I can hear you smile

Hey techmath, The method you've used might get tedious for 3 digit numbers. I've got a much easier way: 1) For 2 digit number: NOTE: we are dealing with 2 digit number, so we must have only one digit entries. Let's take (32)^2. -> Square each digit and write 'em down at both ends. Like--> 9____4. (3^2=9, 2^2=4). -> Now need to find the middle portion of answer, for that just multiply both digit and double the result, That implies, (3x2)x2 = 12. -> So for that, write 2 in blank space and carry 1 to next digit. Why? Refer the NOTE above! That is, 924 and then add 1 to 9. So we get -> 1024. 2) For 3 digit number: Let take (409)^2. NOTE: we are dealing with 3 digit number, so we must have only 2 digit entries and group the number in pair of two from right. -> Group the numbers from right (in pair of two) , so we have two groups as: Group 1:- 4, Group 2:- 09. Now the rest of the procedure is same. -> Follow same procedure. square the two groups and write at both ends. Like--> 16____81. (4^2=16, 9^2=81). -> Now need to find the middle portion of answer, for that just multiply both group and double the result, That implies, (4x9)x2 = 72. -> So answer is, 16 72 81 = 167281. 3) Another 3 digit number: Let's try with a tougher number:- (825)^2. -> Group the numbers from right (in pair of two) , so we have two groups as: Group 1:- 8, Group 2:- 25. Now the rest of the procedure is same. -> Follow same procedure. square the two groups and write at both ends. Like--> 64____25. Why? Refer NOTE for 3 digit numbers. We must have only 2 digit entries for 3 digit number. And (25)^2=625. So write 25 and take 6 as carry.---------------------------------------> (1). -> Now need to find the middle portion of answer, for that just multiply both group and double the result, That implies, (8x25)x2 = 400. But write only 00 and take 4 as carry----------------------------------> (2). Hence, Result will be:- 64_____25 --> 64 00 25, But we are yet to add the carries from equation (1) and (2). Now from (1) add carry 6 to 00. And from (2) add carry 4 to 64. Final answer: 68 06 25. P.S If anyone has any doubt, ask me! I'll try to clarify! Cheers!

I usually do it like this: e.g. for 32 x 34: double one side and half the other side: 32 x 34 = 16 x 68 = 8 x 136 = 4 x 272 = 2 x 544 = 1088 It's easy to do this very quick in the head.

Without watching this, I'll just say that the way I would do 32^2 is (30 + 2)^2 = 900 + 120 + 4 = 1024.

0:36 O thanks! I want to thank my parents, my teachers and especially to my faithful friend, my calculator, because without it, this would not be possible.

I swear, this method is bloody brilliant!!! I'm squaring numbers in their hundreds! Just squared 280 in seconds in my head. After attempting to square numbers between 100 and 200, and finding that numbers such as 120 which end in a zero were easier to do by writing it as (12*10)^2=12^2 * 10^2=144*100=14,400, I found this method gets difficult after 200; and I then realised that ANY number between 100 and 1000 ending with a zero is EASY to square using tecmath's method (I actually JUST now figured out you can even for numbers above those in the 200's, as I just attempted to square 980!). I'm bloody amazed and excited man I swear ^^. You can still use this method for numbers in their hundreds which do not end in zero, however I have found they are quite difficult to do when the difference between the base you choose and the number is greater than 20, as memorising the square of numbers above 20 is infeasible, whilst those below are pretty easy. For example, choosing a base of 200, and doing 185^2: 200*170 + 15^2 = 34000 + 225 = 34225. If you aren't sure what I did, the difference between 185 and 200 is 15, so add and minus 15 from 185 and multiply those two together, and add the square of 15. If you still aren't sure then refer to tecmaths video above. Here it is easy to square 15, but if you try a number such as 167, the distance to 200 is 33, and I don't know about any of you, but I don't know 33^2 off the top of my head! In reflection though, squaring numbers even with a distance from the base greater than 20 seems to be a lot easier and faster than the traditional long multiplication method! I just attempted to square 167, having to square 32 doing the repeat process, and it still is a lot better than the alternative. Thank you tecmath for giving me the ability to square ANY/MOST numbers between 1 and 1000 in seconds/a-minute!

Comes home after school and watch math videos about something I'm not learning while I need to study on what I'm actually learning.

Thanks for the help. I am spending one month studying all these skills to try to improve in my calcula to no speed. This helps! Continue the good work.

O my freaking god, This helped me soo much.. Can't really thank you enough....

Holly crap, I can't understand why we don't learn this at school. I just figured out at 4:44 that n^2=(n-d)(n+d)+d^2. Thanks

I like the duplex method. First, I will state a few duplex patterns and maybe you will notice a pattern. The duplex method can be used to square any number with any number of digits. D(a) = a^2 ( duplex of a single digit ) D(ab) = 2ab ( duplex of a two digit number ) D(abc) = 2ac + b^2 ( duplex of a three digit number ) D(abcd ) = 2ad + 2bc ( duplex of a four digit number ) D(abcde) = 2ae + 2bd + c^2 ( duplex of a five digit number ) so 33^2 = 3^2 I 2*3*3 I 3^2 33^2 = 9 I 18 I 9 perform any carries that are necessary and we get 1089

those were really cool tricks.. it helped me a lot! make more of em. thank you so much!

I have a different method which I thought of in high school which works really well for me. Let's take 32^2 again, then you find the nearest 10 just like you -> 30^2 (=900), then let's call (for the sake of the explanation) the difference between the 2 numbers: 'd' (=2 in this case). Then to get the answer you do: 900 + (30+32)*d = 900+62*2=1024. The great thing about this method is that if you want to calculate the square of broken numbers like 41.5^2, you can do the same thing: 1600 + (40+41.5)*1.5 = 1722.25. Or the square of 0.913? .81 + (.9+.913)*.013 = .81 + .01813 + .005439 = .833569! Okay, maybe that was a bit too hard...

Just found this site and love it! I have no math in my background and am learning as much as I can. This is so much fun!

That was a great video, like most of your others and keep it up! You earned a subbie!

I've said this before and ill say it again. Small channels are always better at explaining stuff. Im subscribing because of this video

Very good, Tecmath... I teach math so I'll have to use it with my students. ThanksX

this was so heelpfull keep on making these vids :)

The method in the video works for three digit, even four digit, numbers as well. Though it is harder to multiply some numbers (for example 463, which is the same as 460*466+9)

The trick can be expressed by the equation x^2 = (x + a)(x - a) + a^2 Technically, we could choose any number "a" to add and subtract from x, then add back its square, and it would still work. But we choose "a" as the difference to the nearest 10 so that one of the (x +/- a) terms come out to round 10's, for easy mental multiplication.

This works. It is faster than doing it by hand most of the time. Of course, you have to be used to "storing" sub-calculations in your head while you "get the rest of it." That's the challenge for me, not so much the computations. There are a several methods that are almost instant: they are a little more sophisticated, BUT require less mental "storage" and computation. They are especially effective if you know your perfect squares up to 24 (not that hard to memorize). If you know those, you can square any number in your head REALLY fast with very little computation--like no more than two seconds. The hardest numbers are between 71 and 74, but they are still faster than the method in the video. PLEASE NOTE, I am not trying to discredit the video. It works and this guy has a bunch of great methods for doing all sorts of fun math computations that I hadn't seen and I think are wonderful. But for squares, there are four methods used for different ranges of numbers that are almost instant and less mentally taxing. The quickest of them is what you could call "base 50 on 25" (NOT the same base 50 method he uses in one of his multiplication videos). So, if someone asked you to square 57 (for example), you almost instantly blurt out "thirty two forty-nine" with almost no mental computation at all (only squaring the one's digit, 7, and adding 7 to 25). That forms in your head as 32, which you affix to 49. If asked 42 squared, you almost instantly say, "seventeen sixty-four" (slightly harder than the above number, but not much). The only thing you actually did computationally in your head is subtracted 8 from 25 (8 is the difference between 50 and 42), squared that difference, and affixed your results side-by-side. That forms in your head as 50-42 is 8. 25-8 is 17. 8 squared is 64. You now affix17 and 64. 1764. That last squared quantity will take the first two place values if it is a two digit number and three place values if it has three digits, meaning you will have to "overlap" a single digit when you affix the two quantities. So you always work with the distance between 50 and your number and "center" the result around 25: you either subtract it from 25 or add to 25. That result will start with the thousands place. You then square the distance from 50 affix it to what you had from the first step. Here's one more: 63 squared. It is "thirty-nine sixty-nine." To get it, you think of the difference between 50 and 63. So, 13. You add that thirteen to 25. So 38. This is actually 3800, but just think of it as 38; it's less to think about and "store" in your head. You now square 13. So 169 (you should have this memorized for these techniques). Now since the 169 is three digits long this time, you "overlap" the last digit of the 38 and the first digit of the 169. The "overlapping" is really addition. What you are doing is adding 3800 to 169. But don't even think of it that way. Only think of adding the overlapping digit: the 8 and the 1. So, you now think 39 rather than 38 and affix it to the last two digits of 169, so 69. You now have a 39 and a 69 joined to 3969. Again, don't even think of the actual big numbers or that you are adding them. You had three very simple calculations and an "affixing" or joining of the results. So you never really have to do any bigger computations, assuming you've memorized your squares up to 24. This technique works well with any number between 26 and 74. It's based on a simple algebra concept. Perhaps I'll do a video sometime... :-) (or maybe tecmath already has this method in another video?)

You helped me so much. I'm in 8th grade and I have to do three one minute 10 problem quizzes of my squares and roots 1-25. Thanks so much! You earned yourself a subscriber!

Thank you for being so smart. You just made me feel a lot more intelligent by explaining this method!! Why aren't taught all this in high school!

My calculator says you're right!! Two thumbs up and thanks for sharing :)

You just helped so many children with their maths tests well done :)

The reason behind this is simple actually You can write the equation of first question like this :- (30+2)(34-2) Let 30 be 'a', 34 be 'b' and 2 be 'c' Now, (a+c)(b-c) Now ab+cb-ac-c^2 => ab + c(b-a) - c^2 => ab + c(2c) - c^2. (b-a = 2c) => ab + 2c^2 - c^2 = ab + c^2 = 30×34 + 2^2 = 1024. (Putting the values of a,b and c)

Great Video! Keep up with the great work! Ok now I gotta show off my new quick maffz. 😀

very helpful ! thank you so much ! this was so good and effective !

I knew what 32^2 was in the first half a second because of binary and stuff.

I think it is also worth mentioning that either method works for numbers on either side of the rounding up line. Like 32^2 using the method of rounding up to the higher place of ten (40), and the difference being 8, 32-8=24. Which multiplying would be (40*20=800)+(40*4=160)=960. Adding the squared difference (8^2=64) 960+64=1024. On the other hand, using 77 and putting it to the nearest ten of 70 at a difference of 7, the multiplier would be 84. Thus it would (70*80=5600)+(70*4=280)=5880. Adding the squared difference (7^2=49), 5880+49=5929. ((Granted this is possible, I never said it was easier than your given on how to do this. I appreciate that this is out there in the first place.))

Very cool . Its a life saver. Thank you Tecmath!

OMG! Thank you so much! This method is really brilliant ^-^

For 32 squared I would do this: Picture it as 32 X 32 Units: 2 X 2 = 4 so you write down 4 Outer and inner: (2X3)+(2X3)=12 so you write down the 2 and carry the 1 Tens: 3 X 3 = 9 plus the carried 1 = 10 so you put down the 0 and carry the 1 Add the carried 1 from the end so you have 1024 It's the quickest way to do it in my opinion and you can do it in your head as long as you can visualise it and know your times tables solidly

Thank you very much it really works. Bye guys I have a BIG EXAM tomorrow. 😉(Edexcel Exam).

thank you very much.i was finding such tricks.keep uploading.

Thank you so much, this trick is making exponents ALOT easier.

Nice video! :) but while doing these questions I found another way to get the answer. Eg 77^2, the difference is 7 so you can also do (70* 84) + (7^2) still give you the same answer!

The best way to find the square of 2 digit number than this method is showing below ?=34^2 step 1= last number is 4 ryt,,square of 4=16,enter last number ".........6" balance is 1 step 2=multiply between 3*4 and double it =12*2=24,,,,,,,,plus balance 1=25,enter 5 ".......56" balance is 2 step 3=square of 3 =9 plus balance 2 =11 enter that 11.......total is "1156"

I have a maths test coming up and this seriously is helping me revise!!!!!! Thank you so much!!!

This is so cool! I love learning. :)

for those looking for the math behind it: Say the number to be squared =A so looking for A*A. Name the nearest multiple of ten = B*10 (so it ends with a zero) So in general we get A = 10*B+C, where C is the difference between the nearest multiple of 10 and the original number; in this case C can be positive or negative. (for instance if A = 71, then B= 7 and C= 7*10-711= MINUS 1) then A squared = A*A = (10*B +C)*(10*B+C) = (10*B)*(10*B) + 2*10*B*C + C*C = 10*B * (10*B + 2C) + C*C As 10*B is the nearest multiple of 10, and as 10*B +2C = (10*B + C)+C = A+C, or the original number plus the difference, it can be rewritten as A*A= (10*B)*(A+C) + C*C

Is it possible to reverse the process and find the original number with the square?

Thanks for the vid.! Your videos are so helpful. Thanks again

For 3-digits, it's the same technique it's just that you need to round that number to it's nearest ten. Ex: 273 = 270 471 = 470 298 = 300 So it's easier to add/subtract. Also that the base multiplying by the sum/difference is the same. It's just a 3 digit number.

Thats too many steps... May as well multiply 32 and 32. If your that fast at multiplying 30 and 34 and adding the square and all that stuff,what is so hard about multiplying 32 and 32 and getting it over with?

〖32〗^2 = 〖30〗^2 + 2(30+32) = 1024

All these tricks are incredible! +1 subscriber!

thnks so much neat little trick, really helpful

I'm sorry but if we have to do 30 x 34 in our head, we might as well just do 32 x 32 in our head

what is the nearest number of 55

Wow this is amazing I didn't think it would be that easy

tecmath ,boss! this method is bloody awesome

Can you be my school math teacher please ?

Why bother multiplying 30x34....it takes just as much work to multiply 32x32

chris hemsworth teaching me maths

this is amazing! thanks for the tip!

use (a+b)^2 or (a-b)^2 instead.. its faster this way.. For 77 it'll be (80-3)^2 = 80^2 + 3^2 - 2×80×3 = 6400+9 -480=5929

32^2 was easy because 32=2^5 32^2=2^10=1024

For 77² I go down from 80² : so, 6400 - 6x80 + 9 My method is from drawing squares on cm/mm paper, and from there simply "visualized". So, to go from 10 to 12 squared you need to add 2 rows of 10 on the right, as well as on top, plus add 2 square to fill the "missing piece", to get the 12 by 12 square

this is like the best thing ever!! I can do it in like ten seconds

It feels like Mr.Bean is talking in a different style...

If it is 32 * 32 You could do 30 * 2 + 30 * 2 = 120 Then you would do 30 * 30 = 900 Add them 120 + 900 = 1020 Then 2 * 2 = 4 Add them 1020 + 4 = 1024

Hello tecmath, can u teach me how to find the answer for a number with big power root by mind trick? For example : a^2018 I found that many math competition involves this. Thank you.

Thank you for your video, it has been really helpful.

32^2 is just 32x32... How is doing all that work to get 2 2 digit numbers which you still have to multiply easier? :S

it would be easier just to multiply the number being squared by itself

how did u get the extra number on top.. for instant, 77 squared and then there u got number 3 on top how did u get it??

Great trick man Your're doing great work!

you can just multiply 32×32

ORRRR.... you can make it simpler: Let's take a number. I'll use 32 like he did in the video. I'll call digit a = 3 and digit b = 2. Take the first digit, square it, and multiply by 100: 100(3^2) = 900 Multiply the two digits together, then by 20: 3 * 2 * 20 = 120 Square the last digit: 2 ^ 2 = 4 Add them up: 900 + 120 + 4 = 1024

Awesome video this method is brilliant now I can square too fast than before

I am a bit late but I'd like to post my theory on squares. Let's say you know a number squared...per say 2^2, but you don't know the next number squared... in this case, 3^2. Take those numbers (2 & 3) and add them (5). Now add it to the number that you know squared. Since you know what 2^2 is, add it to the number you got in result of adding the square roots(5). You get 9...AKA 3^2. This works for any square as long you know the one before. Not too sure how I found this out but hey, I like it.

Truth is: You wrote it down and did not calculate merely in your head.;-)

thanks for helping and uploading these types of video

Thank you so much! I am starting to understand algebra a lot better now!

thanks a lot cause this is not one of those videos guys where you've gotta do huge complications and all....... THIS IS BRILLIANT MAN....... TECMATH ROCKS.............

Currently studying the ASVAB Arithmetic Reasoning subtest. And one of these questions definitely messed me up. This helps out a lot!

This is so cool! I have a different method i made up when i was bored but both work awesome!

Thank you so much for your videos, you are truly a godsend.

This helped me sooooooo much!

thanks for this awesome trick!!

I must say. The teacher is a very good guide. I do not know much maths .I just watched your vedio and calculated all saqures. It is brilliant and it works well. Thanks very much. I have one request that please upload any sort of vedio which shows multiplication in a quick and easy way.

Thx teacher ! Greetings from Germany

my simple trick of 32^2 is 3^2 is written as 09 and 2^2 is written as 04 now mix both i.e 0904 now 3*2 is 6 and multiply with 2 i.e 12 now add 0904 to 12 in this manner 0904 12x -------- 1024

this is a great video great explanation and clear steps good job m8

This channel is freaking awesome.

Woah! Thanks for this cheeky method to square numbers easily! This method is legitimate!

THAT"S AWESOME THANKS!!!!!!!

So, for the last problem, 77 squared, is the distance from the number to the nearest ten counted as a negative? This is what I get from it, but I don't want to assume and get a question wrong by using an incomplete formula.

Thanks a lot sir for the video. Really a time saving method.

Wow thanks for helping me I really appreciate it you made squaring easier for my test

Hi Mathtech what if you have a number like 55? What closest 10s number do you go with? 60? Or 50?

that'd be cool if you had downloadable quizes in the description box to really grind in this method. Other than that, ur awesome. thank you so much!

For 77^2, I was afraid to write 3 (to the nearest 10) worried it was going to be a -3, so I wrote a +7. 70x84=5880+49=5929. It still works. But then I can see why it makes sense to go with the smallest number possible if we're doing this mentally.

best strategy to think square of number fastly .before I have never known of this methon

Good work with your videos tecmath, it is really easy to comprehend the concepts, but had a suggestion to the concept (or trick) in this video though. As per the trick it requires to perform a multiplication of numbers anyway, so why not multiply the number itself two times (using the fast multiplication trick), wouldn't that be neat? For example: 45^2 would be the same as 45*45 Perform this calculation, instead of finding the nearest tens number and following the rest of the trick!

Thx! It was a great help for my homework!

Thank you a ton!

this method is brilliant!!! thank you!!!

Really helpful! Thanks :)