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 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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