SOLUTION: the first two terms of a sequence are 4 and 5, and each subsequent term is the sum of the two preceding terms. how many of the first 1000 terms are odd?

We use the facts that
An even number + an odd number = an odd number
An odd number + an odd number = an even number

The 1st term is 4 which is even
The 2nd term is 5 which is odd
The 3rd term is an even + an odd which is odd
The 4th term is an odd + an odd which is even
The 5th term is an odd + an even which is odd
The 5th term is an odd + an even which is odd
...

So the pattern is:

 1. even 
 2. odd 
 3. odd
 4. even
 5. odd
 6. odd
 7. even 
 8. odd
 9. odd
10. even

It will be easier to find the number of evens than the number of 
odds in that sequence.  Then we can subtract from 1000 to get the
number of odds.

The sequence of term NUMBERS of evens in that sequence is

1,4,7,10,...

That is an arithmetic sequence with first term a1 = 1, 
and common difference d = 3

[Don't get confused here because we have a new ARITHMETIC sequence 
whose terms themselves are the term NUMBERS of evens in the original 
sequence, which is NOT an arithmetic sequence.]

an = a1 + (n-1)d

an = 1 + (n-1)(3)

an = 1 + 3(n-1)

an = 1 + 3n - 3

an = 3n - 2

There are 1000 terms in the original sequence. Therefore
all term numbers of the original sequence are 1000 or less:

    an < 1000 
3n - 2 < 1000
    3n < 1002
     n < 334

So term number a334 = 3(334) - 2 = 1002 - 2 = 1000
That is, term number 1000, the last term in the original sequence
is even, so there are 334 term NUMBERS in the original sequence that
are term NUMBERS of evens.  So there are 334 terms in the original 
sequence that are even.

However we were asked for the number of odd terms in the
original sequence, so the answer is 1000 - 334 = 666.

Answer: 666  [the mark of the beast! :)  ]

Edwin