To find the smallest number of marbles Ali could have, we denote the number of marbles as \( n \).
From the problem, we have the following conditions:
- \( n \) is divisible by 5.
- \( n \) is divisible by 8.
- When \( n \) is divided by 9, the remainder is 1.
From the first condition, we can express \( n \) as: \[ n = 5k \quad \text{for some integer } k. \]
From the second condition, since \( n \) is also divisible by 8, we can express it as: \[ n = 8m \quad \text{for some integer } m. \]
Since \( n \) must satisfy both conditions, we want to find the least common multiple (LCM) of 5 and 8. The LCM of 5 and 8 can be calculated as: \[ \text{lcm}(5, 8) = 5 \times 8 = 40. \] Thus, \( n \) must be a multiple of 40, so we can write: \[ n = 40t \quad \text{for some integer } t. \]
Now, we also need to satisfy the third condition, which states: \[ n \equiv 1 \pmod{9}. \]
Substituting \( n = 40t \) into this condition gives: \[ 40t \equiv 1 \pmod{9}. \]
Now, we first need to reduce \( 40 \mod 9 \): \[ 40 \div 9 = 4 \quad \text{(the quotient is 4)} \] \[ 40 - 9 \times 4 = 40 - 36 = 4. \] So, \[ 40 \equiv 4 \pmod{9}. \] Now we can rewrite the congruence: \[ 4t \equiv 1 \pmod{9}. \]
Next, we need to find the multiplicative inverse of 4 modulo 9. This means we need to find \( t \) such that: \[ 4t \equiv 1 \pmod{9}. \] We can check the integers from 1 to 8 to find the inverse:
- \( t = 1: \quad 4 \cdot 1 = 4 \)
- \( t = 2: \quad 4 \cdot 2 = 8 \)
- \( t = 3: \quad 4 \cdot 3 = 12 \equiv 3 \pmod{9} \)
- \( t = 4: \quad 4 \cdot 4 = 16 \equiv 7 \pmod{9} \)
- \( t = 5: \quad 4 \cdot 5 = 20 \equiv 2 \pmod{9} \)
- \( t = 6: \quad 4 \cdot 6 = 24 \equiv 6 \pmod{9} \)
- \( t = 7: \quad 4 \cdot 7 = 28 \equiv 1 \pmod{9} \)
Thus, \( t = 7 \) is a solution. We can now substitute back to find \( n \): \[ n = 40t = 40 \cdot 7 = 280. \]
Lastly, we should verify if \( n = 280 \) meets all the initial conditions:
- \( 280 \div 5 = 56 \), so it is divisible by 5.
- \( 280 \div 8 = 35 \), so it is divisible by 8.
- \( 280 \div 9 = 31 \) remainder 1 (since \( 280 - 9 \cdot 31 = 280 - 279 = 1 \)), so it leaves a remainder of 1.
All conditions are satisfied, therefore the smallest possible number of marbles Ali had is: \[ \boxed{280}. \]