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 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 . We know that 
, and thus 
, and since 
, it's leftmost digit must be 9, and the other number's leftmost digit is 1.
There are total 4001 elements 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.
- AlexLikeMath
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
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