All Questions
9
questions
0
votes
1
answer
345
views
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
I KNOW THIS IS SOLUTION BUT I DON'T KNOW WHY?
We first find the difference of the numbers and then find the HCF of the got numbers.
183−91=92
183−43=140
91−43=48
Now find HCF of 92, 140 and 48, we get
...
0
votes
2
answers
131
views
Proving that $k|(n^k-n)$ for prime $k$ [closed]
Prove that for any integer $n$,we have $(n^k)- n$ is divisible by $k$ for $k=3,5,7,11,13$
I tried using prime factorization but that does not work here
- 9
1
vote
1
answer
51
views
Is brute force trial the only approach to find smallest k such that (840k + 3) is a multiple of 9?
The following is the answer approach given for the below problem in my old book. I am skeptical about the brute trial approach suggested (though k is found after 2 trials in this case). Is there a ...
- 175
0
votes
2
answers
118
views
Determine the number of positive integers $a$ such that $a\mid 9!$ and gcd $(a, 3600)=180$.
Determine the number of positive integers $a$ such that $a\mid 9!$ and gcd $(a, 3600)=180$.
What I know as of now is that $180\mid 9!$ and that $180\le a\le9!$.
The prime factorization of 180 is $(...
- 23
0
votes
2
answers
66
views
Fractions of the form $\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\frac{n}{k}$
For any $n,k\in\mathbb{Z}^+$, $n\gt k$. I found that the majority of the time, either there are no positive integers, $a,b,c\dots,$ such that
$$\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\...
- 3,557
1
vote
1
answer
40
views
Why $xy=N\Rightarrow (((x−1)p−1\text{ mod }N)−1)\text{ mod} X=0$
Experimentally, I found that for $N=xy;\ 3<x<y$ where $x,y$ are prime numbers and for prime numbers, $p>x$ the below expression is always true:
$$\Big(\big((x-1)^{p-1}\text{ mod }N\big) -1\...
- 1,450
-1
votes
1
answer
32
views
Highest common factor of integers which are the same. [closed]
If $a=-b \in \mathbb{Z}, $
Then $hcf(a, b)=|a|=|b|$, right?
- 1,319
5
votes
1
answer
106
views
Prime factorization and hcf [closed]
For any given integer $n$, we prime factorize it as follows
$$n = p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r}. $$
Let $g = \gcd(k_1, k_2, \ldots, k_r)$ and $m_i = k_i / g$.
The function $F$ is ...
- 1,329
4
votes
3
answers
393
views
If $a \mid c$ and $b \mid c$, but $\gcd(a,b) = 1$, then $ab \mid c$.
If $a | c$ and $b | c$ and $a$ and $b$ are relatively prime prove that
$ab|c$.
What I did was since $(a,b)=1$ then we can find integers $m,n$ such that $ma + nb=1$. Now since $a|c$ then $a = mc$. ...
- 1,089