Difference between revisions of "2006 iTest Problems/Problem 34"
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Then we can compute <math>d = \lambda(2006)</math> (where <math>\lambda</math> is the [[Carmichael function]]) by Carmichael's theorem: it is <math>\text{lcm}(\lambda(2), \lambda(17), \lambda(59)) = \text{lcm}(1, 16, 58) = 2^4 * 29 = 464</math>. | Then we can compute <math>d = \lambda(2006)</math> (where <math>\lambda</math> is the [[Carmichael function]]) by Carmichael's theorem: it is <math>\text{lcm}(\lambda(2), \lambda(17), \lambda(59)) = \text{lcm}(1, 16, 58) = 2^4 * 29 = 464</math>. | ||
− | As for solving <math>P(i) = d</math>, we must have <math>i</math> odd (otherwise it would not be coprime to <math>2</math>), and we must also have <math>i</math> be a primitive root modulo <math>17</math> as well as a primitive root modulo <math>59</math>. There are <math>\phi(\phi(17)) = \phi(16) = 8</math> primitive roots modulo <math>17</math> (where <math>\phi</math> is the [[Euler totient function]]) and <math>\phi(\phi(59)) = \phi(58) = (2-1)*(29-1) = 28</math> primitive roots modulo <math>59</math>. Then we have <math>m = 1 * 8 * 28 = 224</math> by the [[Chinese | + | As for solving <math>P(i) = d</math>, we must have <math>i</math> odd (otherwise it would not be coprime to <math>2</math>), and we must also have <math>i</math> be a primitive root modulo <math>17</math> as well as a primitive root modulo <math>59</math>. There are <math>\phi(\phi(17)) = \phi(16) = 8</math> primitive roots modulo <math>17</math> (where <math>\phi</math> is the [[Totient function|Euler totient function]]) and <math>\phi(\phi(59)) = \phi(58) = (2-1)*(29-1) = 28</math> primitive roots modulo <math>59</math>. Then we have <math>m = 1 * 8 * 28 = 224</math> by the [[Chinese Remainder Theorem]], so our answer is <math>d + m = 464 + 224 = 688</math> and we are done. |
== See also == | == See also == | ||
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{{iTest box|year=2006|num-b=33|num-a=35|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} | {{iTest box|year=2006|num-b=33|num-a=35|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} | ||
− | [[Category:iTest]] [[Category:Number Theory | + | [[Category:iTest]] [[Category:Number Theory Problems]] |
Latest revision as of 17:04, 13 September 2020
For each positive integer let denote the set of positive integers such that is divisible by . Define the function by the rule Let be the least upper bound of and let be the number of integers such that and . Compute the value of .
Solution
We find that the prime factorization of is .
Then we can compute (where is the Carmichael function) by Carmichael's theorem: it is .
As for solving , we must have odd (otherwise it would not be coprime to ), and we must also have be a primitive root modulo as well as a primitive root modulo . There are primitive roots modulo (where is the Euler totient function) and primitive roots modulo . Then we have by the Chinese Remainder Theorem, so our answer is and we are done.
See also
2006 iTest (Problems, Answer Key) | ||
Preceded by: Problem 33 |
Followed by: Problem 35 | |
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