Comments on: Three Tricks for Squaring Numbers

By: ashok

ashok — Sun, 27 Nov 2011 01:03:18 +0000

if u want more short cut just mail me once

By: ashok

ashok — Sun, 27 Nov 2011 01:02:05 +0000

hey let me tel abt the suaring of two digit number ending with 5′s

for ex take 25*25

step 1: multiply 5*5=25
step 2:multiply 2*3=6

the answer is 625

first of all wen u see 25*25 nd 35*35 etc wen ever u fing 2 digit numbers ending with 5 you just write 25 atlast and then multiply the first num with the conseqte num as shown above

if it is 45*45
u just write last 25
and multiply 4*5=20
so answer is 2025

By: priyanka

priyanka — Wed, 07 Sep 2011 10:23:55 +0000

its very interesting

By: priyanka

priyanka — Wed, 07 Sep 2011 10:23:20 +0000

it is a really best . very interesting thanks

By: Shraman N L

Shraman N L — Thu, 18 Nov 2010 07:52:07 +0000

RECTIFICATION OF MY POST DATED NOV 4, 2005

There is another very easy metod to make a sqare:

Suppose you have to find out 43^2, just start writing from the right side
(i) 3*3=9 (unit multiplied by unit)
(ii) (3*4)*2=24 write 4 carry 2 (unit*tens and double it)
(iii) 4*4=16 add carry 2, write 18 (tens*tens)
The answer is 1849

By: tom

tom — Mon, 01 Nov 2010 19:23:33 +0000

You are all thinking WAYYYY too much into this.

lets take 18squared.

18 x 18.

You look at the end of the numbers first 8 x 8. This is 64. So you know that the final number will end up in 4.
Then you go back to the begining. 10 x 10 = 100
Now the middle. 10 x 8 = 80. Plus another 80. This is 160.

Then finish the rest 100 + 160 + 64 = 324.

THIS WORKS WITH EVERYTHING UP TO 99.

This is probably really obvious but what is the point in making it harder for yourself? I am 16 years old.

By: sushant

sushant — Wed, 02 Jun 2010 12:47:37 +0000

43 * 43 s not 1729 .it’s 1849…pls dun post wrong answers

By: Avi Pai

Avi Pai — Sun, 09 May 2010 13:03:15 +0000

Hey Guys, this is great learning. Thanks for contributing. Any discoveries on how to cube?

By: Yam Dak

Yam Dak — Sat, 24 Apr 2010 18:43:14 +0000

I thought of your second 2nd method along time ago too. I improved on it too. It’s (x+a)(x-a)+a^2.
x^2 – a^2 + a^2 = x^2
31^2= (32)(30)+1^2=961
Another example: 55^2=(60)(50)+5^2=3000+25=3025
Another example: 75^2=(80)(70)+5^2=5600+25=5625
Same example: 75^2=(100)(50)+25^2=5000+625=5625

By: Jessica

Jessica — Wed, 03 Feb 2010 04:21:17 +0000

It’s kind of hard to read since you put x2 instead of x^2 or something, but good job.

By: Matt

Matt — Tue, 01 Apr 2008 02:48:28 +0000

Just curious if anyone still reads this, i came up with something, but i don’t want to post if no one is going to read it

By: Case

Case — Wed, 09 Jan 2008 17:41:45 +0000

I too came up with the (a|n)^2 = a0^2 + n(a|n+a0) as described earlier while on my way to work and I must say that for the hard ones this seems to be the quickest mental way for me to do it… that is to say I can typically solve in 30 seconds or less with it now.

By: Ian Luke Kane

Ian Luke Kane — Tue, 21 Mar 2006 12:58:03 +0000

Jenna, unfortunately I do believe that all of these methods have been discovered long ago, although I’m sure it’s the case in mathematics that some very simple conjectures have gone a long time without having been stated. Take the Beal Conjecture, for instance. With all the talk of Fermat’s Last Theorem no one had stated the more generalized conjecture until literally hundreds of years later.

By: Jenna

Jenna — Tue, 21 Mar 2006 09:32:39 +0000

hey; the first method you have here I’ve wondered, is this a published theorum? I came up with it once myself and showed my precal teacher, who said that yes, she did recognize it. Also, though.. it can be extrapolated further if you have the spare time.

(a+n)^2= a^2 + n[a + [a+n)]

22^2 = 484

20^2 + 2(20+22) = 484

For any positive nonzero number a, the square of a+n will equal the square of a plus n[a + (a+n)].

By: Aung Khaing

Aung Khaing — Sun, 05 Mar 2006 23:06:09 +0000

Hi Ian and Nand,

My name is Aung Khaing, a sophomore student in the University of Arkansas, Fayetteville. I also like to play with numbers like you do. I devised a general formula for any squares. Please see how you like that. Because of the limited way to experess my thought on this web, all I can do is to type into Excel and copy it.

If you would have a chance to see it in Microsoft word explanation and Microsoft equation editor, you would see more clearly. If you do that on paper you don’t really need to wirte down all the zeros. You can just leave blanks in places of zeros (eg. four blanks for 10^4)and can fill in with nonzero numbers. In this way, you only need 3 lines on paper to square 3 digit number and add them up. For any 2 digit number, you can square it in mind.

For now, this is the best way I can transfer my thought. You can just copy the following formula into and excel and test it.

For 2 digit numbers, treat them like AB^2.
=(((10^2)*(A1^2))+(B1^2)+(2*(10)*(A1*B1)))

For 3 digits numbers, treat them like ABC^2.
=(((10^4)*(A1^2))+((10^2)*(B1^2))+(C1^2)

+(2*(10^3)*(A1*B1))+(2*(10^2)*(A1*C1))+(2*(10)*(B1*C1)))

For 4 digits numbers, treat it like ABCD^2.
=(((10^6)*(A1^2))+((10^4)*(B1^2))+((10^2)*(C1^2))+(D1^2)

+(2*(10^5)*(A1*B1))+(2*(10^4)*(A1*C1))+(2*(10^3)*(A1*D1))

+(2*(10^3)*(B1*C1))+(2*(10^2)*(B1*D1))+(2*(10)*(C1*D1)))

For 5 digits numbers, treat it like ABCDE^2.
=(((10^8)*(A1^2))+((10^6)*(B1^2))+((10^4)*(C1^2))

+((10^2)*(D1^2))+(E1^2)+(2*(10^7)*(A1*B1))+(2*(10^6)*(A1*C1))

+(2*(10^5)*(A1*D1))+(2*(10^4)*(A1*E1))+(2*(10^5)*(B1*C1))

+(2*(10^4)*(B1*D1))+(2*(10^3)*(B1*E1))+(2*(10^3)*(C1*D1))

+(2*(10^2)*(C1*E1))+(2*(10)*(D1*E1)))

Sincerely,
Aung Khaing
Sophomore, University of Arkansas, Fayetteville