Question 355346
<pre>
The only way 3 digits can have product 24 is 

1x3x8 = 24
1x4x6 = 24
2x2x6 = 24
2x3x4 = 24

So the digits consist of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 

To be divisible by 3 the sum of the digits must be divisible by 3.

1+3+8=12
1+4+6=11
2+2+6=10
2+3+4=9

Of those sums of digits, only 12 and 9 are divisible by 3. 
So we have ruled out all but integers whose digits consist of
1,3,8, and 2,3,4.

Since they must be odd they either must end in 1 or 3.

The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 15<sup>2</sup> which is 225. So the
first digit cannot be 1. So the only way its digits can consist of
1,3,8 and end in 3 is to be 813.

The rest must consist of the digits 2,3,4, and the only way they can 
end in 3 is to be 243 or 423.  

So there are exactly five such three-digit integers: 

381, 831, 813, 243, and 423.

Edwin</pre>