1990 AIME Problems/Problem 13
Since has 3816 digits more than , numbers have 9 as their leftmost digits.
Let's divide all elements of T into sections. Each section ranges from to And, each section must have 1 or 2 elements. So, let's consider both cases.
If a section has one element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be and the section ranges from to . To the contrary, let's assume the number () does have 9 as the leftmost digit. Thus, . But, if you divide both sides by 9, you get , and because , so we have another number () in the same section (). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.
If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be and , and the section ranges from to . So We know . From , we know , and since . The number()'s leftmost digit must be 9, and the other number()'s leftmost digit is 1.
There are total 4001 elements in T and 3817 sections (each esction can only has one or two elements). The two elements section must have 9 as the left most digit. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get numbers that have 9 as the leftmost digit.
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