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ps_dahiya
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Square of a number whose unit digit is 5

Take example of 45^2 = ABCD

Multiply tenth digit and (tenth digit + 1) and then then put 25 in the end.
AB = 4* 5
CD = 25
So ABCD = 2025

This also applies to others like 125 ^2 = ABCDE

ABC = 12* 13 = 156
DE = 25
So ABCDE = 15625
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Finding square of any number

We will do it by example of 96

Steps:
1. Find a multiple of 10 nearest to 96 i.e 100
2. Is that multiple greater than 96 or lower than 96, and how much? If greater then subtract the difference from 96, if lower then add the difference to 96. Here we will add. i.e 96 - 4 = 92
3. Multiply the integer found in step 1 to step 2 i.e 9200
4. Add the square of difference found in step 2 i.e 4^2 = 16 i.e 9200 + 16 = 9616

I guarantee that this is a very easy method.

Suggestion: If you know the squares of 1 to 25 then its always advisable to find the nearest multiple a number so that multiplication with that multipe of 10 is easy.

For example, to find square of 82 you may find 80 as nearest multiple of 10 but why not take 100 as the nearest.

Then square of 82 is (82-18) * 100 + 324 = 6724

Don't think its difficult. Try it and then find yourself.
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Looks like you are doing Vedic Mathematics? That will save you a lot of time. :good
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Squares of numbers having only digit 1:

11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321

and so on....
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Bhai
Looks like you are doing Vedic Mathematics? That will save you a lot of time. :good


I did it in my high school. Just sharing with you guys. They are kind of printed in my mind since my old days.
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goodd job ps_dahiya.

thanks,
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your squares rule directly follows from

x^2 -y^2 = (x+y) (x-Y)

26^2 - 24^2 = (26+24) (26-24) = 50 *2 = 100 - Distance from 25 is 1
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Sum of consecutive squares


The formula is n(n+1)(2n+1)/6.

Where n is the number of terms


For example:

What's the sum of the squares from 3^2 + 4^2 ... 15^2 ?

S(15) - S(2) = 1236
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