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 20:19, 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. ! Undefined control sequence.)
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 |