Difference between revisions of "1990 AIME Problems/Problem 13"
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Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost digit? | Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost digit? | ||
| − | ==Solution== | + | ==Solution 1== |
| + | Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - 3816 = \boxed{184}</math> numbers have 9 as their leftmost digits. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Let's divide all elements of T into sections. Each section ranges from <math>10^{i}</math> to <math>10^{i+1}</math> 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 the 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 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} \leq 10^{i+1}</math>, so we have another number in the same section. Which is a contradiction to our assumption that the section only has 1 element. So, 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 that <math>9^{k} \geq 10^{i}</math>, and thus <math>9^{k+1} \geq 9 \cdot 10^{i}</math>, and since <math>9^{k+1} \leq 10^{i+1}</math>, it's leftmost digit must be 9, and the other number's leftmost digit is 1. | ||
| − | + | There are 3817 sections and 4001 elements. Each section has exactly one element that doesn't 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 | ||
== See also == | == See also == | ||
Revision as of 13:01, 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 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let the 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, 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 3817 sections and 4001 elements. Each section has exactly one element that doesn't 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. 
