1. Given Information:
- The input consists of a non-decreasing sequence of \(N\) numbers: \(a₁, a₂, ..., aₙ\).
- The average of this sequence is 300, represented by the equation: \(\frac{a_1 + a_2 + ... + a_N}{N} = 300\).
2. Modified Sequence:
- The first element, \(a₁\), is replaced by \(6a₁\).
- The average of the new sequence is 400, expressed as: \(\frac{6a_1 + a_2 + ... + a_N}{N} = 400\).
3. Equation Formulation:
4. Derivation of \(a₁\) and \(N\) Relationship:
Multiplying the average equations by \(N\) gives the sums:
Original sum: \(a₁ + a₂ + ... + aₙ = 300N\)
New sum: \(6a₁ + a₂ + ... + aₙ = 400N\)
Subtracting the first sum from the second:
\((6a₁ + a₂ + ... + aₙ) - (a₁ + a₂ + ... + aₙ) = 400N - 300N\).
This simplifies to:
\(5a₁ = 100N\).
Dividing both sides by 5 yields:
\(a₁ = 20N\).
5. Constraints on \(a₁\) and \(N\):
- \(a₁\) must be a positive integer since it is the first term of a non-decreasing sequence of numbers.
- Given \(a₁ = 20N\), \(N\) must also be a positive integer.
- For a sequence to be considered non-decreasing, it must contain at least two elements, implying \(N>1\).
6. Determining Possible Values for \(N\):
- The average of the original sequence is 300. Since the sequence is non-decreasing, \(a₁ \le 300\).
- Substituting \(a₁ = 20N\) into this inequality: \(20N \le 300\).
- Dividing by 20: \(N \le 15\).
- Combining with the constraint \(N>1\), the possible integer values for \(N\) are 2, 3, ..., 15.
- This range of \(N\) values results in the following possible values for \(a₁\):
\((20 \times 2), (20 \times 3), ..., (20 \times 15)\), which are \(40, 60, ..., 300\).
7. Final Count:
- The number of possible integer values for \(N\) from 2 to 15 inclusive is \(15 - 2 + 1 = 14\).
- Each unique value of \(N\) corresponds to a unique value of \(a₁\).
- Therefore, there are 14 possible values for \(a₁\).
The correct answer is 14.