Difference between revisions of "1990 AIME Problems/Problem 13"
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==Solution 2== | ==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. | + | Let's divide all elements of <math>T</math> 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 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 1 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. | ||
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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. | 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 in T and 3817 sections | + | There are total 4001 elements in <math>T</math> 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 <math>\boxed{184}</math> numbers that have 9 as the leftmost digit. |
- AlexLikeMath | - AlexLikeMath | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We know that <math>9</math> is very close to <math>10</math>. But we also know that for all <math>k \in \mathbb{Z}^+</math>, <math>10^k</math> has <math>k+1</math> digits. Hence <math>10^{4000}</math> has <math>4001</math> digits. Now, notice that if a power of <math>9^n</math> has <math>9</math> as the leftmost digit, then <math>9^{n-1}</math> has <math>1</math> as the leftmost digit and will preserve the same number of digits (just due to division). Hence we find that consecutive powers of <math>9</math> are "supposed" to increase by one digit, however, when consecutive powers of <math>9</math> have leading digit <math>1</math> then leading digit <math>9</math> they "lose" this one digit. Therefore, it suffices to find the number of times that <math>9^k</math> has "lost" a digit since the number of times that this occurs shows us the number of pairs where the leading digit of <math>9^k</math> is <math>1</math> and the leading digit of <math>9^{k+1}</math> is <math>9</math>. Hence, we find that <math>9^{4000}</math> is "supposed" to have <math>4001</math> digits, but only has <math>3817</math> digits. Thus, the number of times <math>9^k</math> has lost a digit is <math>4001-3817 = 184</math>. Thus, <math>\boxed{184}</math> elements of <math>T</math> have <math>9</math> as their leading digit. | ||
+ | |||
+ | ~qwertysri987 | ||
== See also == | == See also == |
Latest revision as of 23:35, 21 November 2019
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 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.
- AlexLikeMath
Solution 3
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.
~qwertysri987
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 |
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