1990 AIME Problems/Problem 13
Lemma: For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9
(Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n. If there is exactly 1 n-digit power of 9, then such a number cannot begin with 9 since that would result in also being an n-digits, hence a contradiction. Therefore, this single n-digit power of 9 must not begin with 9. In the case that there are 2 n-digit powers of 9, we have already discovered that one of them must begin with 9, let it be . The integer must then be an n-digit number that begins with .
Using this lemma, out of the 4001 powers of 9 in T, exactly 3816 must not begin with 9 (one for each digit). Therefore, the answer is numbers have 9 as their leftmost digits.
~Edited by Mathandski (with help from a math god)
Let's divide all elements of 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 1 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 and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. 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.
We know that is very close to . But we also know that for all , has digits. Hence has digits. Now, notice that if a power of has as the leftmost digit, then has as the leftmost digit and will preserve the same number of digits (just due to division). Hence we find that consecutive powers of are "supposed" to increase by one digit, however, when consecutive powers of have leading digit then leading digit they "lose" this one digit. Therefore, it suffices to find the number of times that has "lost" a digit since the number of times that this occurs shows us the number of pairs where the leading digit of is and the leading digit of is . Hence, we find that is "supposed" to have digits, but only has digits. Thus, the number of times has lost a digit is . Thus, elements of have as their leading digit.
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