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-   -   Methods for squaring 2-digit numbers (http://mymathforum.com/elementary-math/339316-methods-squaring-2-digit-numbers.html)

 burner830 March 3rd, 2017 11:30 AM

Methods for squaring 2-digit numbers

Hi there,

I recently noticed a pattern in squaring 2 digit numbers. I squared 13 by doing this:

10x10=100, 3+3=6, and 3x3=9
13 squared = 169.

I then tried it out with 14 by doing this:
10x10=100, 4+4=8, and 4x4=16
Then, I added 8 to 1, getting 9 and my final answer was 196.

Over the past two days, I noticed that this method worked with all two digit numbers. So the method goes like this:

Square the first number, add the second number by itself and multiply it by the first number, then square the second number (and make sure to carry everything over to its right place!). This always yields the correct answer and it's simply enough for me that I can do it in my head in seconds.

I was just wondering if anybody else has learned or discovered this method before. If not, how do you square numbers in your head?

 MarkFL March 3rd, 2017 11:43 AM

What you've discovered essentially is:

\$\displaystyle (10A+B)^2=(10A)^2+2(10A)(B)+B^2\$ :D

 JeffM1 March 3rd, 2017 11:49 AM

This is an old trick. But VERY good job discovering it on your own.

But your description is not quite accurate.

$(10n + m)^2= 100n^2 + 20mn + m^2.$

With 13, 100 * 1 = 100.

20 * 3 * 1 = 60.

3^2 = 9. So 100 + 60 + 9 = 169 = 13 * 13.

Here is a more accurate description.

Square the first digit and multiply that square by 100.

Multiply the first and second digits and multiply that product by 20.

Square the second digit.

 burner830 March 3rd, 2017 11:55 AM

Ahh, ok - thank you so much for putting that together! I asked my mother and she didn't remember how to do it, but she remembered that they used to use something similar to what I did.

I appreciate your reply as I've been telling everyone about it and trying to find out how many people do it this way. :)

 skipjack March 3rd, 2017 12:58 PM

Quote:
 Originally Posted by burner830 (Post 563433) Square the first number, add the second number by itself and multiply it by the first number, then square the second number.
This isn't a correct description. If, for example, the 2-digit number is 31, the first and last digits of its square (961) are 9 and 1, which are the squares of the original digits, but the middle digit is 6, which isn't obtained by following your instruction "add the second number by itself and multiply it by the first number".

 burner830 March 3rd, 2017 01:14 PM

Quote:
 Originally Posted by skipjack (Post 563439) This isn't a correct description. If, for example, the 2-digit number is 31, the first and last digits of its square (961) are 9 and 1, which are the squares of the original digits, but the middle digit is 6, which isn't obtained by following your instruction "add the second number by itself and multiply it by the first number".
Actually, when I said "first" and "second" numbers, I wasn't referring to the square. Let me illustrate:
Let's take 31. The first number is 3. We square the first number and get 9. Then, we add the second number (1) by itself to get 2. Then we multiply that by the first number (3) to get 6. Finally, we multiply the second number to get 1. The answer is 961 which is correct.

Sorry for the confusion!

 JeffM1 March 3rd, 2017 03:01 PM

Burner

You have found an algorithm (look it up). Good for you.

But you have not perfected it, or if you have, you are not doing a good job describing it. I gave you an accurate description.

\$87^2 = 7569.\$

\$8^2 = 64.\$

\$7 + 7 = 14.\$

\$8 * 14 = 112.\$

\$7^2 = 49.\$

The correct answer is NOT 6411249.

Nor is the correct answer 64 + 112 + 49 = 225.

The proper description

Multiply 100 times the square of the first digit.

Multiply 20 times the product of the first and second digits.

Square the second digit.

\$8^2 * 100 = 6400.\$

\$20 * 8 * 7 = 20 * 56 = 1120.\$

\$7^2 = 49.\$

\$6400 + 1120 + 49 = 7569.\$

 skipjack March 3rd, 2017 04:21 PM

Quote:
 Originally Posted by burner830 (Post 563440) Then, we add the second number (1) by itself to get 2.
Ah, you meant "add the second number (1) to itself to get 2."
You also need to align the results appropriately before adding them. For example,

\$\ \$ 9
\$\ \$ 42
\$\ \ \ \ \$49
-----
1369

calculates the square of 37. The intermediate result, 42, would be more easily calculated as (3 × 7) × 2 than as (7 + 7) × 3.

 burner830 March 3rd, 2017 04:47 PM

Yeah, I've perfected it (I've successfully used it with numbers 11-99) but I know my description is poorly lacking. I'm not really a math buff so it's not super easy for me to explain what I know. It's just one of those situations where I understand it and see it in my mind's eye but it comes out overly complicated when I try to explain it. =/
That's why I mentioned that bit about carrying everything to the right places.

 Timios March 23rd, 2017 06:42 PM

Here's another trick for squaring special two digit numbers:

To square any two-digit number ending in 5, multiply the tens digit by 1 greater than the tens digit and place 25 to the right of it.

For example to obtain 65² , multiply 6 × 7 to get 42 and place 25 to the right to get 4225.

87²
8×9 = 72 and so 87²=7225.

It works with three-digit numbers too, but in cases such as 425², it isn't easy to multiply 42×43 mentally (at least not for me).

Interestingly, this method can be applied to multiplying two two-digit numbers in which the tens digits are the same and the ones digits add to 10. Do the same with the tens digits and simply multiply together the ones digits.

Examples:

To work 43×47, the tens digits give us 4×5 =20 and the ones digits give us 3×7=21
Thus 43×47=2021
62×68 = 4216
71×79=5609