Question 1186630
<br>
(1) True, there is no algebraic method for solving this problem.<br>
(2) But just counting them is not a very efficient method.  A formal mathematical solution does not require an algebraic solution; logical analysis is perfectly good mathematics, and it can make this problem easy.<br>
(3) There is nothing particularly brilliant or elegant, as one tutor called it, about the method for solving the problem; only a very little logical analysis is required.<br>
(4) A response simply saying "I got 28 for the answer", without any justification, is of no use to the student -- especially since that answer is wrong.<br>
(5) Finally, asking for the "fraction of the integers between 0 and 1000" makes it unclear what the answer should be, because of the ambiguity of the meaning of "between 0 and 1000"; we can't be sure whether the denominator is 999 or 1000, or even 1001.  To make the problem (and expected answer) clear, the problem should simply ask how many numbers between 0 and 1000 have exactly two 6's.<br>
The relatively simple solution:<br>
(1) integers of the form 66A, where A is not 6: 9 choices for A (any digit except 6)
(2) integers of the form 6A6, where A is not 6: 9 choices for A (any digit except 6)
(3) integers of the form A66, where A is not 6: 9 choices for A (any digit except 6)<br>
Note for the last case we can have 0 as the leading digit, representing the integer 66, which is between 0 and 1000.<br>
ANSWER: There are 27 integers between 0 and 1000 that contain exactly two 6's.<br>