To solve the problem of finding the 4-digit number $7a3b$ that is divisible by both 5 and 9, we need to consider the following conditions:

1. Divisibility by 5: For a number to be divisible by 5, its last digit $b$ must be 0 or 5.
2. Divisibility by 9: For a number to be divisible by 9, the sum of its digits must be a multiple of 9.

Step 1: Apply the Divisibility Rule for 5

Since $7a3b$ must be divisible by 5, $b$ can only be 0 or 5.

Step 2: Apply the Divisibility Rule for 9

The sum of the digits in $7a3b$ is $7 + a + 3 + b$. This sum must be a multiple of 9.

So, we have: $7 + a + 3 + b = 10 + a + b$

This sum, $10 + a + b$, must be divisible by 9.

Step 3: Consider Each Case for $b$

We consider the two possibilities for $b$:

1. If $b = 0$: $10 + a + 0 = 10 + a$ $10 + a$ must be divisible by 9. The possible value for $10 + a$ that is a multiple of 9 is 18 (since $9 \times 2 = 18$). Therefore, $a = 8$.

   In this case, the number is $7830$.

2. If $b = 5$: $10 + a + 5 = 15 + a$ $15 + a$ must be divisible by 9. The possible value for $15 + a$ that is a multiple of 9 is 18 (since $9 \times 2 = 18$). Therefore, $a = 3$.

   In this case, the number is $7435$.

Step 4: Conclusion

The possible 4-digit numbers that meet the conditions are 7830 and 7435.

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Would you like more details on how these conditions work or have any further questions?

Here are some related questions:

1. How do you check divisibility by other numbers, such as 6 or 12?
2. What if the number was required to be divisible by both 3 and 8?
3. Could there be more than one number if different digits were allowed?
4. Why are $b = 0$ and $b = 5$ the only possibilities for divisibility by 5?
5. What if $b$ was restricted to non-zero digits only?

Tip: Always check the smallest condition first when working with multiple divisibility rules. This can narrow down possibilities quickly!