The average of a non-decreasing sequence of $N$ numbers $a_{1}, a_{2}, \ldots, a_{N}$ is $300$. If $a_{1}$ is replaced by $6 a_{1}$, the new average becomes $400$. Then, the number of possible values of $a_{1}$ is

We need to find how many different values $a_1$ can take given the constraints about averages and the non-decreasing sequence.

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Let us translate the given information into equations.

Original sequence: The average of $a_1, a_2, \ldots, a_N$ is 300

$a_1 + a_2 + a_3 + \ldots + a_N = 300N$

Modified sequence: When $a_1$ is replaced by $6a_1$, the new average becomes 400

$6a_1 + a_2 + a_3 + \ldots + a_N = 400N$

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We'll find the relationship between $a_1$ and $N$ by subtracting the first equation from the second:

$(6a_1 + a_2 + a_3 + \ldots + a_N) - (a_1 + a_2 + a_3 + \ldots + a_N) = 400N - 300N$

$5a_1 = 100N$

$a_1 = 20N$

This tells us that $a_1$ must be a multiple of 20, specifically $a_1 = 20N$ where $N$ is the number of terms.

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Since the sequence is non-decreasing, we have $a_1 \leq a_2 \leq a_3 \leq \ldots \leq a_N$.

Let us check different values of $N$:

For N = 1:

$a_1 = 20(1) = 20$

But then the average would be 20, not 300 as required

Not possible

For N = 2:

$a_1 = 20(2) = 40$

From the original sum: $40 + a_2 = 600$, so $a_2 = 560$

Since $40 \leq 560$, the non-decreasing condition is satisfied

Possible

For N = 3:

$a_1 = 20(3) = 60$

From the original sum: $60 + a_2 + a_3 = 900$, so $a_2 + a_3 = 840$

Since $a_2 \geq 60$ and $a_3 \geq a_2$, this is achievable

Possible

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The key insight is that since all terms are at least $a_1$, the minimum possible sum is:

$\text{Sum} \geq a_1 + a_1 + \ldots + a_1 = Na_1 = N(20N) = 20N^2$

But we also need the sum to equal $300N$. Therefore:

$20N^2 \leq 300N$

$20N \leq 300$

$N \leq 15$

For N = 15:

$a_1 = 20(15) = 300$

Since the average is 300 and all terms are at least 300 (non-decreasing), all terms must equal 300

Possible

For N = 16:

$a_1 = 20(16) = 320$

Since all terms are at least 320, the average would be at least 320 > 300

Not possible

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The valid values of $N$ are: $2, 3, 4, 5, \ldots, 15$

This gives us possible values of $a_1$ as: $40, 60, 80, 100, \ldots, 300$

Counting these values: from $N = 2$ to $N = 15$ inclusive, we have $15 - 2 + 1 = 14$ possible values.

Answer: 14