Answer:
The first term in the sequence is odd-valued and so there are m choices for it.2020 Gauss Contest Solutions Page 19
The second term in the sequence is even-valued and so there are m − 1 choices for it.
Thus, there are a total of m × (m − 1) FT sequences that begin with an odd-valued term
followed by an even-valued term.
Finally, we consider the FT sequences that begin with an even-valued term followed by an
odd-valued term (Parity #4).
Again, there are exactly twice as many odd-valued terms as there are even-valued terms in the
first 2019 terms (since the pattern repeats even, odd, odd).
However in this case, the 2020th term is even and so there are fewer than twice as many odd valued terms as there are even-valued terms.
Thus, there are m2 + m × (m − 1) FT sequences that satisfy the required conditions.
Since there are 2415 such FT sequences, we may solve m2 + m × (m − 1) = 2415 by trial and
error.
Evaluating m2 + m × (m − 1) when m = 30, we get 302 + 30 × 29 = 1770, and so m is greater
than 30.
When m = 33, we get 332 + 33 × 32 = 2145.
When m = 34, we get 342 + 34 × 33 = 2278.
When m = 35, we get 352 + 35 × 34 = 2415, as required.
Answer: (D)
Step-by-step explanation:
