Fun with Mathematics

 

                                                                             (2010)²⁰⁰⁹

What would be the last two digits in (2011)

If the root number ends with 1, whatever may its power, it will always end with 1. So the problem is to determine the last but one digit.

11² = 121, when the power ends with 2, the last but one digit is 2

11³ = 1331

11⁴= 14641
11⁵ = 161051
11⁶ = 1771561
11⁷ = 19487171
11⁸= 214358881
11⁹ = 2357947691
11¹⁰ = 25937424601

Since the power number of 2011 is 2010 ²⁰⁰⁹   [=201²⁰⁰⁹x 10²⁰⁰⁹]  ends with zero(zeros), the last but one digit will be 0.

If the power number of 2010 is changed from 1 to any real number , our prediction will remain unchanged.

                                                    (2009)²⁰¹⁰

If the given number is  (2011)                      the result will be quite different. It is noted that the power of any numbers ending with 9 will always end either with 9 (for odd power) and 1 for even power

[for example 9² = 81;9³ = 729;9⁴ = 6561; 9⁵ =59049 and so on.].Since the power of 2009 is even, the power of 2011 will end with 1. Hence the, last two digits will be 11 and this is true for all even powers of 2009. In the case of odd power ,the power of 2011 will ends with 9, and hence the last two digits will be 91.