this trick, but it’s there for you if you want it.

Is that all?

…Nope.

If you end up having to deal with remainders on your SAT, you’ll almost definitely have to do more than just divide two integers and find the remainder. You’ll probably be asked (as you are in the problem featured in the video above) to figure out something about unknown constants given some information about remainders. When that happens to you, here are the things it’s important to know:

If n is divided by k and leaves a remainder of r, then n must be r greater than a multiple of k. For example, if a number divided by 8 leaves a remainder of 3, then that number must be 3 greater than a multiple of 8. You’ll do well to plug in a nice, low number that fits that description, like 8 + 3 = 11.

The greatest possible remainder when dividing by k is k – 1. For example, if you’re dividing by 10, then the greatest possible remainder you can get is 9.

When you divide a bunch of consecutive integers by the same divisor k, the remainders will form a repeating pattern of consecutive integers from 0 to k – 1. For example, when you divide a bunch of consecutive integers by 3, you’ll get a repeating pattern like: 0, 1, 2, 0, 1, 2, 0, 1, 2, … The pattern might begin with any of those numbers, depending on which consecutive integer you begin dividing by 3, but the pattern will be the same.

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I don’t know if you followed the kerfuffle between Jonathan Coulton and Glee a few weeks back. It’s old news now, but I watched it unfold at the time with great interest, and I’ve been thinking about it again the last few days. The incredibly short version: Jonathan Coulton is a fairly popular musician (on the Internet, anyway) who recorded a cover of Sir Mix-a-Lot’s “Baby Got Back” in 2005 (above). Glee did a note-for-note recreation of his cover without crediting him. Then Fox’s lawyers told him he should be thankful for the exposure he didn’t get because nobody credited him.

This is a case of morality and legality not completely overlapping, and that’s all very interesting if you’re into intellectual property law (which I know is very popular among high school students these days) but that’s not where I want to go with this. The reason I bring it up is that Mr. Coulton ended up announcing that rather than pursue recourse through the courts, he’d completely change direction and try to turn this into something positive for him, and for some great charities. And there’s an SAT lesson there: know when you’re beat, and do something about it.

Coulton’s indignation was justified, but he recognized early on that he’s not going to beat an army of Fox’s lawyers, so he shifted tactics. If what you’re doing isn’t working, try something else. This is what I’m talking about when I implore you to be nimble. It’s pretty good advice for life in general, and it’s particularly germane to the SAT, on which many of the most difficult questions are vulnerable to techniques that will allow you to sidestep the math solution, if you let them. Like this one, for example:

  1. Yesterday, a group of y friends went to the mall and each purchased p pairs of gym socks. If y > x > 1 and p is a positive multiple of 3, how many fewer pairs of gym socks would they have purchased if x of the members of the group had purchased only a third as many socks as they actually did?
    (A)
    (B)
    (C)
    (D)
    (E)

If you’re looking for a top score on SAT math, you should be able to solve this with algebra, and you should also be able to solve it by plugging in. Being nimble in this way is how you work around the fact that you’re likely to see at least one problem on test day that thwarts your first attempt to solve it. Being comfortable solving a question like this two ways is also the best way to avoid careless errors—check your work by solving the way that you didn’t solve it the first time. If you get the same answer both ways, you’re almost certainly right. Both solutions below.

Let’s start with plug in

Say 10 friends go to the mall (y = 10) and each buy 3 pairs of gym socks (p = 3). So what actually happened yesterday is that the group purchased 10 × 3 = 30 pairs of socks. Now say 2 of the friends (x = 2) purchased a third as many socks as they really did. So 2 friends bought only 1 pair of gym socks each. 8 friends buy 3 pairs: 8 × 3 = 24, and 2 friends buy 1 pair each: 2 × 1 = 2. Total pairs of socks purchased: 24 + 2 = 26, or 4 fewer pairs of socks than were actually purchased. Look to your answer choices, and see which one gives you 4 when you plug in y = 10, p = 3, and x = 2.

(A) Does:
(B) doesn’t:
(C) doesn’t:
(D) doesn’t:
(E) doesn’t:

So (A) is clearly the answer.

Now let’s do the algebra

Note that, since you’ve already spent some time working through the problem logically with plug in, the algebra should be a bit more intuitive now than it might have seemed at first. First, create an expression for what was actually purchased. That’s easy: y people purchased p pairs of socks each. Total pairs purchased: yp. Now, figure out how many would be purchased in the alternate scenario where x of the friends purchased a third as many socks as they really did. y – x purchased p pairs of socks, and x purchased p/3 pairs of socks. Total pairs purchased in alternate scenario:

Now just do some subtraction to find out how many fewer pairs of socks would have been purchased:

Unsurprisingly, we arrive at the same answer either way. Again, if you’re shooting for an 800, you really should be able to breeze through this question (and ones like it) both ways.

Bonus solution

As you might have deduced from the fact that the correct answer doesn’t contain y at all, y totally doesn’t matter. All that really matters is how many fewer socks the x friends would have purchased. If they were to purchase 1/3 of what they did, they would purchase 2/3 less than they did. Since the x friends purchased xp socks in real life, they would purchase 2/3 of that is . [See also: “Is there a math way?“]

When a student asks me how to solve a math problem, my default response is to show, if possible, how to solve it by plugging in, backsolving, or guesstimating. I do this because I figure if the “math way” was obvious, the student wouldn’t be asking me for help in the first place. Besides, problem solving—in life, or on the SAT—isn’t about following a circumscribed set of procedures. It’s about creativity and flexibility. I’ve written before about the importance of being nimble. Consider this post a sequel.

It’s fun to be good at math, and it’s nice to understand how the underlying algebra on a tough word problem works. But if you’re aiming for top scores, it’s imperative that you cast a critical eye on your own ability to tease the “math way” of solving a problem out of the problem during the fairly tight time constraints imposed by the SAT.

If x + y = p and x – y = q, what is p2 + q2 in terms of x and y?

(A) 2(x + y)2
(B) 4xy
(C) 2x2 – 2y2
(D) 2(x2 + y2)
(E) 2(x2 – 4xy + y2)

Like all questions, there’s a “math way” to do this, but unlike all questions, this one is a prime candidate for plugging in. There will be some students who can breeze through the algebra in their head and identify the correct answer almost instantly. If that’s you, then great. You needn’t plug in. But if that’s not you, or if you only kinda think that’s you, then you should probably just plug in. It’s fast, it’ll get you the right answer, and then, later on you can go home, make an awesome couch fort, and figure out the algebra when you’re not pressed for time.

The plugging in solution

Say x = 3 and y = 2. Then 3 + 2 = p = 5, and 3 – 2 = q = 1. 52 + 12 = 26, so you’re looking for an answer choice to give you 26. Type the answer choices into your calculator carefully, substituting 3 for x and 2 for y, and you’ll be done in a hot second:

(A) 2(x + y)2 = 2(3 + 2)2 = 50
(B) 4xy = 4(3)(2) = 24
(C) 2x2 – 2y= 2(3)2 – 2(2)2 = 10
(D) 2(x2 + y2) = 2(32 + 22) = 26
(E) 2(x2 – 4xy + y2) = 2(32 – 4(2)(3) + 22) = –22

The algebra
Now add ’em up:
Not impossible, right? Totally doable. But arguably more involved than the plug-in solution above.
The bottom line

Look, I really just want you to be happy. If you want the algebra, I’ll give you the algebra. But I really think it’s a good idea for you to know how to plug in, too. Because if you have to ask me for the algebra on a question like this, that means it wasn’t obvious to you right away when you encountered it on the test. And that means there’s a good chance that when you sit down for the real thing, the algebra isn’t going to be obvious to you for every single question. And if, when the algebra isn’t obvious, you don’t have a backup plan, then you’re doing yourself a disservice.

Try the algebra first, if that’s your bent. But you should have a few other tricks up your sleeve for the questions where the “math way” isn’t jumping off the page onto your lap.

I went back to my old high school last night to attend the final concert of the choir director that presided over so many of my formative moments. I got to hang out with some of the same people with whom I used to have the kind of deep, meaning-of-life conversations that only happen in movies or in real life when you’re between the ages of 15 and 17. A good number of them are teachers now, as it turns out. Their students are lucky to have them.

I went back with a few of the people who helped make me into me, to honor a man who helped shape all of us. I saw a fantastic concert. I had a lot of hugs and handshakes. I found a plaque that still has my name on it. And then I drove home listening to a record I listened to a lot back then. Now I’m sitting in my apartment back in New York, wistful and weary. HOLD ON TO THESE MOMENTS WITH BOTH HANDS. DRINK FULLY AND RICHLY FROM THE CUP OF YOUTH. AND RETURN FOR SECONDS.

Anyway, here’s a challenge question! First correct response gets a free copy of the Math Guide.

Louis and Rebecca each had x dollars on Monday morning. On Monday afternoon Louis paid Rebecca 20% of his money for a computer that she was selling. On Tuesday morning, Rebecca paid Louis 20% of her money for a lawnmower that he was selling. On Wednesday morning, both Louis and Rebecca paid 20% of their money to Steven for…something. I don’t know. Steven was selling something. Assuming they did not spend or receive any other money in that period, how much money did Steven receive, in total, from Louis and Rebecca in terms of x?

Leave your answers in the comments. I’ll post a solution on Monday (assuming someone gets it by then).

In order to win a book, you must not comment anonymously. I need to be able to get in touch with you to get your shipping info, etc. Also, while you can still win a free book if you don’t live in the US, you’re going to have to pay for shipping. So the book is only sorta free if you’re international.

Good luck!

UPDATE: Congratulations to Jason, who nailed it with alarming speed, and to whom a Math Guide will be en route. Solution below the cut.

There’s a fast way to do this problem, and then there’s a very fast way to do this problem. I’m gonna give you both.

First, the fast way. Plug in. Say x = 100. Here’s how the money flows:

Day and timeLouisRebeccaSteven
Monday morning$100$100
Monday afternoon$80$120
Tuesday morning$104$96
Wednesday morning$83.2$76.8$20.80+$19.20=$40

$40 is 40% of x (which, as you’ll recall, we said was 100), so the answer is 40% of x, or 0.4x.

As you look at the table, you might notice that each row adds up to $200. Interesting, no? Which brings us to the very fast way to do the question. Regardless of the money that changes hands between Louis and Rebecca, the total amount of money they have together never changes. So we can just, from the get-go, say that the amount of money that gets handed to Steven is 20% of the total money Louis and Rebecca begin with.

They start with, in total, 2x dollars. 20% of 2x is 0.4x.

Honestly, this makes me really nervous because I have no idea how it’s is going to work and also because I hate looking at a moving image of my face while I talk—my mouth moves weirdly—but I’m gonna dive in.

So mark your calendars, homies. Monday, June 11 at 8 PM EST: SAT prep Google+ hangout numero uno. For this first one, I want to talk about plugging in on math questions. We’ll talk briefly about the technique, and work through a few questions from my book together. You won’t need the book to participate, but it’d make me feel good to see a few people hanging out with Math Guides in hand. (I have this fantasy of one day getting on the subway and seeing some kid working through my book as he heads downtown. I’ll be like, Hey, is that book any good? And he’ll be like, Who are you, weirdo? It’s OK I guess. And then I’ll be all, AWWWWW YEAAAAAHHHHHH.)

If you want to participate, make sure you get a Google+ account if you don’t already have one, and add PWN the SAT to your circles on by Sunday, so I can invite you when things get rolling.

Important notes
  • There are a limited number of seats. Google+ limits Hangouts to ten people at a time, and one of them is me. That means there can only be nine of you. I cannot reserve seats, so if you get locked out, you get locked out.
  • This is for high school students only. Due to the above, you’d be occupying a seat that a student could use. Sorry.
  • Please test out the hangout feature with your friends ahead of time so we can avoid tech issues. I like you guys a ton, but I don’t want to be your tech support, and if this thing devolves into a bunch of I can’t get the sound to work! then it’s probably not worth doing. So please make sure you can get all that working before Monday at 8.
  • There’s no guarantee that this happens more than once. If it goes well, I’m excited about doing it again. I just don’t want to commit myself to doing something crappy all summer. I’ll do my best to make it good. You do the same? Cool.

In the Blue Book, about this common:

Screengrab from the PWN the SAT Math Guide

I’m not saying you need to plug in numbers for each (or even most) of these. But I am saying you should be aware of how often you have the option.

Hey all. Here’s your weekend challenge question. The prize this weekend: any $5 album from the Amazon mp3 store. I can only give you the prize if I can get in touch with you (using your email or Disqus/Google/Yahoo/Twitter/Facebook account), so please don’t be completely anonymous if you want the prize.

(m + n + p + 180)(qbd – 2eac + r) = 3x + y

Based on the figure and equation above, what is x in terms of y?

Good luck, and have a great weekend! I’ll post the answer Monday.

UPDATE: solution below the cut.

Nobody got this, although I did get a few emails with valiant attempts. The answer is: x = –y/3.

Why? Well, it’s a geometry plug-in. You’re given a bunch of shapes and not a single angle to call your own, so MAKE THEM UP. You just need to make sure you don’t break anything in the process — all your triangles need to add up to 180°, all your straight lines need to add up to 180° and in this case your pentagon needs to add up to 540° (because that’s how a pentagon rolls).

Easiest way to go is just to pretend your pentagon is regular (even though it’s clearly not) and plug in 108° for m, n, p, q, and r. That’s going to make all the base angles of the triangles come out to 72°, and a, b, c, d, and e all equal 36°.

The (m + n + p + 180) part of the question is horse-hockey. I only put it in there because I didn’t use those variables elsewhere in the problem, but I named them in the diagram.

The (q – b – d – 2e – a – c + r) part, using the numbers I just plugged in above (or any numbers that don’t break triangles, straight lines, or pentagons) comes out to ZERO. (Since nobody won the prize for this question, I’ll award it to anyone who feels like writing a general proof of this. I won’t hold my breath.)

So you’re left with:

0 = 3x + y

Which is why x = –y/3.

Come at me, bro. (Source.)

OK, so you know how I’m always saying that the SAT is not a math test? This is one of the primary reasons why. On the SAT, it’s often completely unnecessary to do the math that’s been so carefully laid out before you. A lot of the time (and on a lot of the most otherwise onerous problems), all you need to do is make up numbers.

Sounds crazy, right? Well it’s not. It would be crazy to just make up numbers on just about any other pain-in-the-ass task (for instance, it would be bad just to make up numbers on your taxes), but you’ll be dumbstruck by how often it works on the SAT. Of course, you have to practice doing it to get good at it, so that when an opportunity to do it on the real test pops up, you don’t panic and blow it. That’s what your old buddy Mike is here for.

I’m thinking we should start with a more obvious plug-in. If you would consider trying to solve this one with pure algebra, you’re probably out of your mind. Still, it’s a great illustration of the technique:

  1. If m and n are divided by 8, the remainders are 3 and 5, respectively. What is the remainder when mn is divided by 8?
     
    (A) 0
    (B) 1
    (C) 3
    (D) 5
    (E) 7

What we want to do with a question like this is plug in values for m and n so that we’re not dealing with abstracts. Of course, there are infinite possibilities for the values of both m and n, but we’re just going to pick values and stick with them.

Since the problem stipulates that m divided by 8 gives me a remainder of 3, and n divided by 8 gives me a remainder of 5, let’s pick m = 11 and n = 13 (because 8 + 3 = 11, and 8 + 5 = 13). That will keep our numbers nice and low, and make the division we’ll have to do in the next step less arduous.

Since I’ve plugged in 11 for m and 13 for n, I need to find the remainder when 11×13=143 is divided by 8. Remember long division? That’s going to be the easiest way to calculate a remainder, so let’s do it:

Bam. Remainder 7. That’s choice (E). I feel so alive right now.

Note that if we picked different numbers for m and n (like, say, 83 and 85), we’d still get the same answer (try it yourself to see). That’s the beautiful thing about plugging in!

Let’s do another, slightly tougher one:

  1. If the inequality above is true for integer constant k > 1, which of the following could be a value of x?
     
    (A) k – 3
    (B) k – 1
    (C) k
    (D) 3k – 4
    (E) 3k – 2

OK. Forget for a minute that this can be solved with algebra and think about how to solve it by plugging in. Remember, if you don’t practice plugging in on problems you know how to do otherwise, you won’t be able to plug in well when you come to a problem you don’t know how to solve otherwise!

We know k is a positive integer greater than 1, so let’s say it’s 2. If k = 2, then we can do a little manipulation to see that x has to be greater than or equal to 3:



Note that we don’t just make up a number for x! Once we’ve chosen a value for k, we’ve constrained the universe of possible x‘s. When we have an equation (or an inequality), we usually can’t plug in values for both sides; we have to choose one side on which to plug in, and then see what effects our choices have on the other side.

So, which answer choice, given our plugged in value of k = 2, gives us a number greater than or equal to 3 for x?

(A) k – 3 = 2 – 3 = -1 (too low!)
(B) k – 1 = 2 – 1 = 1 (nuh-uh)
(C) k = 2 (nope)
(D) 3k – 4 = 6 – 4 = 2 (still no good)
(E) 3k – 2 = 6 – 2 = 4 (yes!)

Rock. On. Note again that if we had picked a different number for k, we still would have been OK. Try running through this with k = 10 to see for yourself.

So…when do you plug in?
  • When you see variables in the question and the answer, you might want to try plugging in.
  • On percent questions, you’ll probably benefit from plugging in (and using 100 as your starting value).
  • On triangle questions where no angles are given, you might try plugging in 60 for all angles.
    • If you’re plugging in on a geometry question, just make sure that all the angles in your triangles and straight lines add up to 180°.
  • Anytime you don’t know something that you think it would be helpful to know, try making it up!
Anything else I should know?
  • As a general rule, DO NOT plug in 0 because when you multiply things by 0, you always get 0, and when you add 0 to anything, it stays the same. Usually, that will make too many answer choices work.
  • Similarly, DO NOT plug in 1, since when you multiply things by 1, they don’t change. This will also often make more than one choice seem correct.
  • DO try to keep your numbers small. There’s no need to plug in 2545 when 2 will do.
  • DO think for a minute before picking your numbers. Will the numbers you’re choosing result in messy fractions or negative numbers? We plug in to make our lives easier, so practice avoiding these scenarios!
  • You always have to check every single choice when there are variables in the answers and you plugged in, because there’s a small chance that more than one answer will work. If that happens, don’t panic…just try new numbers. You can greatly mitigate this by following rules #1 and 2 above.
Let’s try some more problems!

Note: all of these problems can be solved without plugging in, but you’re not here to do that right now, you’re here to practice plugging in. Don’t be intractable in your methods. If you’re amenable to change, it’s more feasible that you’ll improve your scores.

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Want to see more plugging in? Browse the “plug in” tag on my Tumblr for some recently posted examples!