Difference between revisions of "1990 AIME Problems/Problem 13"

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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 that <math>10^{i} \leq 9^{k} </math>, and thus <math>9 \cdot 10^{i} \leq 9^{k+1} </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.
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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

Problem

Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$. Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit?

Solution 1

Since $9^{4000}$ has 3816 digits more than $9^1$, $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits.

Solution 2

Let's divide all elements of T into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ 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 $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$. To the contrary, let's assume the number ($9^{k}$) does have 9 as the leftmost digit. Thus, $9 \cdot 10^i \leq 9^k$. But, if you divide both sides by 9, you get $10^i \leq 9^{k-1}$, and because $9^{k-1} < 9^{k} < 10^{i+1}$, so we have another number ($9^{k-1}$) in the same section ($10^i \leq 9^{k-1} < 10^{i+1}$). 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 $9^{k}$ and $9^{k+1}$, and the section ranges from $10^{i}$ to $10^{i+1}$. So We know $10^{i} \leq 9^{k} < 9^{k+1} < 10^{i+1}$. From $10^{i} \leq 9^{k}$, we know $9 \cdot 10^{i} \leq 9^{k+1}$, and since $9^{k+1} < 10^{i+1}$. The number($9^{k+1}$)'s leftmost digit must be 9, and the other number($9^{k}$)'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 $\boxed{184}$ numbers that have 9 as the leftmost digit.

- AlexLikeMath

See also

1990 AIME (ProblemsAnswer KeyResources)
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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