Methods for squaring 2digit 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? 
What you've discovered essentially is: $\displaystyle (10A+B)^2=(10A)^2+2(10A)(B)+B^2$ :D 
This is an old trick. But VERY good job discovering it on your own. But your description is not quite accurate. 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. Add the three numbers. 
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. :) 
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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! 
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. Add those three numbers. $8^2 * 100 = 6400.$ $20 * 8 * 7 = 20 * 56 = 1120.$ $7^2 = 49.$ $6400 + 1120 + 49 = 7569.$ 
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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. 
Yeah, I've perfected it (I've successfully used it with numbers 1199) 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. 
Here's another trick for squaring special two digit numbers: To square any twodigit 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 threedigit 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 twodigit 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 Now you try 44×46 and check your answer with your calculator. 
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