#### Solution

We shall generalize this problem. Let one gold coin worth $A$ silver, and one silver coin worth $B$ bronze coins. The price of the item is thus $3AB$ bronze coins. We will next consider the number of gold coins used to pay.

If Ali uses 3 gold coins, there is only one way to pay, that is, using 3 gold coins.

If Ali uses 2 gold coins, then he must use a combination of silver and bronze to pay for the remaining price that's equivalent to $AB$ bronze coins. He could use $0,1,cdots, A$ silver coins to pay for this. Once he decides the number of silver coins used, the remaining price must be paid all by bronze coins. So there are $(A+1)$ ways to pay with exacty 2 gold coins.

Similarly, there are $(2A+1)$ ways to pay with exactly 1 gold coin, and $(3A+1)$ ways to pay without any gold coins.

In total, there are $1 + (A+1) + (2A+1) + (3A+1) = 6A+1 = 2008$ ways to pay.

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