Difference between revisions of "2005 Alabama ARML TST Problems/Problem 13"
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For the first one, the sum of the reciprocals of the divisors of <math>n</math> is therefore <math>1+\dfrac{1}{p_1}+\dfrac{1}{p_1^2}+\dfrac{1}{p_1^3}+\dfrac{1}{p_1^4}+\dfrac{1}{p_1^5}</math>. The smallest prime (2) makes that less than 2, and if <math>p_1</math> gets bigger, then that expression gets smaller, so there is absolutely no way that <math>n=p_1^5</math>. So the second case is true. | For the first one, the sum of the reciprocals of the divisors of <math>n</math> is therefore <math>1+\dfrac{1}{p_1}+\dfrac{1}{p_1^2}+\dfrac{1}{p_1^3}+\dfrac{1}{p_1^4}+\dfrac{1}{p_1^5}</math>. The smallest prime (2) makes that less than 2, and if <math>p_1</math> gets bigger, then that expression gets smaller, so there is absolutely no way that <math>n=p_1^5</math>. So the second case is true. | ||
| − | < | + | <cmath>\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}{p_1p_2^2}=2\\ | \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 | ||
| − | \end{eqnarray}</ | + | \end{eqnarray*}</cmath> |
Therefore, <math>p_1\equiv -1\bmod{p_2}</math>. Now the only way that that is possible is when <math>p_2=2</math>. Solving for <math>p_1</math>, we get that <math>p_1=7</math>. Checking, the sum of the reciprocals of the divisors of <math>\boxed{28}</math> indeed sum to 2, and 28 does have 6 factors. | Therefore, <math>p_1\equiv -1\bmod{p_2}</math>. Now the only way that that is possible is when <math>p_2=2</math>. Solving for <math>p_1</math>, we get that <math>p_1=7</math>. Checking, the sum of the reciprocals of the divisors of <math>\boxed{28}</math> indeed sum to 2, and 28 does have 6 factors. | ||
| + | |||
| + | Furthermore, if we rearrange <math>\textbf{(2)}</math>, we can see that the number would be the sum of all its divisors other than itself, making it a perfect number. Checking the perfect numbers, we find that the second perfect number, <math>28</math>, fits the required form and is indeed the solution. | ||
==See also== | ==See also== | ||
Latest revision as of 18:14, 10 March 2015
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.
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.
Furthermore, if we rearrange 
, we can see that the number would be the sum of all its divisors other than itself, making it a perfect number. Checking the perfect numbers, we find that the second perfect number, 
, fits the required form and is indeed the solution.
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 | ||
