Sour Swinger Predominantly follows my Army & Iraq Deployment. Also has some Geek, School, and other talk.

There are two different methods to quickly square any two digit number. The one I’m about to show you is by far the easier since it eliminates any complex multiplications. I’ll do the other method in a different post. Lets look at our first example:

43² =
4 (3²) =
(4 × 3 × 2) 9 =
4² (24) 9 =
(16 + 2) 49 =
1849

We should always work from right to left since we need to worry about carrying. First we squared the last digit, 3. This gave our last digit of the answer, 9. Next multiply the original two digits 4 and 3 and double the answer. That gave 24 which means we had to carry the 2 into the next step. Finally, we squared the original first digit, 4, and added anything carried over from the previous step, 2. Our final answer 1849 has been found.

Thats basically it. The only thing that could happen different is the first step will end up needing to carry. Lets try another example for that situation:

39² =
3 (9²) =
[(3 × 9 × 2) + 8] 1 =
3² (62) 1 =
1521

Once again, we started on the right and squared the 9. This gave us 81 which means we had to carry the 8 over to the next step. The remaining 1 became the last digit in our answer. Then we multiplied the original two digitis 3 and 9, doubled it, and added what was carried over from the first step, 8. This gave us 62 which meant we had to carry again, leaving the 2 as the next digit in our answer. Finally we squared the 3 and added the carry which gave 15. Our final answer 1521 has been found. With a little practice, you can get pretty quick at this.

Surprisingly there’s a quick method to multiplying numbers by 11.  Of course the larger the number gets, the harder it becomes.  There’s a simple pattern to follow when totally up the answer.  It’s best to slowly work up to exposing the pattern then to dive straight in.  It’s also easier to show rather then explain the pattern.  Lets start with a two digit number:

24 × 11 =
24 =
2 __ 4 =
2 (2 + 4) 4 =
264

We first drop the whole “× 11″ as that part is no longer important to remember.  Looking directly at the 24, split it apart and place the sum of its digits in between.  That’s it.  The pattern isn’t noticeable yet.  Before we move on to three digits lets look at a special case:

49 × 11 =
49 =
4 __ 9 =
4 (4 + 9) 9 =
4 (13) 9 =
(4 + 1) 39 =
539

That looks worse then what it is.  We did the same thing as before.  Ignored the “× 11″ part, split the 49 and summed the digits.  However this time the sum yielded a two digit number, 13.  All you do is carry the one over to the four.  Just like you would in normal addition.  Lets try a three digit number:

314 =
3 __  __ 4 =
3 (3 + 1) (1 + 4) 4 =
3454

I didn’t bother writing the “× 11″ since we always ignore that anyway.  We split the number up again, but this time left space for two more numbers.  How does one know the number of additional digits to add?  Its always 1 minus the total number of digits.  So if we had a number with 10 digits, we’d be adding 9 more numbers.

Here’s where the pattern starts to show.  All you do is work your way across the original number starting from the right.  Add all the pairs of numbers that are next to each other.  Still don’t see it?  Lets try a longer number a little bit slower:

12345 =

We’ll start from the right.  Remember, add all the pairs of numbers whom are next to each other.  So the first pair is (4 + 5).  Now move one and add the next pair (3 + 4).  Then (2 + 3) and finally (1 + 2).  All together, it’ll look like this:

1 (1 + 2) (2 + 3) (3 + 4) (4 + 5) 5 =
135795

See the pattern now?  Once numbers start getting this long, its hard to do in your head especially if carrying is involved.  However this makes for a fast and easy computation on paper.

A bit ago I talked about a multiplication trick that involved halving one number and doubling the other.  The only stipulation was that one of the numbers had to be even.  What if both your numbers are odd?  Is it still possible?  Of course!  Its just a little more involved and probably not something you can strictly do in your head.  However its still fun to see it in action.  Lets try one:

113 × 27 =
56  × 54 =
28 × 108 =
14 × 216 =
7 × 432 =
3 × 864 =
1 × 1,728 =

Notice I did the same thing as if one of the numbers were even.  Just pick one to half and the other to double.  When you half an odd number, simply ignore the remainder.  We’re not done yet as 1,728 is not the correct answer.  Remove those numbers whose halved number is even.

113 × 27 =
56  × 54 =
28 × 108 =
14 × 216 =
7 × 432 =
3 × 864 =
1 × 1,728 =

Now, sum all the remaining numbers that were doubled.  Do not include those numbers that were halved.

         27
       432
       864
+ 1,728
   3,051

And that’s your answer!

Of all the multiplication techniques I’ve discuss thus far, this one is probably most commonly done. Its based on simply breaking down a given number(s) into its factors and multiplying from there. Knowing a numbers divisibility may help. Example:

20 × 16 =
10 × 2 × 8 × 2 =
80 × 2 × 2 =
160 × 2 = 320

I broke up the 20 into 10 and 2. I also broke the 16 into 8 and 2. Since 2 is an easy multiplication, I multiplied the 8 and 10 first. Note you don’t have to break up both numbers and you don’t need to break a given number all way down. Notice the 10 and 8 I left as is. It all depends on what would make the problem easier to solve for you.

When multiplying two numbers, you can half one and double the other until a more suited set of numbers are formed.  This can be accomplished whether the numbers are odd or even.  Since odd numbers are handled slightly differently, lets take a look at strictly even.  Keep in mind both numbers do not need to be even, only one.  First example:

24 × 36 =
12 × 72 =
6 × 144 =
3 × 288 = 864

In that example, I decided to half 24 and double 36.  Since both numbers are even it doesn’t matter which I chose.  Then continue to half and double.  I eventually got down to three and from there finished the multiplication.  Lets try another:

23 × 32 =
46 × 16 =
92 × 8 =
184 × 4 =
368 × 2 = 736

This time around, there was only one even number, 32, so that had to be the number to half.  The odd, 23, was then doubled.  This is repeated until an easier equation is reached.  Notice the halved number eventually went down to 2.  Thats because 32 is a power of 2.  This brings up a further trick for this trick.

If your good with the powers of two, you can save some time by ignoring the halving number.  Thirty-two is 2 to the fifth or 2^5.  That means you double the other number 5 times.  In the previous example, that’s exactly what we did.  Twenty-three was doubled 5 times which brought the answer 736.

This technique revolves around the whole idea of rounding one number up/down to an easier number to calculate. This could potential save you time. It all comes down to what you round to, usually a multiple of 10, 100, 1000, etc. First example:

99 × 4 =

Here, you’ll notice that 99 is just one away from 100 which is a lot easier to multiple. Lets round up by adding 1 and multiply.

100 × 4 = 400

Now, multiply the rounded difference with the unrounded number and subtract. Why subtract? Since we rounded up, 400 is larger then the actual answer. This means we’ll need to subtract to compensate.

1 × 4 = 4
400 – 4 = 396

Lets try another example:

304 × 9 =
300 × 9 = 2,700

Notice this time, we rounded down by 4 so our answer 2700 is less then the actual number. We’ll need to add in order to compensate.

4 × 9 = 36
2,700 + 36 = 2,736

Here’s another way of looking at it. It allows you to see more of the math behind this little trick:

24 × 8 =
(25 – 1) × 8 =
(25 × 8) – (1 × 8) =
200 – 8 = 192

Also notice the rounding was done to 25 since multiplying by 25 can be as easy to some as multiplying by 10. Remember, one can round to any number.

I briefly mentioned FOIL in my previous posting and since my next multiplication post will mention it again, I thought it best to touch on it. FOIL is taught in schools as a way to help remember how to multiply binomials. Most people probably know it though I’ve heard of some schools not teaching it. FOIL stands for: First, Outer, Inner, Last. Where each word describes a terms specific location.

(a + b) × (c + d)

F: (a × c)
O: (a × d)
I: (b × c)
L: (b × d)

This yields the following:

ac + ad + bc + bd