The solution is: 14
Given:
The average of a non-decreasing sequence of N numbers, \(a_1, a_2, ..., a_N\), is 300.
When \(a_1\) is replaced by \(6a_1\), the new average is 400.
Recall the formula for average:
Average = \(\frac{\text{Sum of elements}}{N}\)
Original sum = \(300N\)
New sum = \(400N\)
The change in the sum is due to the replacement of \(a_1\) with \(6a_1\). Therefore:
New sum = \(300N - a_1 + 6a_1 = 300N + 5a_1\)
Equating the two expressions for the new sum:
\(300N + 5a_1 = 400N\)
\(5a_1 = 100N\)
\(a_1 = 20N\)
Since \(a_1\) must be a positive integer in a non-decreasing sequence, N must also be a positive integer. The original average implies:
\(\frac{a_1 + a_2 + \cdots + a_N}{N} = 300 \Rightarrow a_1 + a_2 + \cdots + a_N = 300N\)
Substituting \(a_1 = 20N\):
\(20N \leq 300 \Rightarrow N \leq 15\)
Additionally, for a sequence to be considered non-decreasing, it must contain at least two elements, thus \(N \geq 2\).
Therefore, the possible integer values for N range from 2 to 15, inclusive. This gives 14 possible values for N.
Each value of N corresponds to a unique value of \(a_1 = 20N\): 40, 60, 80, ..., 300.
∴ The number of possible values for \(a_1\) is 14.