1990 AIME Problems/Problem 13
Contents
Problem
Let 
. Given that 
has 3817 digits and that its first (leftmost) digit is 9, how many elements of 
have 9 as their leftmost digit?
Solution 1
Since 
has 3816 digits more than 
, 
numbers have 9 as their leftmost digits.
Solution 2
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 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 T 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.
- AlexLikeMath
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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