Difference between revisions of "2005 Alabama ARML TST Problems/Problem 13"
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<center><math>\begin{eqnarray}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\ | <center><math>\begin{eqnarray}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\ | ||
| − | \dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1 | + | \dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=2\\ |
p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\ | p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\ | ||
p_1(p_2^2-p_2-1)=p_2^2+p_2+1 | p_1(p_2^2-p_2-1)=p_2^2+p_2+1 | ||
Revision as of 11:17, 17 April 2008
Problem
There is one natural number with exactly 6 positive divisors, the sum of whose reciprocals is 2. Find that natural number.
Solution
Let the number be 
, and let 
and 
be primes. Therefore, one of the following is true:
For the first one, the sum of the reciprocals of the divisors of is therefore 
. The smallest prime (2) makes that less than 2, and if gets bigger, then that expression gets smaller, so there is absolutely no way that 
. So the second case is true.
\dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=2\\ p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\ p_1(p_2^2-p_2-1)=p_2^2+p_2+1
\end{eqnarray}$ (Error compiling LaTeX. Unknown error_msg)Therefore, 
. Now the only way that that is possible is when 
. Solving for , we get that 
. Checking, the sum of the reciprocals of the divisors of 
indeed sum to 2, and 28 does have 6 factors.
See also
| 2005 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 12 |
Followed by: Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
