asthagupta wrote:
isnt there any other way, other than finding all the initial values for both statement, as it took a lot of time?
looking forward for OE
I think there is another approach possible to this question. For every base and exponent of the form a^n, where a and n are positive integers, there always exists a certain cyclicity in powers for units digit, tens digit, hundreds digit etc.
Eg, consider 2^n. For various values of 'n', 2^n will take various values of last digit (units digit) but at a certain point it will start getting repeated. 2^1, 2^2, 2^3, 2^4 end in 2, 4, 8, 6 respectively. After that, it starts getting repeated - so 2^5, 2^6, 2^7, 2^8 also end in 2, 4, 6, 8 respectively.
Similarly 11^n will take various values of tens digit depending on the value of n. So 11^1, 11^2, 11^3, 11^4, 11^5, 11^6, 11^7, 11^8, 11^9, 11^10 respectively take the last two digits as 11, 21, 31, 41, 51, 61, 71, 81, 91, 01. After that it starts getting repeated. So 11^11, 11^12, 11^13.... also end in 11, 21, 31.. respectively.
Similarly 5^n, once it starts getting into 3 digits, will take its hundreds place digit as 1 and 6 only. 5^3 = 125, 5^4 = 625, 5^5 = 3125... and so on.
So, the approach is as follows: Knowing the fact that units or tens or hundreds digit for a^n will start repeating after a certain fixed value of power 'n' (depending on the value of base 'a' also),
we can never uniquely determine a single value for which the tens place digit will be 4 or for which the hundreds place digit will be 6. We are sure to get various values for which the same tens digit and hundreds digit will occur.
Since there is no unique value, answer has to be straight
E.