2023 AMC 10 A Problem 23

Difference of Squares via Averages

Express each pair of divisors in terms of their average. This naturally converts both products into perfect squares, transforming the Diophantine equation into a standard difference of squares that can be factored.

Step 1: Express the products using the average of the divisors

For any two numbers, their product can be written as the square of their average minus the square of their half-difference. Let the first pair of divisors be $a$ and $a+20$. Their half-difference is $10$, so we can define their average as $x=a+10$. We write $N$ in terms of $x$. Similarly, let the second pair of divisors be $c$ and $c+23$. Their half-difference is $11.5$. Defining their average as $y=c+11.5$, we write $N$ in terms of $y$.

Step 2: Equate the two expressions and scale to integers

Since both expressions equal $N$, we set them equal to each other. To eliminate the decimal and create an integer equation, we multiply the entire equation by $4$. Notice that since $y=c+11.5$, multiplying $y$ by $2$ results in an integer. The same is true for $2x$.

Step 3: Factor as a difference of squares

We apply the difference of squares identity to factor the left side. Then, we substitute $2y$ and $2x$ back in terms of the original variables $a$ and $c$.

Step 4: Constrain the factors using logical bounds

Since $a$ and $c$ are positive integers, the smallest they can be is $1$. This allows us to find the minimum possible value for the sum factor. The right side is a constant, and we must find a factor pair of this constant that satisfies our minimum bound. The only factor pair of $129$ large enough for the sum is $129$ and $1$.

Step 5: Solve for the divisors

We simplify the two factor equations to find a system of linear equations in terms of $a$ and $c$. Adding the two simplified equations allows us to solve for $a$, and subtracting them gives $c$.

Step 6: Calculate the target number and its digit sum

We substitute $c$ back into the original expression to find the value of $N$. Finally, we sum the digits of $N$ to find our answer.

Variable Substitution and Bounding

Instead of factoring, recognize that the divisor pairs must be very close in value to produce the same product. By substituting $a=c+k$ and analyzing the resulting equation’s signs, $k$ is forced to be exactly $1$.

Step 1: Establish a size relationship between the divisors

Let the first pair of divisors be $a$ and $a+20$, and the second pair be $c$ and $c+23$. We set their products equal. Since both $c$ and $c+23$ are strictly increasing functions for positive $c$, and the gap between $c$ and $c+23$ is larger than the gap between $a$ and $a+20$, the starting value $a$ must be strictly greater than $c$ to compensate and make the products equal.

Step 2: Introduce a difference variable

Since $a$ is an integer strictly greater than $c$, we can define a new positive integer variable $k$ representing their difference. We substitute this into our equation and expand the terms.

Step 3: Isolate the original variable

We simplify the equation by canceling common terms and moving all terms containing $c$ to one side. We then factor $c$ out to isolate the relationship between $c$ and $k$.

Step 4: Apply positivity constraints to find the exact difference

Because $k$ is a positive integer, the left side of the equation is strictly positive. For the right side to be positive as well, the expression in the parentheses must be positive. This creates a strict inequality that limits the possible values of $k$. The only positive integer that satisfies this inequality is $1$.

Step 5: Calculate the divisors, the target number, and the digit sum

We substitute $k=1$ back into our factored equation to solve for $c$. With $c$ known, we compute $N$ and then sum its digits.