N heart

We define nn\heartsuit recursively as follows. 1=1; n=((n1))n+11\heartsuit =1;\ n\heartsuit = ((n-1)\heartsuit)\cdot n +1 Find the largest n<1000n<1000 such that the last two digits of nn\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=(1)2+1=12+1=3 2 \heartsuit = (1\heartsuit) \cdot 2 + 1 = 1 \cdot 2 + 1 = 3 and 3=(2)3+1=33+1=10 3 \heartsuit = ( 2 \heartsuit) \cdot 3 + 1 = 3 \cdot 3 + 1 = 10 .

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