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November 24, 2014

November 24, 2014

Posted by **Anonymous** on Monday, March 11, 2013 at 2:30am.

- math -
**Steve**, Monday, March 11, 2013 at 2:41amfirst digit: 9 choices (1-9)

then, 2nd digit: 9 choices (now 0 is allowed)

then, 3rd digit: 8 choices

so, 9*9*8

- math -
**Anonymous**, Monday, March 11, 2013 at 1:33pmIt has to be odd, so the 3rd digit has to be 1, 3, 5, 7, or 9. So there are only 5 options. Moreover, you do not consider the cases in which the digits do not repeat.

- math -
**Steve**, Monday, March 11, 2013 at 3:20pmyou got me on the odd value, but I do account for the non-repeatability of digits.

If they can be repeated, then there would be 9*10*10 possibilities for 3 digits.

My counting forces the preceding choices not to be used again.

Since the number must be odd, things get a bit more complicated, because we don't know whether the previous choices were odd or even.

You seem to have a handle on the problem. What is your reasoning?

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