Prove that n is not divisible by 105.

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In summary, to prove that a number is not divisible by 105, we need to show that it leaves a remainder when divided by 105. This is based on the concept of remainders. An example of a number that is not divisible by 105 is 37. The steps to prove this are: choose a number to be tested, divide it by 105, and if it leaves a remainder, it is not divisible by 105. Other methods include checking for factors of 105 or using the rule of divisibility for 105.
  • #1
lfdahl
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Problem: Suppose that a natural number $n$ is an odd perfect number, i.e.:

$n$ is odd and $n$ is equal to the sum of all its positive divisors (including $1$ and excluding $n$).Prove that $n$ is not divisible by $105$.P.S.: To this day, no one knows, whether such a number exists. Here
is a comment on the subject from Wolfram Mathworld.
 
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  • #2
Suggested solution:
Suppose that $n$ is divisible by $105 = 3 \cdot 5 \cdot 7$. Consider the prime factorization of $n$:

\[n = 3^{\alpha_1}\cdot 5^{\alpha_2 } \cdot 7^{\alpha_3}\cdot p_4^{\alpha_4}\cdot ...\cdot p_k^{\alpha_k},\: \: \: \alpha_1,\alpha_2 , \alpha_3 \geq 1.\]

Let $S(n)$ be the sum of all positive divisors of $n$ (including $1$ and $n$):

\[S(n)= n\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )\left ( 1+\frac{1}{5}+...+\frac{1}{5^{\alpha_2}} \right )\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )\left ( 1+\frac{1}{p_4}+...+\frac{1}{p_4^{\alpha_4}} \right )...\left ( 1+\frac{1}{p_k}+...+\frac{1}{p_k^{\alpha_k}} \right )\]

Since $n$ is an odd perfect number $S(n) = 2n$ and $S(n)$ is not divisible by $4$. Since all primes in the decomposition of $n$ are odd, $\alpha_1$ and $\alpha_3$ are at least $2$, - otherwise

$\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )=\frac{4}{3}$ and $\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )=\frac{8}{7}$ and $S(n)$ is divisible by $4$. Finally:
\[2 = \frac{S(n)}{n}= \left ( 1+\frac{1}{3}+\frac{1}{3^{2}} \right )\left ( 1+\frac{1}{5}\right )\left ( 1+\frac{1}{7}+\frac{1}{7^2} \right )=\frac{13}{9}\cdot \frac{6}{5} \cdot \frac{57}{49}= \frac{4446}{2205}>2.\]
Contradiction.
 
  • #3
lfdahl said:
Suggested solution:
Suppose that $n$ is divisible by $105 = 3 \cdot 5 \cdot 7$. Consider the prime factorization of $n$:

\[n = 3^{\alpha_1}\cdot 5^{\alpha_2 } \cdot 7^{\alpha_3}\cdot p_4^{\alpha_4}\cdot ...\cdot p_k^{\alpha_k},\: \: \: \alpha_1,\alpha_2 , \alpha_3 \geq 1.\]

Let $S(n)$ be the sum of all positive divisors of $n$ (including $1$ and $n$):

\[S(n)= n\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )\left ( 1+\frac{1}{5}+...+\frac{1}{5^{\alpha_2}} \right )\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )\left ( 1+\frac{1}{p_4}+...+\frac{1}{p_4^{\alpha_4}} \right )...\left ( 1+\frac{1}{p_k}+...+\frac{1}{p_k^{\alpha_k}} \right )\]

Since $n$ is an odd perfect number $S(n) = 2n$ and $S(n)$ is not divisible by $4$. Since all primes in the decomposition of $n$ are odd, $\alpha_1$ and $\alpha_3$ are at least $2$, - otherwise

$\left ( 1+\frac{1}{3}+...+\frac{1}{3^{\alpha_1}} \right )=\frac{4}{3}$ and $\left ( 1+\frac{1}{7}+...+\frac{1}{7^{\alpha_3}} \right )=\frac{8}{7}$ and $S(n)$ is divisible by $4$. Finally:
\[2 = \frac{S(n)}{n}= \left ( 1+\frac{1}{3}+\frac{1}{3^{2}} \right )\left ( 1+\frac{1}{5}\right )\left ( 1+\frac{1}{7}+\frac{1}{7^2} \right )=\frac{13}{9}\cdot \frac{6}{5} \cdot \frac{57}{49}= \frac{4446}{2205}>2.\]
Contradiction.
why $S(n)=2n\, ?$ ,can you give me an example ?
 
  • #4
Albert said:
why $S(n)=2n\, ?$ ,can you give me an example ?

Yes: Take $n = 6$, which is an even perfect number, because the sum of its divisors (except for $n$) is:

$1+2+3 = 6$. The proof defines $S(n)$ as the sum of all divisors of $n$ including $n$ itself.

Thus: $S(6) = 1+2+3+6 = 12 = 2 \cdot 6$.

I would like to give you an example with $n$ odd, but this would be a hard task, since math researchers have shown, that the
smallest possible odd perfect number $n > 10^{1500}$.
 

FAQ: Prove that n is not divisible by 105.

How do you prove that a number is not divisible by 105?

To prove that a number, n, is not divisible by 105, we will show that n leaves a remainder when divided by 105.

What is the mathematical concept behind proving that a number is not divisible by 105?

The concept behind proving that a number is not divisible by 105 is the concept of remainders. If a number, n, leaves a remainder when divided by 105, then it is not divisible by 105.

Can you provide an example of a number that is not divisible by 105?

One example of a number that is not divisible by 105 is 37. When 37 is divided by 105, it leaves a remainder of 37, which means it is not divisible by 105.

What are the steps to prove that a number is not divisible by 105?

The steps to prove that a number is not divisible by 105 are:

  • Step 1: Choose a number, n, to be tested.
  • Step 2: Divide n by 105.
  • Step 3: If n leaves a remainder, then it is not divisible by 105.
  • Step 4: If n does not leave a remainder, then it is divisible by 105.

Are there any other methods to prove that a number is not divisible by 105?

Yes, there are other methods to prove that a number is not divisible by 105. One method is to factorize the number and check if it contains any factors of 105. If it does not, then it is not divisible by 105. Another method is to use the rule of divisibility for 105, which states that a number is divisible by 105 if it is divisible by both 3 and 5. If the number is not divisible by either 3 or 5, then it is not divisible by 105.

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