MHB How many nine-digit numbers meet these criteria?

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To determine the number of nine-digit numbers divisible by 225 with all different digits and the hundred's digit as 7, it is essential to understand the divisibility rules for 225, which factors into 3, 5, and 17. A number is divisible by 3 if the sum of its digits is a multiple of 3, and it must end in either 0 or 5 to be divisible by 5. The challenge lies in ensuring divisibility by 17, which complicates the calculation. Participants in the discussion seek strategies to set up the problem correctly, particularly focusing on the constraints of unique digits and the specific placement of the digit 7. Overall, the conversation emphasizes the mathematical intricacies involved in solving this problem.
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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!
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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