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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
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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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In the subtraction below, what is the sum of the digits in the result?
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A competition has $$25$$ questions and is marked as follows.
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$$A,M,T,I$$ are positive integers such that $$A+M+T+I=10$$. The maximum possible value of $$A\times M\times T\times I+A\times M\times T+A\times M\times I+A\times T\times I+M\times T\times I+A\times M+A\times T+A\times I+M\times T+M\times I+T\times I$$ is
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Let $$x$$ and $$y$$ be real numbers satisfying $$x^4y^5+y^4x^5=810$$ and $$x^3y^6+y^3x^6=945$$. Then the value of $$2x^3+x^3y^3+2y^3$$ is
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