Let U be the set of all three-digit integer positive numbers (an "universal set").


Let A be a subset in U consisting of all three-digit integer positive numbers having "5" as the left-most digit.

Let B be a subset in U consisting of all three-digit integer positive numbers having "5" in the second position.

Let C be a subset in U consisting of all three-digit integer positive numbers having "5" in the third position.



The set A contains 10*10 = 100 numbers;

The set B contains 9*10 = 90 numbers (zero is prohibited as the left-most digit);

The set C contains 9*10 = 90 numbers (zero is prohibited as the left-most digit).



The sets A and B have intersection comprised of all positive integer numbers of the form  55X.  There are 10 numbers in this intersection.

The sets A and C have intersection comprised of all positive integer numbers of the form  5X5.  There are 10 numbers in this intersection.

The sets B and C have intersection comprised of all positive integer numbers of the form  X55.  There are 9 numbers in this intersection.


The sets A, B and C have TRIPLE intersection comprised of ONE positive integer number  555.  There is 1 number in this intersection.



Now, we want to determine the number of elements in the union  (A U B U C).

Use the Inclusion-Exclusion formula 


    n(A U B U C) = n(A) + N(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C) = 

                 = 100  +  90  + 90   - 10 - 10 - 9 + 1 = 252.


ANSWER.  In all, there are 252 such numbers.