To determine the values of $a$ and $b$ such that the number $8726a2415b$ is divisible by 9, we need to use the rule for divisibility by 9: a number is divisible by 9 if the sum of its digits is divisible by 9.

Let's start by calculating the sum of the digits in $8726a2415b$:

$8 + 7 + 2 + 6 + a + 2 + 4 + 1 + 5 + b = 35 + a + b$

For the number to be divisible by 9, the sum $35 + a + b$ must be divisible by 9. Let's find the values of $a$ and $b$ that make this sum divisible by 9.

First, let's look at the possible remainders when $35$ is divided by 9:

$35 \div 9 = 3 \quad \text{remainder} \quad 8$

So, $35 \equiv 8 \pmod{9}$. Therefore, we need:

$35 + a + b \equiv 0 \pmod{9} \implies 8 + a + b \equiv 0 \pmod{9} \implies a + b \equiv 1 \pmod{9}$

This implies that $a + b$ must be equal to 1, 10, or 19. However, since $a$ and $b$ are digits (from 0 to 9), the only valid sums are those between 0 and 18. Hence, we have:

$a + b = 1 \quad \text{or} \quad a + b = 10$

Now, let's find the pairs $(a, b)$ that satisfy these equations:

1. For $a + b = 1$:

   • $(a, b) = (0, 1)$
   • $(a, b) = (1, 0)$

2. For