I agree that the above solution does not really solve anything.
Since this question appears in a chapter of the binomial theorem, we will use that. (If you mention that this problem comes from a binomial theorem chapter, it is a huge hint.)
Observe that
(√3 + √2)2002 + (√3 - √2)2002
will be an integer. Since if you look at the binomial expansion of both expressions, the odd terms will cancel out and then the even terms will have even exponents, turning the square-roots into whole integers.
Next, observe that (√3 - √2) = (√3 + √2)-1 (you can find this by rationalizing a denominator.)
So combining these gives that
(√3 + √2)2002 + (√3 + √2)-2002
is an integer.
Since √3 + √2 is clearly larger than 1, we have that (√3 + √2)-1 is less than one and so
(√3 + √2)-2002 is really small (between 0 and 1) but positive. That is, it will look like
(√3 + √2)-2002 = 0.0000....000x... (for some digits x)
If this adds to (√3 + √2)2002 to make an integer, then it must mean that
the expression (√3 + √2)2002 must have a bunch of 9s after the decimal place:
So the digit immediately to the right is a 9.
(You can easily verify this with a calculator as well: add √3 and √2 to get
(√3 + √2) = 3.146
And just keep squaring... you'll see that "2002" has no real meaning in the problem besides being a large even number:
(√3 + √2)2 = 9.899
(√3 + √2)4 = 97.990
(√3 + √2)8 = 9601.999...
(√3 + √2)16 = 92198401.999...
And you'll see that in your textbook, the following question involving √11 + √10 is similar since 11 and 10 differ by 1 (allowing for the reciprocal property) and the exponent is again even (allowing for the first sum to be an integer.)