Here is an Easy way to find Squares of numbers which contain 5 as their last digit..Consider the digits written before 5 as a single number and multiply it be its succcessor and then write it and put 25 as the last two digits of it...For example:
to find square of 15:: (1*2)=2 and then write 25 at the end ,ie. 225
similarly to find square of 25:: (2*3)=6 and then write 25 at the end i.e, 625
for finding square of 105:: (10*11)=110 and then write 25 at the end ,i.e, 11025
PLS RATE IT IF YOU LIKE IT !!
Some of The Related Articles:
What is PSEUDO FORCE ?
Image formation in Plane Mirror @ Infinity Distance
Kaprekar Transformation for 3-Digit Numbers
i will show how it works.
the numbers are of the form 10n + 5.
(10n+5)^2 = 100n^2 + 100n + 25.
or, 100n(n+1) + 25.
since we are multiplying by 100, at least 2 zeroes must be at the end , so adding 25 will result in the last two digits be 25.
n(n+1) is also justified.
its very nice