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**Answer:**

Let the number of balls, wickets, and bats purchased be A, B, and C, respectively.

Thus,

20A + 5B + C = 1000 and A + B + C = 100

Solving the above two equations by eliminating C, we Get

19A + 4B = 900

\Rightarrow B = 225 - \frac{19}{4}A

Now, as B is the number of wickets and 0 < B < 99,

So, putting these limiting values of B in the above equation will provide the value of A as

27 < A < 47. Since A has to be the multiple of 4, possible values of A are 28, 32, 36, 40, and 44. Now, for A = 28 and 32; A + B > 100, so these values of A can be rejected.

For all other values of A, we get the desired solution:

A = 36, B = 54, C = 10

A = 40, B = 35, C = 25

A = 44, B = 16, C = 40

Thus, there are three possible solutions.

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