Find the value of $h$ such that the following equation in base $h$ is true: $$\begin{array}{c@{}c@{}c@{}c@{}c@{}c} &&6&4&5&3_h\\ &+&7&5&1&2_h\\ \cline{2-6} &1&6&1&6&5_h.\\ \end{array}$$
The two rightmost columns do not result in any carrying; however, in the third column there is a residue of $1$. Subtracting $1$ from the sum of $4$ and $5$, we see that $h$ is $\boxed{8}$.
8
Level 3
Number Theory
Problem: Find the value of $h$ such that the following equation in base $h$ is true: $$\begin{array}{c@{}c@{}c@{}c@{}c@{}c} &&6&4&5&3_h\\ &+&7&5&1&2_h\\ \cline{2-6} &1&6&1&6&5_h.\\ \end{array}$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the base \( h \), we need to convert the numbers from base \( h \) to base 10. The first number \( 6453_h \) can be expressed in base 10 as:

\[
6h^3 + 4h^2 + 5h + 3
\]

The second number \( 7512_h \) can be expressed as:

\[
7h^3 + 5h^2 + 1h + 2
\]

The sum \( 16165_h \) can be expressed as:

\[
1h^4 + 6h^3 + 1h^2 + 6h + 5
\]

Setting up the equation based on the addition, we have:

\[
(6h^3 + 4h^2 + 5h + 3) + (7h^3 + 5h^2 + 1h + 2) = 1h^4 + 6h^3 + 1h^2 + 6h + 5
\]

Combining the left side:

\[
(6h^3 + 7h^3) + (4h^2 + 5h^2) + (5h + 1h) + (3 + 2) = 13h^3 + 9h^2 + 6h + 5
\]

Now, equate this to the right side:

\[
13h^3 + 9h^2 + 6h + 5 = 1h^4 + 6h^3 + 1h^2 + 6h + 5
\]

Subtract \( 1h^4 + 6h^3 + 1h^2 + 6h + 5 \) from both sides:

\[
13h^3 + 9h^2 + 6h + 5 - (1h^4 + 6h^3 + 1h^2 + 6h + 5) = 0
\]

This simplifies to:

\[
-1h^4 + 7h^3 + 8h^2 = 0
\]

Factoring out \( h^2 \):

\[
h^2(-h^2 + 7h + 8) = 0
\]

This gives us \( h^2 = 0 \) or \( -h^2 + 7h + 8 = 0 \). The first case \( h^2 = 0 \) is not valid since \( h \) must be a positive integer. 

Now, solving the quadratic equation:

\[
h^2 - 7h - 8 = 0
\]

Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
h = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}
\]
\[
h = \frac{7 \pm \sqrt{49 + 32}}{2}
\]
\[
h = \frac{7 \pm \sqrt{81}}{2}
\]
\[
h = \frac{7 \pm 9}{2}
\]

Calculating the two possible values:

1. \( h = \frac{16}{2} = 8 \)
2. \( h = \frac{-2}{2} = -1 \) (not valid)

Thus, the only valid solution is \( h = 8 \).<|END_OF_LATENT|> The final answer is $8$.