The equation $x^2 - (1A)x + A0 = 0$ has positive integer solutions where $A$ is a positive  single digit. How many such $A$s exist? (Since $A$ is representing a digit, if $A = 2$ then   $A0$ represents the integer 20.)
We need to find two numbers with a product of $A0$ and a sum of $1A$, where $A$ is a positive single digit. There are only 9 digits to try for $A$. Suppose we have a product of 10 and a sum of 11, then the two numbers could be 1 and 10. Suppose we have a product of 20 and a sum of 12, then the two numbers are 2 and 10. This will work for all values of $A$ from 1 to 9, so there are $\boxed{9\text{ values}}$ of $A$ that work.
9values
Level 5
Algebra
Problem: The equation $x^2 - (1A)x + A0 = 0$ has positive integer solutions where $A$ is a positive  single digit. How many such $A$s exist? (Since $A$ is representing a digit, if $A = 2$ then   $A0$ represents the integer 20.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many values of A yield positive integer solutions for the equation x^2 - (1A)x + A0 = 0, we first recognize that A is a single-digit positive integer, meaning A can take values from 1 to 9. 

Next, we rewrite the equation in standard quadratic form: x^2 - (10 + A)x + 10A = 0. Here, we substitute 1A with (10 + A) and A0 with 10A. 

For the quadratic equation to have positive integer solutions, the discriminant must be a perfect square. The discriminant D is given by D = b^2 - 4ac, where a = 1, b = -(10 + A), and c = 10A. Thus, we compute:

D = (10 + A)^2 - 4(1)(10A) = (10 + A)^2 - 40A.

Expanding this, we have:

D = 100 + 20A + A^2 - 40A = A^2 - 20A + 100.

We require D to be a perfect square, say k^2, leading to the equation:

A^2 - 20A + 100 = k^2.

Rearranging gives us:

A^2 - 20A + (100 - k^2) = 0.

The discriminant of this new quadratic in A must also be a perfect square for A to yield integer solutions. The discriminant is:

D_A = (-20)^2 - 4(1)(100 - k^2) = 400 - 400 + 4k^2 = 4k^2.

This means k must be an integer, and thus k can take values such that 4k^2 is a perfect square. 

Next, we analyze the original equation's roots using the quadratic formula:

x = [(10 + A) ± √D] / 2.

Both roots must be positive integers. This leads us to check integer values of A from 1 to 9, ensuring that both roots derived from the quadratic formula yield positive integers.

By testing each value of A, we find the following:

- For A = 1: Roots are positive integers.
- For A = 2: Roots are positive integers.
- For A = 3: Roots are positive integers.
- For A = 4: Roots are positive integers.
- For A = 5: Roots are positive integers.
- For A = 6: Roots are positive integers.
- For A = 7: Roots are positive integers.
- For A = 8: Roots are positive integers.
- For A = 9: Roots are positive integers.

Each value of A from 1 to 9 results in positive integer solutions for x. Therefore, the total count of valid A values is 9. 

The final answer is 9values.<|END_OF_LATENT|> The final answer is $9values$.