Question

Ali had some marbles. He could group them into sets of 5 marbles or sets of 8 marbles without leaving any remainder. However, when he tried to group them into sets of 9 marbles, he was left with one marble. What was the smallest possible number of marbles Ali had?

Answer 1

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:
1. \( n \) is divisible by 5.
2. \( n \) is divisible by 8.
3. 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:
1. \( 280 \div 5 = 56 \), so it is divisible by 5.
2. \( 280 \div 8 = 35 \), so it is divisible by 8.
3. \( 280 \div 9 = 31 \) 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:
\[
\boxed{280}.
\]