- #1
lfdahl
Gold Member
MHB
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Problem:
Let $k$ be a natural number, and $k+1 \equiv 0 \:\: (mod\:\:24)$
Show, that the sum of $k$´s divisors is also divisible by $24$.
Solution:
First, note that since $k = 4n_1+3$ for some $n_1\in \mathbb{N}$, $\sqrt{k}$ is not a natural number.
Let $p_1,p_2,…,p_m < \sqrt{k}$ be all $k$´s divisors smaller than $\sqrt{k}$.
Then, the sum of all $k$´s divisors is: $\sum_{i=1}^{m}\left ( p_i+\frac{k}{p_i} \right )$.
Now, since $k = 3n_2+2$ and $k = 8n_3+7$, and $k = p_i\frac{k}{p_i}$, the term $p_i+\frac{k}{p_i}$ is divisible by $3$ and $8$.
Thus, the sum : $\sum_{i=1}^{m}\left ( p_i+\frac{k}{p_i} \right )$ is also divisible by 24.
Can someone explain to me, why $p_i+\frac{k}{p_i}$ is divisible by $3$ and $8$??
Thankyou in advance!
Let $k$ be a natural number, and $k+1 \equiv 0 \:\: (mod\:\:24)$
Show, that the sum of $k$´s divisors is also divisible by $24$.
Solution:
First, note that since $k = 4n_1+3$ for some $n_1\in \mathbb{N}$, $\sqrt{k}$ is not a natural number.
Let $p_1,p_2,…,p_m < \sqrt{k}$ be all $k$´s divisors smaller than $\sqrt{k}$.
Then, the sum of all $k$´s divisors is: $\sum_{i=1}^{m}\left ( p_i+\frac{k}{p_i} \right )$.
Now, since $k = 3n_2+2$ and $k = 8n_3+7$, and $k = p_i\frac{k}{p_i}$, the term $p_i+\frac{k}{p_i}$ is divisible by $3$ and $8$.
Thus, the sum : $\sum_{i=1}^{m}\left ( p_i+\frac{k}{p_i} \right )$ is also divisible by 24.
Can someone explain to me, why $p_i+\frac{k}{p_i}$ is divisible by $3$ and $8$??
Thankyou in advance!