\(\overline{{ABC0ABC}}\) be a 7 digit number.

Let the number of values of \(\overline{{ABC0ABC}}\) divisible by 11 be \(x\).

Let the number of values of \(\overline{{ABC0ABC}}\) divisible by 121 be \(y\).

Let the number of values of \(\overline{{ABC0ABC}}\) divisible by 1331 be \(z\).

Then find the value of \(x+y+z\).

**Details and Assumptions**

The letters A,B,C do not necessarily stand for distinct digit.

The fourth digit of the number is zero.

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