0Any three digit number is of the form ax2+bx+c for x = 10, a ¹ 0, a, b, c • 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 same result as in (a)
for P=10a + b, the ‘previous’ of 5.
