Algebra-basic
Algebra
Level
3
If x + 1/x = 3,find the last two digit of the equation (x²)^2013 + 1 / (x²)^2013.
47
13
07
17
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1 solution
(x+(1/x))=1, square both sides and we have x^2 + 2x/x + 1/(x^2) = 9 which can be simplified as x^2^1 + 1/(x^2^1) = 7. --- square both sides again we have x^2^2 + 2x^2/x^2 +1/( x^2^2) = 49 x^2^2 + 2 +1/( x^2^2) = 49 x^2^2 + 1/( x^2^2) = 47
---square both sides again we have x^2^3 + 2x^2^/x^2^2 +1/( x^2^3) = 2209 x^2^3 + 2 +1/( x^2^3) = 2209 x^2^3 + 1/( x^2^3) = 2207 continue the process and we get * x^2^4 + 1/( x^2^4) = 4870847 * x^2^5 + 1/( x^2^5) = 23725150497407
we are only ask for the last two digit of x^2^2013 + 1/(x^2^2013) and in x^2^a where a is a whole number we get when a=0, the last two digits are 03 a=1, the last two digits are 07 a=2, the last two digits are 47 a=3, the last two digits are 07 a=4, the last two digits are 47 a=5, the last two digits are 07
and we can see a pattern that for every odd value of a, a>o is 07 and for every even value of a, a>0, is 47 and since 2013 is odd we have 07 as the answer.