A divisor of a number is a proper divisor if it is not equal to the number. What is the sum of the proper divisors of $432$?
The prime factorization of $432 = 2^4 \cdot 3^3$. It follows that the sum of the divisors is equal to $(1 + 2 + 2^2 + 2^3 + 2^4)(1 + 3 + 3^2 + 3^3)$, as each factor of $432$ is represented when the product is expanded. We have to subtract $432$ so that we only count the proper divisors, so the answer is \begin{align*}
(1 + 2 + 4 + 8 + 16)(1 + 3 + 9 + 27) - 432 &= (31)(40) - 432\\
&= 1240 - 432\\
&= \boxed{808}.\\
\end{align*}
808
Level 4
Number Theory
Problem: A divisor of a number is a proper divisor if it is not equal to the number. What is the sum of the proper divisors of $432$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the proper divisors of 432, we first need to determine the divisors of 432. We start by finding the prime factorization of 432. 

432 can be divided by 2: 
432 ÷ 2 = 216 
216 ÷ 2 = 108 
108 ÷ 2 = 54 
54 ÷ 2 = 27 

Now, 27 is not divisible by 2, so we switch to the next prime number, which is 3: 
27 ÷ 3 = 9 
9 ÷ 3 = 3 
3 ÷ 3 = 1 

Thus, the prime factorization of 432 is:
432 = 2^4 × 3^3.

Next, we use the formula for finding the sum of the divisors based on the prime factorization. The sum of the divisors σ(n) for n = p1^k1 × p2^k2 is given by:
σ(n) = (p1^(k1+1) - 1) / (p1 - 1) × (p2^(k2+1) - 1) / (p2 - 1).

Applying this to our factorization:
For 2^4:
(2^(4+1) - 1) / (2 - 1) = (2^5 - 1) / 1 = 31.

For 3^3:
(3^(3+1) - 1) / (3 - 1) = (3^4 - 1) / 2 = (81 - 1) / 2 = 40.

Now, we multiply these results to find the sum of all divisors of 432:
σ(432) = 31 × 40 = 1240.

Since we are interested in the sum of the proper divisors, we need to subtract 432 from the total sum of the divisors:
Sum of proper divisors = σ(432) - 432 = 1240 - 432 = 808.

Thus, the final answer is 808.<|END_OF_LATENT|> The final answer is $808$.