MHB How many nine-digit numbers meet these criteria?

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The number of nine-digit numbers divisible by 225 in which all digits are different and in which the digit of hundred is 7 is?
I don't know how to set the number to be divisible by 225, so if anyone can help
 
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255= 3(5)(17) so any number divisible by 255 must be divisible by 3, 5, and by 17. Any number divisible by 3 has digits that sum to a multiple of 3. Any number divisible by 5 has one's place 5 or 0. 17 is the hard one!
 
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I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
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