Squares of numbers ending with 5

Squares of numbers ending in 5 : 

Consider the example 252.(consider it as square(25))

Here the number is 25. We have to find out the square of the number.

For the number 25, the last digit is 5 and the 'previous' digit is 2.

Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in

this context, gives the procedure 'to multiply the previous digit 2 by

one more than itself, that is, by 3'. It becomes the L.H.S (left hand

side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand side)

of the result is 52, that is, 25.

Thus 252 = 2 X 3 / 25 = 625.

In the same way, 

352= 3 X (3+1) /25 = 3 X 4/  25 = 1225;

652= 6 X 7 / 25 = 4225;

1052= 10 X 11/25 = 11025;

1352= 13 X 14/25 = 18225;



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Algebraic proof: 

 a) Consider (ax + b) 2 Ξ a2. X2 + 2abx + b2.

    This identity for x = 10   and b = 5

   becomes  (10a + 5) 2 = a2. 102  + 2. 10a. 5 + 52                  

                        =  a2. 102 + a. 102 + 52               

                        =  (a 2+ a ) . 102 + 52             

                        =  a (a + 1) . 10 2 + 25.

  Clearly 10a + 5 represents two-digit numbers 15, 25, 35,-------, 95

  for the values a = 1, 2, 3, -------,9 respectively.

  In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1)

  and R.H.S is 25, that is, a (a + 1) / 25.

   Thus any such two-digit number gives the result in the same fashion.

   Example:    45 = (40 + 5)2,

  It is of the form (ax+b)2 for a = 4, x=10 and b = 5,

  giving the answer a (a+1) / 25

   That is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.





 b)  Any three digit number is of the form ax2+bx+c

 for x = 10, a ≠ 0,a, b, c Є W.   

 Now (ax2+bx+ c) 2 =  a2 x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2 

                   = a2 x4+2ab. x3+ (b2 + 2ca)x2+2bc . x+ c2.  

  This identity for x = 10, c = 5 becomes  (a . 102 + b .10 + 5) 2  

                   = a2.104 + 2.a.b.103 + (b2 + 2.5.a)102 + 2.b.5.10 + 52 

                   = a2.104 + 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52   

                   = a2.104 + 2ab.103 + b2.102 + a . 103  + b 102 + 52  

                   = a2.104 + (2ab + a).103 + (b2+ b)102 +52         

                   = [ a2.102 + 2ab.10 + a.10 + b2 + b] 102+ 52      

                   = (10a + b) ( 10a+b+1).102 + 25       

                   = P (P+1) 102 + 25,  where P = 10a+b.



   Hence any three digit number whose last digit is 5 gives

  the sameresult as in (a) for P=10a + b, the 'previous' of  5.

  Example :    1652 = (1 . 102 + 6 . 10 + 5) 2. 

  It is of the form (ax2 +bx+c)2

  for a = 1, b = 6, c = 5 and x = 10.

  It gives the answer P(P+1)  / 25,

  where P = 10a + b = 10 X 1 + 6 = 16,the 'previous'.

  The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.