Difference between revisions of "1996 AHSME Problems/Problem 29"
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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. | 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. | ||
− | 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. | + | 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 fourth case, < | + | 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. |
− | Thus, < | + | 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. |
− | In this case, < | + | 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}</math> |
==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 21:19, 20 August 2011
Problem
If is a positive integer such that
has
positive divisors and
has
positive divisors, then how many positive divisors does
have?
Solution
Working with the second part of the problem first, we know that has
divisors. We try to find the various possible prime factorizations of
by splitting
into various products of
or
integers.
$5\cdot 6 \rightrarrow p^4q^5$ (Error compiling LaTeX. Unknown error_msg)
The variables are different prime factors, and one of them must be
. We now try to count the factors of
, to see which prime factorization is correct and has
factors.
In the first case, is the only possibility. This gives
, which has
factors, which is way too many.
In the second case, gives
. If
, then there are
factors, while if
, there are
factors.
In the second case, gives
. If
, then there are
factors, while if
, there are
factors.
In the third case, gives
. If
, then there are
factors, while if
, there are
factors.
In the third case, gives
. If
, then there are
factors, while if
, there are
factors.
In the fourth case, gives
. If
, then there ar
factors. This is the factorization we want.
Thus, , which has
factors, and
, which has
factors.
In this case, , which has
factors, and the answer is
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 |