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Difference between revisions of "1996 AHSME Problems/Problem 29"
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<math> \text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38 </math>
 
<math> \text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38 </math>
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==Solution==
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Working with the second part of the problem first, we know that <math>3n</math> has <math>30</math> divisors.  We try to find the various possible prime factorizations of <math>3n</math> by splitting <math>30</math> into various products of <math>1, 2</math> or <math>3</math> integers.
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<math>30 \rightarrow p^{29}</math>
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<math>2 \cdot 15 \rightarrow pq^{14}</math>
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<math>3\cdot 10 \rightarrow p^2q^9</math>
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<math>5\cdot 6 \rightrarrow p^4q^5</math>
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<math>2\cdot 3\cdot 5 \rightarrow pq^2r^4</math>
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The variables <math>p, q, r</math> are different prime factors, and one of them must be <math>3</math>.  We now try to count the factors of <math>2n</math>, to see which prime factorization is correct and has <math>28</math> factors.
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In the first case, <math>p=3</math> is the only possibility.  This gives <math>2n = 2\cdot p^{28}</math>, which has <math>2\cdot {29}</math> factors, which is way too many.
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In the second case, <math>p=3</math> gives <math>2n = 2q^{14}</math>.  If <math>q=2</math>, then there are <math>16</math> factors, while if <math>q\neq 2</math>, there are <math>2\cdot 15 = 30</math> factors.
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In the second case, <math>q=3</math> gives <math>2n = 2p3^{13}</math>.  If <math>p=2</math>, then there are <math>3\cdot 13</math> factors, while if <math>p\neq 2</math>, there are <math>2\cdot 2 \cdot 13</math> factors.
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In the third case, <math>p=3</math> gives <math>2n = 2\cdot 3\cdot q^9</math>.  If <math>q=2</math>, then there are <math>11\cdot 2 = 22</math> factors, while if <math>q \neq 2</math>, there are <math>2\cdot 2\cdot 10</math> factors.
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In the third case, <math>q=3</math> gives <math>2n = 2\cdot p^2\cdot 3^8</math>.  If <math>p=2</math>, then there are <math>4\cdot 9</math> factors, while if <math>p \neq 2</math>, there are 2\cdot 3\cdot 9<math> factors.
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In the fourth case, </math>p=3<math> gives </math>2n = 2\cdot 3^3\cdot q^5<math>.  If </math>q=2<math>, then there ar </math>7\cdot 4= 28<math> factors.  This is the factorization we want.
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Thus, </math>3n = 3^4 \cdot 2^5<math>, which has </math>5\cdot 6 = 30<math> factors, and </math>2n = 3^3 \cdot 2^6<math>, which has </math>4\cdot 7 = 28<math> factors.
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In this case, </math>6n = 3^4\cdot 2^6<math>, which has </math>5\cdot 7 = 35<math> factors, and the answer is </math>\boxed{C}$
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=28|num-a=30}}
 
{{AHSME box|year=1996|num-b=28|num-a=30}}

Revision as of 20:18, 20 August 2011

Problem

If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?

$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution

Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.

$30 \rightarrow p^{29}$

$2 \cdot 15 \rightarrow pq^{14}$

$3\cdot 10 \rightarrow p^2q^9$

$5\cdot 6 \rightrarrow p^4q^5$ (Error compiling LaTeX. Unknown error_msg)

$2\cdot 3\cdot 5 \rightarrow pq^2r^4$

The variables $p, q, r$ are different prime factors, and one of them must be $3$. We now try to count the factors of $2n$, to see which prime factorization is correct and has $28$ factors.

In the first case, $p=3$ is the only possibility. This gives $2n = 2\cdot p^{28}$, which has $2\cdot {29}$ factors, which is way too many.

In the second case, $p=3$ gives $2n = 2q^{14}$. If $q=2$, then there are $16$ factors, while if $q\neq 2$, there are $2\cdot 15 = 30$ factors.

In the second case, $q=3$ gives $2n = 2p3^{13}$. If $p=2$, then there are $3\cdot 13$ factors, while if $p\neq 2$, there are $2\cdot 2 \cdot 13$ factors.

In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$. If $q=2$, then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$, there are $2\cdot 2\cdot 10$ factors.

In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$. If $p=2$, then there are $4\cdot 9$ factors, while if $p \neq 2$, there are 2\cdot 3\cdot 9$factors.

In the fourth case,$ (Error compiling LaTeX. Unknown error_msg)p=3$gives$2n = 2\cdot 3^3\cdot q^5$. If$q=2$, then there ar$7\cdot 4= 28$factors. This is the factorization we want.

Thus,$ (Error compiling LaTeX. Unknown error_msg)3n = 3^4 \cdot 2^5$, which has$5\cdot 6 = 30$factors, and$2n = 3^3 \cdot 2^6$, which has$4\cdot 7 = 28$factors.

In this case,$ (Error compiling LaTeX. Unknown error_msg)6n = 3^4\cdot 2^6$, which has$5\cdot 7 = 35$factors, and the answer is$\boxed{C}$

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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
All AHSME Problems and Solutions
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