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Squaring numbers

  1. Mar 28, 2013 #1
    Ok, I don't know if this method is already known or not, but I found this all by myself after some observations... so here it is...

    Suppose we want to square a number, say 67.

    What i have found is this:

    1. First get 652 which is [6*7][5*5] = 4225
    2. Forget the digit in the unit's place, ie, 5.
    3. Take the remaining digits. In this case it's 422
    4. This is my observation. For every 5, add 1 as many times as the original number is greater than the nearest 5 multiple. Since 65 is 5*13, we add 13 (67-65)=2 times.ie, we add 13*2=26 to the 'remaining number'. That is we do 422+26=448
    5. Put the last digit of the square of the digit in the unit's place of the original number, after the sum obtained ie, put the last digit of 7*7 ie, 9, after 448, getting 4489 as the required square!

    Another example:

    Finding 822...

    1. 802=6400
    2. 80=5*16
    3. 640+16*(82-80)=640+32=672
    4. required square is 6724

    The nearest multiple of 5 has to be taken. For 33, we must take 35 and then subtract instead of adding.

    1. 352=1225
    2. 35=5*7
    3. 122 + 7*(33-35) = 122 - 14 = 108
    4. required square is 1089

    So what do you people think of this?

    This is equally applicable for 3 digit numbers, though then the initial squaring of 5 multiples become another thing.
    But we can still apply this.

    2022 is:

    1. 2002=40000
    2. 200=5*40
    3. 4000+40*2=4000+80=4080
    4. required square is 40804.
    Last edited: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2


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    Dearly Missed

    Well, I'll look it over, but your first step looks right anyway:

    Writing a number X as y*10+5, with "y" some number, we have:
    X^2=(y*10+5)^2=y^2*100+2*y*10*5+25=y^2*100+y*100+25=(y+1)*y*100+25, which general formula is in agreement with your observations.
    I'll give the rest of your recipe a check, I don't think I've seen this particular algorithm before.
  4. Mar 28, 2013 #3


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    But don't you first have to apply your method to the closest multiple of 5 (unless all digits are zero except the first)?

    I think what you're doing is a variant of the binomial:






  5. Mar 29, 2013 #4
    squaring a 5's multiple is the simplest thing! if don't know it, know it now... to get 45^2 place these two products side by side... 4*(4+1) and (5^5) ie 4*5 and 5*5 ie 2025...
    similarly 105^2 = (10*11)(5*5)=11025

    arildno, of course this has an algebraic counter part, but we really don't need to know that do we... this stuff is arithmetic and as long as observations hold true, there's no need to prove it using formal maths... take Ramanujan's number for instance, 1729... did he do algebra while lying in death bed to say that its the smallest number to be expressed as sum of two different cubes? most likely not... he merely 'observed' it to be true...
  6. Mar 29, 2013 #5


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    Well, for that matter, there are methods that are way simpler than yours for squaring numbers:


    105^2 =(100)(110)+25^2 ,

    and even for larger numbers, it is easy:

    988^2 =(988+12)(988-12)+12^2=(976)(1000)+144=976144

    But I still believe you're using a variant of the binomial.
    Last edited by a moderator: Mar 29, 2013
  7. Mar 29, 2013 #6


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    Sorry, I realized I came off as an *** , please ignore my post.
  8. Mar 30, 2013 #7
    ...........i know there are simpler methods, i'm just excited cause this is "my discovery"... as i said, i dunno if this is already known or not, but i found it on my own :)
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