Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N) +
S(N)$. What is the units digit of $N$?
Suppose $N=10a+b$. Then $10a+b=ab+(a+b)$. It follows that $9a=ab$, which implies that $b=9$, since $a \neq 0$. So the units digit of $N$ is $\boxed{9}$.
9
Level 4
Algebra
Problem: Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N) +
S(N)$. What is the units digit of $N$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the units digit of the two-digit number \( N \), we start by expressing \( N \) in terms of its digits. Let \( N = 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit. The product of the digits is \( P(N) = ab \) and the sum of the digits is \( S(N) = a + b \). 

The problem states that \( N = P(N) + S(N) \). Substituting the expressions for \( P(N) \) and \( S(N) \) into this equation gives:

\[
10a + b = ab + (a + b)
\]

Simplifying this equation, we have:

\[
10a + b = ab + a + b
\]

Subtracting \( b \) from both sides results in:

\[
10a = ab + a
\]

Next, we can isolate terms involving \( a \):

\[
10a - a = ab
\]

This simplifies to:

\[
9a = ab
\]

Assuming \( a \neq 0 \) (since \( N \) is a two-digit number), we can divide both sides by \( a \):

\[
9 = b
\]

This indicates that the units digit \( b \) must be \( 9 \). 

To confirm, we can substitute \( b = 9 \) back into the expression for \( N \):

\[
N = 10a + 9
\]

Calculating \( P(N) \) and \( S(N) \):

\[
P(N) = a \cdot 9 = 9a
\]
\[
S(N) = a + 9
\]

Now substituting these back into the original equation:

\[
10a + 9 = 9a + (a + 9)
\]

This simplifies to:

\[
10a + 9 = 9a + a + 9
\]

Both sides are equal, confirming that our solution holds true. Therefore, the units digit of \( N \) is \( 9 \).<|END_OF_LATENT|> The final answer is $9$.