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# Squaring of Integers in 2 easy steps!

I just figured out this amazing way of squaring numbers with ones digit 5. Suppose you need to find out the square of 35. Then simply square 5 and write down 25 at last and then multiply 3*3 and add 3.That is square the remaining numbers except 5 and add the same value.You will get 12.Since you can also write it as 3 (3) + 3 = 3 (1 + 3) = 3 *4. So,you may just as well multiply the next integer. You can now easily write 35^2 = 1225. In case of another no.,say 125,you take 25 at last and multiply 12 * 13.So,you have 15625. If you are more comfortable in squaring,simply square 12,you'll get 144 and then add 12,you will get 156.So,125^2 = 15625. Rule works for no. whose ones digit is 5. You will probably not find this anywhere else.It was something I came up with while randomly squaring integers.

Note by Titas Biswas
1 year, 9 months ago

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Very good that you $$\color{red}{rediscovered}$$ this yourself. You may be interested in knowing that square of any two and three digit numbers can be found as under. $$(10a\pm b)^2$$ write square of a and then of b. (if b is 1, 2 , or 3, write its square as 01, 04 or09) to this $$\pm$$ 20 times a*b. Say $$37^2 = (30+7)^2=\color{red}{9} \color{blue}{49} +20*21 = 1369. \\37^2 = (40-3)^2=\color{red}{16} \color{blue}{09} -20*12=1369 ~\\ Three ~digit~ number ~the ~same ~way~~\\127^2 =(120+7)^2= \color{red}{144} \color{blue}{49} +20*84=16129 \\127^2 = (130-3)^2=\color{red}{169} \color{blue}{09} -20*39=16129\\With~5~at~ unit ~place ~it ~is ~ easy.~~~ (10a +5)^2= \color{red}{a*(a+1)} \color{blue}{25}.\\125^2=(120 +5)^2= \color{red}{(12*13)} \color{blue}{25}= \color{red}{156} \color{blue}{25} .$$ · 1 year, 9 months ago

Amazing! Thank you. · 1 year, 9 months ago