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Difference between revisions of "1996 AHSME Problems/Problem 29"

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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 <math>2\cdot 3\cdot 9</math> factors.
 
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 <math>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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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 are <math>7\cdot 4= 28</math> factors.  This is the factorization we want.
  
 
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.
 
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.

Revision as of 13:08, 21 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, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$. If $q=2$, then there are $7\cdot 4= 28$ factors. This is the factorization we want.

Thus, $3n = 3^4 \cdot 2^5$, which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$, which has $4\cdot 7 = 28$ factors.

In this case, $6n = 3^4\cdot 2^6$, 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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