Difference between revisions of "1990 AIME Problems/Problem 13"
Alexlikemath (talk | contribs) m |
Alexlikemath (talk | contribs) m |
||
| Line 10: | Line 10: | ||
If a section has one element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be <math>9^{k}</math> and the section ranges from <math>10^{i}</math> to <math>10^{i+1}</math>. To the contrary, let's assume the number (<math>9^{k}</math>) does have 9 as the leftmost digit. Thus, <math>9 \cdot 10^i \leq 9^k</math>. But, if you divide both sides by 9, you get <math>10^i \leq 9^{k-1}</math>, and because <math>9^{k-1} < 9^{k} < 10^{i+1}</math>, so we have another number (<math>9^{k-1}</math>) in the same section (<math>10^i \leq 9^{k-1} < 10^{i+1}</math>). 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 one element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be <math>9^{k}</math> and the section ranges from <math>10^{i}</math> to <math>10^{i+1}</math>. To the contrary, let's assume the number (<math>9^{k}</math>) does have 9 as the leftmost digit. Thus, <math>9 \cdot 10^i \leq 9^k</math>. But, if you divide both sides by 9, you get <math>10^i \leq 9^{k-1}</math>, and because <math>9^{k-1} < 9^{k} < 10^{i+1}</math>, so we have another number (<math>9^{k-1}</math>) in the same section (<math>10^i \leq 9^{k-1} < 10^{i+1}</math>). 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 <math>9^{k}</math> and <math>9^{k+1}</math>, and the section ranges from <math>10^{i}</math> to <math>10^{i+1}</math>. We know | + | 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 <math>9^{k}</math> and <math>9^{k+1}</math>, and the section ranges from <math>10^{i}</math> to <math>10^{i+1}</math>. |
| + | So We know <math>10^{i} \leq 9^{k} < 9^{k+1} < 10^{i+1}</math>. From <math>10^{i} \leq 9^{k} </math>, we know <math>9 \cdot 10^{i} \leq 9^{k+1} </math>, and since <math>9^{k+1} < 10^{i+1}</math>. The number(<math>9^{k+1}</math>)'s leftmost digit must be 9, and the other number(<math>9^{k}</math>)'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 <math>\boxed{184}</math> numbers that have 9 as the leftmost digit. | + | 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 <math>\boxed{184}</math> numbers that have 9 as the leftmost digit. |
- AlexLikeMath | - AlexLikeMath | ||
Revision as of 14:37, 27 June 2019
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 .
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.
- AlexLikeMath
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
