We define \(n\heartsuit \) recursively as follows. \[1\heartsuit =1;\ n\heartsuit = ((n-1)\heartsuit)\cdot n +1\] Find the largest \(n<1000\) such that the last two digits of \(n\heartsuit\) are zeroes.

**Details and assumptions**

Just to make it clear: unlike "n-factorial," "n-heart" is NOT an official mathematical terminology.

Clarification: There is no restriction / requirements on the third last digit.

Clarification: We can calculate that \( 2 \heartsuit = (1\heartsuit) \cdot 2 + 1 = 1 \cdot 2 + 1 = 3\) and \( 3 \heartsuit = ( 2 \heartsuit) \cdot 3 + 1 = 3 \cdot 3 + 1 = 10 \).

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