Question 394328
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Since the number is an odd multiple of 3, it must be an odd number.  That is because an odd multiple of 3 cannot have a factor of 2.


List the possible all single digit factorizations of 24:


8 X 3 X 1
6 X 4 X 1
6 X 2 X 2
4 X 3 X 2


We can exclude 6 X 2 X 2 because it does not contain an odd digit hence you cannot form an odd 3 digit number.


We can exclude 6 X 4 X 1 because 6 + 4 + 1 is not divisible by 3, hence no three digit number formed from those digits will be divisible by 3.


8, 3, and 1 can be arranged in 6 different orders:


831, 813, 381, 318, 183, 138.


Exclude 318 because it is even and exclude 183 and 138 because they are less than 225.


So, 831, 813, and 381 are all possible.


4,3, and 2 can be arranged in 6 different ways (finding the ways is left as an exercise for the student.  All of the 6 ways will be larger than 225, but only two of the ways are odd numbers.


All together you have 5 different possibilities.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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