Difference between revisions of "2004 AIME I Problems/Problem 15"
m (Undo revision 162893 by Physicsninja 275 (talk)) (Tag: Undo) |
m (→Solution 2) |
||
Line 30: | Line 30: | ||
First, we note that <math>\mathcal{A}_{1,1}=\{1\}</math>, so by the disjointness of the <math>\mathcal{A}_{n,k}</math>'s, we know that <math>1</math> is not in any of the other sets. Also, we note that <math>\mathcal{A}_{1,k}=\emptyset</math> for <math>k=0,2,3,4,5,6,7,8,9</math>. | First, we note that <math>\mathcal{A}_{1,1}=\{1\}</math>, so by the disjointness of the <math>\mathcal{A}_{n,k}</math>'s, we know that <math>1</math> is not in any of the other sets. Also, we note that <math>\mathcal{A}_{1,k}=\emptyset</math> for <math>k=0,2,3,4,5,6,7,8,9</math>. | ||
− | We claim that if <math>2\le k\le 9</math> and <math>n\ge2</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. To prove this, we show that <math>f</math> (the given function) maps <math>\mathcal{A}_{n,k}</math> bijectively onto <math>\mathcal{A}_{n-1,k+1}</math>. If <math>x\in \mathcal{A}_{n,k}</math>, then <math>x\equiv k\pmod{10}</math>, and <math>x\ne 1</math>, so <math>f(x)=x+1\equiv k+1\pmod{10}</math>. Also, <math>f^{(n)}(x)=1</math>, where <math>n</math> is the smallest positive integer for which this is true. Therefore, <math>f^{{n-1}}(f(x))=1</math>, where <math>n-1</math> is the smallest integer for which this is true. Thus <math>f(x)\in\mathcal{A}_{n-1,k+1}</math>. Also, since <math>f(x)=x+1</math> on this set, we see that <math>f(x)=f(y)</math> implies that <math>x=y</math>. Hence <math>f</math> is an injection. If <math>y\in\mathcal{A}_{n-1,k+1}</math>, then <math>y-1\equiv k\pmod{10}</math>, where <math>2\le k\le 9</math>, so we know that <math>f(y-1)=y</math>, and <math>y-1\in \mathcal{A}_{n,k}</math>. Therefore, <math>f</math> is a surjection, so it must be a bijection. Therefore, if <math>2\le k\le 9</math> and <math>n\ge2</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. | + | We claim that if <math>2\le k\le 9</math> and <math>n\ge2</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. To prove this, we show that <math>f</math> (the given function) maps <math>\mathcal{A}_{n,k}</math> bijectively onto <math>\mathcal{A}_{n-1,k+1}</math>. If <math>x\in \mathcal{A}_{n,k}</math>, then <math>x\equiv k\pmod{10}</math>, and <math>x\ne 1</math>, so <math>f(x)=x+1\equiv k+1\pmod{10}</math>. Also, <math>f^{(n)}(x)=1</math>, where <math>n</math> is the smallest positive integer for which this is true. Therefore, <math>f^{{n-1}}(f(x))=1</math>, where <math>n-1</math> is the smallest integer for which this is true. Thus <math>f(x)\in\mathcal{A}_{n-1,k+1}</math>. Also, since <math>f(x)=x+1</math> on this set, we see that <math>f(x)=f(y)</math> implies that <math>x=y</math>. Hence <math>f</math> is an injection. If <math>y\in\mathcal{A}_{n-1,k+1}</math>, then <math>y-1\equiv k\pmod{10}</math>, where <math>2\le k\le 9</math>, so we know that <math>f(y-1)=y</math>, and <math>y-1\in \mathcal{A}_{n,k}</math>. Therefore, <math>f</math> is a surjection, so it must be a [[bijection]]. Therefore, if <math>2\le k\le 9</math> and <math>n\ge2</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. |
We also claim that if <math>k=1</math>, <math>n\ge 2</math> and <math>n\ne 11</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. The proof is the same as in the previous paragraph, though some additional clarification is needed to prove that <math>f</math> is a surjection. If <math>y\in\mathcal{A}_{n-1,k+1}</math>, or rather <math>y-1\equiv k\equiv 1\pmod{10}</math>, then if <math>y=2</math>, we see that <math>y-1=1</math>, and then <math>f(y-1)=1</math>, not <math>y</math> as in the prior argument. However, this does not happen if <math>n\ne 11</math>. It is easy to check that <math>2\in\mathcal{A}_{10,2}</math>. Therefore, the only time that the above argument could fail is if <math>n-1=10</math> and <math>k+1=2</math>. But in every other case, <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. | We also claim that if <math>k=1</math>, <math>n\ge 2</math> and <math>n\ne 11</math>, then <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. The proof is the same as in the previous paragraph, though some additional clarification is needed to prove that <math>f</math> is a surjection. If <math>y\in\mathcal{A}_{n-1,k+1}</math>, or rather <math>y-1\equiv k\equiv 1\pmod{10}</math>, then if <math>y=2</math>, we see that <math>y-1=1</math>, and then <math>f(y-1)=1</math>, not <math>y</math> as in the prior argument. However, this does not happen if <math>n\ne 11</math>. It is easy to check that <math>2\in\mathcal{A}_{10,2}</math>. Therefore, the only time that the above argument could fail is if <math>n-1=10</math> and <math>k+1=2</math>. But in every other case, <math>|\mathcal{A}_{n,k}|=|\mathcal{A}_{n-1,k+1}|</math>. |
Revision as of 00:17, 26 December 2022
Problem
For all positive integers , let and define a sequence as follows: and for all positive integers . Let be the smallest such that . (For example, and .) Let be the number of positive integers such that . Find the sum of the distinct prime factors of .
Solution
Solution 1
We backcount the number of ways. Namely, we start at , which can only be reached if , and then we perform operations that either consist of or . We represent these operations in a string format, starting with the operation that sends and so forth downwards. There are ways to pick the first operations; however, not all of them may be otherwise we return back to , contradicting our assumption that was the smallest value of . Using complementary counting, we see that there are only ways.
Since we performed the operation at least once in the first operations, it follows that , so that we no longer have to worry about reaching again.
However, we must also account for a sequence of or more s in a row, because that implies that at least one of those numbers was divisible by , yet the was never used, contradiction. We must use complementary counting again by determining the number of strings of s of length such that there are s in a row. The first ten are not included since we already accounted for that scenario above, so our string of s must be preceded by a . There are no other restrictions on the remaining seven characters. Letting to denote either of the functions, and to indicate that the character appears times in a row, our bad strings can take the forms:
There are ways to select the operations for the s, and places to place our block. Thus, our answer is , and the answer is .
Note: This solution is quick and most similar to the official solution; however, neither this nor the official solution prove that the final results of these inverted operations are all distinct. A more sophisticated argument, such as the one below, is needed to do so.
Solution 2
We approach the problem by recursion. We partition the positive integers into the sets First, we note that , so by the disjointness of the 's, we know that is not in any of the other sets. Also, we note that for .
We claim that if and , then . To prove this, we show that (the given function) maps bijectively onto . If , then , and , so . Also, , where is the smallest positive integer for which this is true. Therefore, , where is the smallest integer for which this is true. Thus . Also, since on this set, we see that implies that . Hence is an injection. If , then , where , so we know that , and . Therefore, is a surjection, so it must be a bijection. Therefore, if and , then .
We also claim that if , and , then . The proof is the same as in the previous paragraph, though some additional clarification is needed to prove that is a surjection. If , or rather , then if , we see that , and then , not as in the prior argument. However, this does not happen if . It is easy to check that . Therefore, the only time that the above argument could fail is if and . But in every other case, .
Next, we claim that if , then If , then , which is clearly an injective map. Also, , where is the smallest positive integer for which this is true. Therefore, , where is the smallest integer for which this is true. Thus for some . Conversely, if , then , so , so .
Based on these bijections, if we let , then Let . Then by adding the above equations (valid if ), we find that Now by using the above relations repeatedly, we find The first relation will only be valid if . Therefore, for . For smaller values, it is easy to use the relations to compute that the terms are powers of , although we note that because of the failure of the above argument. After this, we can simply use the recurrence relation for , finding Therefore, there are positive integers with . This factors as , where is prime. Thus the answer is .
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.