Question 1204981
<pre>
The triples of consecutive integers allowable are (1,2,3) through (96,97,98).
So there are 96 of them

Of the integers from 1 to 98, 49 are even and 49 are odd.

There are only 2 types of triples of 3 consecutive integers, the
(even,odd,even)-type and the (odd,even,odd)-type.  

Case 1.  (even,odd,even)-type.  Since every other even integer is divisible by
4, their product will always be divisible by 8. 

These are the triples (1,2,3), (3,4,5), ..., (95,96,97). There are 48 of these.
[That's because the middle numbers 2,4,...,96 are such that if you divide all 
of them by 2 you get 1,2,...,48.]

or 

Case 2.  (odd,even,odd)-type. These will have a product divisible by 8 if and
only if the middle number is divisible by 8.

These are the triples (7,8,9), (15,16,17),..., (95,96,97). There are 12 of these. 
[That's because the middle numbers 8,16,...,96 are such that if 
you divide all of them by 8, you get 1,2,...,12.]

So there are 48+12 = 60 triples of consecutive integers whose product is
divisible by 8.

The desired probability is 60/96 which reduces to 5/8

Edwin</pre>