Difference between revisions of "2014 AIME II Problems/Problem 15"
(→Solution) |
(→Solution) |
||
Line 5: | Line 5: | ||
Note that for any <math>x_i</math>, for any prime <math>p</math>, <math>p^2 \nmid x_i</math>. This provides motivation to translate <math>x_i</math> into a binary sequence <math>y_i</math>. | Note that for any <math>x_i</math>, for any prime <math>p</math>, <math>p^2 \nmid x_i</math>. This provides motivation to translate <math>x_i</math> into a binary sequence <math>y_i</math>. | ||
− | Let the prime factorization of <math>x_i</math> be written as <math> | + | Let the prime factorization of <math>x_i</math> be written as <math>p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots</math>, where <math>p_i</math> is the <math>i</math>th prime number. Then, for every <math>p_{a_k}</math> in the prime factorization of <math>x_i</math>, place a <math>1</math> in the <math>a_k</math>th digit of <math>y_i</math>. This will result in the conversion <math>x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots</math>. |
Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1, and y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090, </math>t = 10010101<math> (base 2) = </math>\boxed{149}$. | Multiplication for the sequence <math>x_i</math> will translate to addition for the sequence <math>y_i</math>. Thus, we see that <math>x_{n+1}X(x_n) = x_np(x_n)</math> translates into <math>y_{n+1} = y_n+1</math>. Since <math>x_0=1, and y_0=0</math>, <math>x_i</math> corresponds to <math>y_i</math>, which is <math>i</math> in binary. Since <math>x_{10010101} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090, </math>t = 10010101<math> (base 2) = </math>\boxed{149}$. |
Revision as of 21:04, 10 January 2016
Problem
For any integer , let be the smallest prime which does not divide Define the integer function to be the product of all primes less than if , and if Let be the sequence defined by , and for Find the smallest positive integer such that
Solution
Note that for any , for any prime , . This provides motivation to translate into a binary sequence .
Let the prime factorization of be written as , where is the th prime number. Then, for every in the prime factorization of , place a in the th digit of . This will result in the conversion .
Multiplication for the sequence will translate to addition for the sequence . Thus, we see that translates into . Since , corresponds to , which is in binary. Since t = 10010101\boxed{149}$.
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.