## How To Multiply Faster Than A Calculator

I’m sorry to tell you that you’ve been duped! I didn’t want to be the one to inform you that nobody can multiply faster than a calculator. It just isn’t possible. I’ve misled you. However, those of you who did not take my title literally, I do have good news.

Consider the following scenario. The process of using a calculator is not nearly as fast as it is portrayed to be. First, you need to get it out and turn it on. Then, you need to enter in the numbers. Finally, it will display the answer for you. Think about it. How long does it really take to do all this? I claim that I can do this task faster in my head, and that I can teach you to do the same. I’ll teach you how to be lazy, save time, and avoid the hassle of getting out the calculator all at the same time!

There are two methods that I would like to share with you: Round And Adjust, and Multiply In Parts. Along with these two techniques, I’ll share with you a couple of tricks that I apply to simplify both of these methods.

Trick #1: Ignore the trailing zeros

What this means is that if you have a number such as 250, pretend it is just 25. Once you do the multiplication, you just add the zero back to the end of your answer. This works for any number of zeros. 250000, 250, and 25000000 should all be treated the same as 25. Just remember to add the respective amount of zeros to the end of your answer. However, you cannot simplify a problem when the zeros are in the middle. You can only use this for trailing zeros (zeros at the far right side). For example, 205 must be thought of as 205. This cannot be reduced to anything smaller. On the contrary, 2050 can be treated as 205, because there was one trailing zero. This helps quite a bit.

Trick #2: Ignore the decimals

Now for the main course, my methods!

I’d like to explain this method by using an example. Let’s say you want to multiply 98 by 52, and try and do it quickly in your head. The answer is 5096, and I did that in around 8-10 seconds – much faster than getting out a calculator. So, what did I do? I rounded, did the multiplication, and then adjusted my result. I rounded 98 up to 100. Then, it is extremely easy to multiply 100 by 52: 5200. Then, I think about how much I rounded by. I rounded up by 2. 100 x 52 is 2 x 52 more than 98 x 52. This means, that 98 x 5equals 100 x 52 minus 2 x 52. Therefore, I find a quick answer by multiplying 100 by 52, then I subtracted 2 times 52 from that. 2 times 52 is 104. 5200 – 104 is 5096. Simple as that.

Now, what if you round down? What is 72 times 45? 3240. In my head, this was the process that I followed. 72 is close to 70, so let’s start with that. 70 times 45 is like 7 times 45. 7 times 45 is 315. Since I turned 70 into 7, I’ve got to add a zero to the end. 3150. I rounded down by 2, this means that I now have to add this 2 back in. To do so I multiply 2 by 45, and add this to my answer. 2 times 45 is 90, and 3150 + 90  is 3240. I know that took a little while to read, but it goes much faster in your head once you understand the method.

Why does this work?
98 x 52 = (100 – 2) x 52 = (100 x 52) – (2 x 52) = 5200 – 104 = 5096.

See? This is the algebra that allows for this method to be used. The same proof can be applied to rounding down. Basically, you look to round one of the numbers to a multiple of ten. It needs to be somewhat close to that number already, in order for this method to really be useful. Remember that you can only round ONE of the numbers. Do not round both numbers. Only one. Try it out =)

Multiply In Parts

What the heck does this one mean? Well, actually, it is very similar to my first method, however it is a bit more universal. It works because of the same logic, but it does not require rounding or adjusting. It is simply breaking down a number piece by piece. What do I mean by that? I’ll show you an example, using algebra to solve it.

1234 x 32 = (1000 + 200 + 30 + 4) x 32 = (1000 x 32) + (200 x 32) + (30 x 32) + (4 x 32)
= 32000 + 6400 + 960 + 128 = 39488

Could you have done that in your head? I think you can =) You need to break up the larger number into it’s parts. For example, think of 345 as a 3, 4, and 5 separately. Multiply them individually, and then add the results. Remember that 3 is really 300, so you must add two zeros to that result, and add one zero to the result from the 4. This is the same way we do multiplication when we do it on paper. The difference is the perspective in which you are thinking about it. You are multiplying each number individually, and then adding the results. It’s not all that hard one you get used to it. 64 x 55. 6 x 55 is 330, and 4 times 55 is 220. The answer is simply 3300 + 220. The extra zero on the 3300 was inserted because the 6 is really a 60.  The answer is just 3520!

What do you think? I promise that doing multiplication in your head is much faster than doing it on paper, and can even be faster than using a calculator. If you have a calculator out already, then by all means use it. However, if it is out of the way to go get it, you should try to do it in your head! I’ll do a couple of examples using all the tricks I showed you above.

31 x 4.5?
31 x 4.5 is like 30 x 45, which is like 3 times 45. 135. 1350. When you add back in the 1 that we rounded down from, this becomes 1395. There was 1 digit to the right of the decimal. The final answer is 139.5. This is exactly what I thought in my head. I know it takes a while to read, however it is much faster when I’m thinking it. Try it out! It is really easy to make up an example for yourself.

2.40 x 314?
2.40 x 314 is like 24 x 314. 24 x 3 is 72. 24 x 1 is 24. 24 x 4 is 96. 7200 + 240 + 96 = 7536. When you add back the zero and decimal, you get 753.60.

To be honest, I don’t add all those large numbers in my head all at once. I keep a running tally. This helps when the numbers get much, much larger. What does that mean? In the example above, I say 24 x 3 is 72, which is 7200. I repeat that number in my head. 7200. 24 x 1 is 24, which is really 240. 7440. I then repeat that in my head, so that I remember it better. 7440. I continue this, until I get to the last number, 24 x 4 which is 96. 7536. Then I add in my zeros and decimal, and get the final result.

The most common use for this is shopping. Do you ever go to a store and see that the price is 30% off? Is that really a good deal? What if it is 30% off, but then take an additional 20% off? Doesn’t that get a bit more complicated? Of course, cell phones can act as a calculator on the go, but is it worth the time to get it out and type in the numbers? I think not.

You go into a store and see a pair of shoes you’d like to purchase. The price for them is 49.99, and they are onsale for 30% off. However, they are having a special 1 day only sale, that allows you to take an additional 20% off the original sale price.

So, you need to take 30% off of the original price. Then, take 20% off of that.

First, sales need to be thought of in an opposite way of thinking. If something is 30% off, that means you are paying 70%. If you multiply the price by 0.7, this will give you the sale price. So, find 49.99 x 0.7.

49.99 x 0.7 is like 5 x 7, or 35. With the zeros, 35000. Subtract 1 x 7, from rounding. 34993. Now, add back in the decimal. 34.993. This is the new price! 34.993, or just 34.99! Now, you can take 20% off of that price. So, we will multiply by 0.8, because you are really paying 80% of the price when you take off 20%.

34.99 x 0.8 is like 35 x 8, 280. 28000. Subtract 1 x 8. 28992. The answer is 28.992, with the decimal. The final cost of the shoes will be 28.99! That is a great deal =)

How else could we have done this problem? Well, you are taking off 30%, and then another 20%. This is the same as paying 70%, and then paying 80% of that. We could just do 0.7 x 0.8 = 0.56. Multiply this by 49.99, and you will get the same answer. Just another way to think about it =)

I know I’m really geeky and all, but when I’m bored I multiply things in my head sometimes. It keeps my mind sharp, I believe. I also like to do some calculus in my head. I work a part-time job after school, and this helps pass the time when things are slow. I know it seems useless to do this sort of thing randomly, however when the time comes when you need to use it, you can do it much, much faster. I hope this has been somewhat helpful!

If there is anything you would like to be explained further, please comment below or email me at admin@calculustricks.com. Thanks!

## Calculus I: The Chain Rule

The chain rule is used extremely often, and is a concept you can’t do without.

If it’s so important, why? What is it, exactly? Before I get to the teaching, I need to warn you. This is one of the most common things that people forget to do. The chain rule is needed every single time you take a derivative. I’ll show you why I haven’t mentioned it until now, but from now on, you need to consider the effects of the chain rule EVERY time you take a derivative. Up until now, you haven’t been able to take the derivative of anything complicated. I’ve only given you things that are fairly simple. I’m sorry to tell you that things can get much, much harder! Here are a couple of examples, to show you what I mean.

Before now, you wouldn’t be able to take the derivative of the functions above. They are too complex, and require the chain rule. With some problems, the effects of the chain rule are negligible, and those are the functions we’ve been working with thus far.

What’s the chain rule?

First, I’ll give you the formal definition. I know it is going to not make much sense, but I’ll thoroughly explain it afterward.

If you don’t understand that, don’t worry. Nobody really expects you to at first. So, what the heck does that mean? Simply, this means when you have one function inside of another. Here’s an example:

The inside function is , and the outside function is . One function is inside of the other. That’s the best way to think about a complicated function. You should think of it in parts, like I just showed. Now, how do we take the derivative of this? To begin, you pretend that the inside of the function is just a simple x, and you don’t do anything with it. You leave it alone completely. For this part, pretend that we are taking the derivative of . It doesn’t matter what the “something” is. Just leave it alone, and take the derivative as if the “something” is just an “x”.

That is exactly what I just told you to do, however the problem isn’t complete. Next, you must multiply that answer by the derivative of the “something”. In the first part, we completely ignore what the “something” is. Now, you look at ONLY the “something”, and take the derivative of that. This is going to be our final answer, unless we can simplify something.

As you can see, I didn’t leave the word “something” in the middle of my answer. I just left it alone completely in the first part. I didn’t touch it. I just rewrote it. I’m being deliberately redundant here. I want to say this as many ways as possible, so that you completely understand what I am saying. This is extremely important, and I would venture to guess that it is the most common mistake made in calculus. Everyone messes up the chain rule at some point.

This is an extremely fundamental topic in calculus. This rule should be applied to every single derivative that you are taking. Sometimes, it will not affect your answer, but most of the time it will. If you forget to use the chain rule, you will get the answer wrong! This is a rule. This is how you take derivatives. It is as important as the order of operations. It is needed. This is how you take a derivative properly. You need to practice this thoroughly!

Before I get to another example for the chain rule, I want to show you why the chain rule was not needed for all the derivatives we’ve been taking until now. The reason is simply that every time we take the derivative, the inside function has just been an “x”. For example,

The outside function is “sin”, and the inside is just “x”. What happens if we use the chain rule here?

The only difference from what we used to do is when we multiply the derivative of the inside “something”. The “something” has always been just an “x”. The derivative of x is just 1, and when you multiply by 1 nothing changes! The chain rule, the part where we multiply by the derivative of the inside, does absolutely nothing. It is completely negligible. That is what we’ve never needed it! However, now that you know about the chain rule, you should always use it. Even when it isn’t going to change the answer. You need to make it a habit, or you will forget to do it at a time where you really need to. It’s inevitable that you will mess this up a few times, but hopefully not much more than that.

Remember this:

To take a derivative, you look at the most “outside” function. In most cases, it will be an exponent on parenthesis. You pretend that everything on the inside is just an x. It could be absolutely anything, and it doesn’t matter. Take the derivative as usual, but leave the inside completely untouched. Just copy it from one line to the next. After you do that, you must multiply that answer by the derivative of the inside stuff. The same stuff you previously skipped over and ignored. That is the most simple way to describe and understand the chain rule.

Want to try another example?

The most outside part is the “sin”, so start there. Take the derivative, and then move to the inside part.

See how that works?

One more basic problem, before I show you some hard ones?

I have two things to admit to you. First of all, you don’t have to actually write “something”, like I’ve been doing in every problem. That is just the way I suggest you think about these problems. Visualize that step as you are using the chain rule, but don’t write it down. Second, you really don’t need to show the step where I write the symbol for the derivative. You can just take the derivative. If you look in this last example, you should skip the entire first step. You can go directly to the second step that I’ve shown. I’m just showing extra so that you understand it a little bit better.

I have two more examples for you. These are a bit more challenging. If you can do this type of problem on your own, you will be able to do any chain rule problem. That said, I suggest you try doing the following two examples yourself. Don’t look at my solution, and work it out on your own. Only scroll further to see my answer when you get stuck. That is the best way for you to learn! (I promise)

These can get pretty complicated, but I hope you could follow me! This answer is technically correct, however many teachers will want you to go further than this. When you have to use both the product rule and chain rule, often times you can factor and simplify things a great deal. I’ll show you what to do with this answer.

See how much nicer that looks? I think so at least, and many of your teachers will require you to do this simplification! Be on alert whenever you are using the product rule to take the derivative of something complicated.

For my last example, I will give you my favorite challenge problem. Whenever I tutor calculus I, I always ask this type of function. Like I said before, try to do it without my help. If you can, that is extremely impressive! This is like the test of all tests. If you can figure this one out, consider yourself an expert of the chain rule! I’m going to break it down step by step, and explain why it needs to be done this way.

The way you should always start taking a derivative is by asking yourself, “What is the most outside part?” This is where we always begin. If you remember, you need to ignore the part on the inside, and take the derivative of only the outermost part of the function. Work your way from the outside to the inside. So, what’s the answer? What is the outermost part of this function? What should we start taking the derivative of?

To be honest, I’m being sneaky. Most of you would say, “cosine of course!” That is exactly what I expect you to say, however that is wrong! Let me rewrite the problem. If you don’t know what I am talking about, you will in just a second.

The reason this may have tricked you, is that I placed the exponent right next to the “cos”. This is a very common way to write exponents on trigonometric functions, but it always leads to confusion! If you get a problem with a trig function that has a power on it, rewrite it like I just did. Be careful though, the parenthesis around the “cos” are extremely important. Otherwise, the exponent isn’t in the correct place, and you will confuse yourself even further.

Now, I’ll ask you one more time. What is the most outside part? Of course, you can now see the most outside part is the exponent. For the first part, you should ignore everything inside the parenthesis with the exponent on it. I’m going to show you the full solution now, and then give a brief explanation afterward.

There was a chain rule needed inside of a chain rule! Isn’t that really neat? Basically, you work your way in from the most outside piece of the function. Then you take one step inside. Then another step inside. Until you finally reach the innermost basic function that was trapped inside everything else. As you go along, you need to respect the chain rule, and multiply by the derivative of the respective next inside piece of the function. You multiply by that derivative. Then you need to determine if you will need to use the chain rule on the that derivative as well. You just keep multiplying them!

Do you see how I simplified my answer? Make sure you bring all of the coefficients to the very front of the expression. Thats is very important for style points, and you might get points off on a test if you don’t do that!

That was my last example. I really like teaching that one. These problems can get pretty complicated, and I suggest you practice the chain rule. It needs to be second nature, and a habit for every time you take a derivative. Always write it! This is the most common mistake in calculus. Don’t forget the chain rule! Remind yourself before you even begin a problem.

Good Luck!