Miscellaneous-2

If  x = (9+4√5)^48 = [x] + f, where [x] is defined as integral part of x and f is a fraction, then find the value of x(1-f)?

SOLUTION-

x = (9+4√5)^48
x = 48c0*(9)^48 + 48c1*(9)^47 *(4√5) + 48c2*(9)^46 *(4√5)^2+........+48c47*9*(4√5)^47 +                  48c48*(4√5)^48----------------(i)
All the terms, that have even power of (4√5) will be integers and the other terms will be irrational numbers.
Now, x = [x] + f
Let x' = (9 - 4√5)^48
Now (9 - 4√5) < 1
Hence, (9-4√5)^48 < 1
x' = (9-4√5)^48 = f'
x' = 48c0*(9)^48 - 48c1*(9)^47 *(4√5) + 48c2*(9)^46 *(4√5)^2 - 48c3*(9)^45 *(4√5)^3+.......+             48c48*(4√5)^48------------(ii)
Adding eqn (i) and (ii), we get
x + x' = [x] + f +f'
          =2*[48c0*(9)^48 + 48c2*(9)^46 *(4√5)^2 + .................+ 48c48*(4√5)^48]
From above result we can see that [x] + f + f' is an Integer but f and f' are fractions,
Hence, the only possible value of (f + f') = 1
Now, f' = (1 - f) = (9 - 4√5)^48

∴ x(1 - f) = (9 + 4√5)^48*(9 - 4√5)^48 = (81 - 80)^48 = 1 ANS
   

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