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Difference between revisions of "2007 iTest Problems/Problem 14"

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== Solution ==
 
== Solution ==
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If <math>n=p_1^{k_1}\cdots p_s^{k_s}</math> then <math>\phi(n)=p_1^{k_1-1}\cdots p_s^{k_s-1}\cdot (p_1-1)\cdots
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(p_s-1)</math>. Hence every prime factor of <math>n</math> is contained in <math>\{2,3,5,7,13\}</math>. Let <math>p</math> be the
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largest prime dividing <math>n</math>. If <math>p=13</math> then <math>n=13,26</math>. If <math>p=7</math> then <math>n=7k</math> with <math>7\not |k,
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\phi(k)=2</math> giving solutions <math>n=21,28,42</math>. If <math>p=5</math> then <math>n=5k</math> with <math>5\not |k,\phi(k)=3</math>,
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no solutions since <math>\phi(k)</math> is always even unless it is 1. Otherwise <math>n=2^a\cdot 3^b</math> where
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<math>a,b\ge 1</math> and <math>\phi(n)=2^a\cdot 3^{b-1}</math>. Hence <math>a=2,b=2,n=36</math>. Therefore all solutions
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are given by <math>n=13,26,21,28,42,36</math> and the answer is G.
  
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See https://artofproblemsolving.com/community/q2h598845p3554139.
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~Solution by [https://artofproblemsolving.com/community/user/209000| Alexheinis] but not posted on the wiki by Alexhienis
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==See Also==
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{{iTest box|year=2007|num-b=13|num-a=15}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 20:23, 4 February 2023

Problem

Let $\phi(n)$ be the number of positive integers $k< n$ which are relatively prime to $n$. For how many distinct values of $n$ is $\phi(n)=12$?

$\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4 \quad \text{(F) } 5 \quad \text{(G) } 6 \quad \text{(H) } 7 \quad \text{(I) } 8 \quad \text{(J) } 9 \quad \text{(K) } 10 \quad \text{(L) } 11 \quad \text{(M) } 12\quad \text{(N) } 13\quad$

Solution

If $n=p_1^{k_1}\cdots p_s^{k_s}$ then $\phi(n)=p_1^{k_1-1}\cdots p_s^{k_s-1}\cdot (p_1-1)\cdots (p_s-1)$. Hence every prime factor of $n$ is contained in $\{2,3,5,7,13\}$. Let $p$ be the largest prime dividing $n$. If $p=13$ then $n=13,26$. If $p=7$ then $n=7k$ with $7\not |k, \phi(k)=2$ giving solutions $n=21,28,42$. If $p=5$ then $n=5k$ with $5\not |k,\phi(k)=3$, no solutions since $\phi(k)$ is always even unless it is 1. Otherwise $n=2^a\cdot 3^b$ where $a,b\ge 1$ and $\phi(n)=2^a\cdot 3^{b-1}$. Hence $a=2,b=2,n=36$. Therefore all solutions are given by $n=13,26,21,28,42,36$ and the answer is G.

~Solution by Alexheinis but not posted on the wiki by Alexhienis

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 13 		Followed by:
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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