# Fastest way you have found to multiply two numbers in your head?

1. Mar 19, 2006

### 3trQN

This is a mental arithmatic question really, but what is the fastest way you have found to multiply two numbers in your head?

How many steps does it take you and what shortcuts do you use?

I apologise if this is the wrong forum.

Examples:
33x24
123x321
9999x6666
etc..

2. Mar 20, 2006

### 3trQN

Wow, do you all consider this too simple a question to answer?

3. Mar 20, 2006

### arildno

Well the first one is perhaps most easily calculated as 660+132=792
the last one as 66660-6666=60000-6=59994

4. Mar 20, 2006

### Hurkyl

Staff Emeritus
I thought the first was better calculated as 720 + 72 = 792.

5. Jul 7, 2009

### mcreager

Re: Multiplication

Maybe a better description of what they are trying to say is by using the distributive property.

Ex. 33(24) = 33(20 + 4) = 660 + 132 = 792, you could also have done
24(30 + 3) = 720 + 72 = 792. When you do it often you pick up which is easiest for you sometimes people are better at certain multiples like 6's versus 8's. Also, don't forget it works over subtraction.

Ex. 66666(99999) = 66666(100000 - 1) = 6666600000 - 66666 = 6666533334, although I highly doubt that many people would be able to do it quickly in their head, but its still faster that way then doing via the method you learn in school.

Also, you can split both numbers, but that's rarely as efficient because of the increased amount of work you'd need to do since you would FOIL.

Ex. 33(24) = (30 + 3)(20 + 4) = 600 + 120 + 60 + 12 = 792

6. Jul 8, 2009

### pzona

Re: Multiplication

This isn't the most practical way to multiply, and it certainly isn't something you'll probably use, but I thought it was cool.

By the way, has anyone ever seen this before? I'm interested in why it actually works. It's youtube so the video comments were no help.

7. Jul 8, 2009

### Count Iblis

Re: Multiplication

What is convenient when doing computations in your head is to get rid of the number of carry digits that you need to keep track of even if that translates to a method that looks less efficient from an algorithmic point of view. You can also subtract a round number from both numbers to simplify the multiplication.

The subtraction method works as follows. We want to compute a*b. If N is some arbitrary number, we can define:

a' = a - N

b' = b - N

Then we have:

a*b = (N + a')(N + b') =

N^2 + N(a' + b') + a' b' =

N(N + a' + b') + a' b' =

N( a + b') + a' b'

The trick is to choose N such that it is an easy round number to work with, while a' and b' are small and/or round. If a and b are large numbers, then the best you will be able to do is get a' and b' that are an order of magnitude less than a and b. But then you can iterate this procedure using some other round number M to write:

a' b' = M(a' +b'') + a'' b''

where a'' = a' - M and b'' = b' - M

Multiplying in this way requires more steps than your calculator uses, however the difficulty you face when doing computations in your head is not really a lack of computing power, as your brain has vastly more computing power than the most powerful supercomputer, it is simply that you only have access to some limited functions of your brain to do arithmetic. So, what you must do is make sure you can easily keep track of numbers that appear in the various stages of the computation.

Simple example:

998 x 983

Subtract 1000:

998 x 983 = 1000 x 981 + 2x17 = 981000 + 34 = 981034

Ok, this was a contrived example, let's look at a more realistic example:

538 x 721

Let's subtract 500:

538 x 721 = 500x(721 + 38) + 38x221 =500x759 + 38x221

500x759 = 500x760 - 500 = 760000/2 - 500 = 380000 - 500

To compute 38x221, let's subtract 40:

38 x 221 = 40x(221 - 2) - 2x181 = 40x(220-1) - 362 =

8800 - 402

380000 - 500 + 8800 - 402 = 388000 - 100 - 2 = 387898

So, even this could be done in your head as the computation only involves easy to work with round numbers.

If you multiply directly without any tricks, you can group together the same powers of ten to minimize the number of carry digits.

So, if we multiply

[c3*10^3 + c2 * 10^2 + c1*10 + c0]x
[d3*10^3 + d2 * 10^2 + d1*10 + d0]

you should compute the digits of this multiplication by computing the last digit first from c0*d0, remember the carry digit.

Then you evaluate c1*d0 + d1*c0 and add the previous carry digit to find the next digit and remember the new carry digit. then you evaluate c2*d0 + d1*c1 + d2*c0 and add the previous carry digit, etc. etc. This way, you only have to remember one carry digit in each step, and the digits of the answer. With some practice you can multiply two five digit numbers in your head this way.

8. Jul 9, 2009

### Count Iblis

Re: Multiplication

The subtraction method can be generalized by subtracting some small multiple of N:

If we put

a' = a - p N

b' = b - q N

we get:

a*b = (p N + a')(q N + b') =

N(q a + p b') + a' b'

Example:

853 x 238

Subtract multiples of 200:

853 x 238 = 200[853 + 38x4] + 38x53

853 + 38x4 = 853 + 160 - 8 = 1005

853 x 238 = 200x1005 + 38x53 = 201000+38x53

Compute 38x53 by subtracting 40:

38x53 = 40x51 - 26 = 2040 - 26 = 2014

So, we have:

853 x 238 = 201000 + 2014 = 203014