To determine which numbers of the form $8726a2415b$ are divisible by 9, we use the rule that a number is divisible by 9 if the sum of its digits is divisible by 9.

First, let's express the sum of the digits:

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

Calculating the sum of the known digits:

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

Now we include $a$ and $b$:

$35 + a + b$

For the number to be divisible by 9, the total sum $35 + a + b$ must be divisible by 9.

Since $a$ and $b$ are digits from 0 to 9, we need to find values of $a$ and $b$ such that:

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

First, calculate $35 \mod 9$:

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

So,

$35 \equiv 8 \pmod{9}$

We need:

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

Which simplifies to:

$a + b \equiv 1 \pmod{9}$

Next, we find pairs of $a$ and $b$ that satisfy this condition:

a = 0, & \quad b = 1 \\ a = 1, & \quad b = 0 \\ a = 2, & \quad b = 8 \\ a = 3, & \quad b = 7 \\ a = 4, & \quad b = 6 \\ a = 5, & \quad b = 5 \\ a = 6, & \quad b = 4 \\ a = 7, & \quad b = 3 \\ a = 8, & \quad b = 2 \\ a = 9, & \quad b = 1 \\ \end{align*}$$ So, the pairs $$(a, b)$$ that make $$8726a2415b$$ divisible by 9 are: - $$(0, 1)$$ - $$(1, 0)$$ - $$(2, 8)$$ - $$(3, 7)$$ - $$(4, 6)$$ - $$(5, 5)$$ - $$(6, 4)$$ - $$(7, 3)$$ - $$(8, 2)$$ - $$(9, 1)$$ Each pair provides a different number $$8726a2415b$$ that is divisible by 9. Would you like more details or have any questions? Here are 8 related questions for further practice: 1. What is the rule for a number to be divisible by 3? 2. Find all numbers of the form $$1234x$$ that are divisible by 9. 3. How many digits are there in the number $$99999 \times 9$$? 4. What are the factors of 81? 5. If $$y = 5$$ and $$z = 4$$, find a number of the form $$7yz8$$ that is divisible by 9. 6. Find the sum of the digits of the number $$987654321$$. 7. How many pairs of digits (a, b) satisfy $$a + b \equiv 2 \pmod{9}$$? 8. Prove that any rearrangement of the digits of 123456789 is divisible by 9. **Tip:** Always double-check your results by substituting values back into the original equation or rule to ensure correctness.