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The number of $$6$$ digit numbers of the form "$$ABCABC$$", which are divisible by $$13$$, where $$A$$, $$B$$ and $$C$$ are distinct digits, $$A$$ and $$C$$ being even digits is
Since $$ABCABC=1001\times ABC$$ and $$1001=13\times77$$, every number of this form is divisible by $$13$$. The digit $$A$$ can be chosen in $$4$$ ways from $$2,4,6,8$$. If $$C=0$$ there are $$8$$ choices for $$B$$, and if $$C$$ is nonzero there are $$3$$ choices for $$C$$ and $$8$$ choices for $$B$$. Thus the total is $$4\times8+4\times3\times8=128$$.
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