SOLUTION: He chose a random positive integer factor of 10^99, what is the probability that it's also a positive integer multiple of 10^88 ?

Since 1099 = (2*5)99 = 299599,

Every factor of of 1099 is of the form 2p5q 

p and q can each be 0,1,2,...,99 which is 100 choices each.

So there are 100x100 = 10000 factors of 1099 

That's the denominator of the desired probability.

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Next we calculate the numerator of the probability.

They are the positive integers of the form 2p5q
which are multiples of 1088.

So p and q can can each be chosen as 88,89,90,...,99 

There are 99 integers from 1 through 99, from which we must subtract the
87 integers from 1 through 87, so there are 99-87 = 12 choices for
each of p and q. 

So there are 12x12 = 144 factors of 1099 which are multiples of 1088.

Therefore the desired porobability is .

Edwin