New answers tagged divisibility
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$13\mid4^{2n+1}+3^{n+2}$
By induction: the base case $n=0$ is obvious. Assume true for $n=k$, so all that remains to show is for $n=k+1$.
Since we know the claim is true for $n=k$:
$$A_k : = 4^{2k+1}+3^{k+2}$$
is divisible by ...
- 2,516
0
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If $n$ is an integer then $4$ does not divide $n^2 - 3$
We have to show that
$\tag 1 n \ge 0 \text{ implies } 4\nmid n^2 -3$
This can be accomplished using Fermat's method of descent.
We check the first two cases:
$\; \text{For }n = 0,\, n^2 -3 = -3 \text{ ...
- 11.2k
1
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Assuming $N$ is not a perfect square and that $D(k)$ is the set of positive divisors of $k$, then $|D(N)\cap [0, \sqrt{N}]| = |D(N)\cap [\sqrt{N},N]|$
A variant of what Brian writes is the following:
Let $D(n)$ be the set of divisors of $n$. The map $\sigma\colon D(n)\to D(n)$ given by $\sigma(k) = n/k$ is an involution (since $\sigma(n/k) = n/(n/k) ...
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Accepted
Assuming $N$ is not a perfect square and that $D(k)$ is the set of positive divisors of $k$, then $|D(N)\cap [0, \sqrt{N}]| = |D(N)\cap [\sqrt{N},N]|$
There're a few issues with your argument, each of which have simple fixes, but as it stands I wouldn't personally be able to accept it. Specifically,
You start by assuming $n_1 \in D_1$ and $n_2 \in ...
- 19.8k
0
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Proving an upper bound for solutions with divisibility
You've made a great start. Your conjecture is correct, i.e., for each $k \ge 2$, with $m \ge n$, the largest value of $m$ is $m = k^3$.
With your last equation, and using some of the reasoning in your ...
- 45.8k
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If $10|abc$, prove that either $10|ab$, $10|bc$, or $10|ac$.
if $10|abc$ then 2 and 5 must be a factor of a,b or c. Thus at least one of ab, bc or ac will have 10 as a factor.
For example if a has 2 as a factor and b has 5 as a factor then 10|ab or if c has 2 ...
- 21
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If $p^{n-1} + \cdots + p + 1$ divides $\sum\limits_{k=1}^m p^{j_k}(p^{i_k - 1} + \cdots p + 1)$, then $n$ divides $\sum\limits_{k=1}^m i_k$?
Alex Ravsky has already provided a good answer which shows a counterexample to your claim which is equivalent to the claim that if $(p^n-1)\mid\sum\limits_{k=1}^m p^{j_k}(p^{i_k}-1)$, then $n\mid \sum\...
- 134k
5
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Simultaneous divisibility condition
This is a boring case-by-case analysis after reducing to a finite set of cases.
We will first note that neither $m,n$ can be odd, because $m-1,n-1$ would be even and thus cannot be a divisor of $4n-1, ...
- 175k
6
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If $p^{n-1} + \cdots + p + 1$ divides $\sum\limits_{k=1}^m p^{j_k}(p^{i_k - 1} + \cdots p + 1)$, then $n$ divides $\sum\limits_{k=1}^m i_k$?
Multiplying both $p^{n-1} + \cdots + p + 1$ and $\sum\limits_{k=1}^m p^{j_k}(p^{i_k - 1} + \cdots p + 1)$ by $p-1$, we obtain that $p^n-1$ divides $\sum\limits_{k=1}^m p^{j_k}(p^{i_k} - 1)$. In ...
- 82.3k
-2
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Find all the positive integers such that $n+7$ is divisible by $3n-1$
$\!\bmod\,\color{#c00}{3n}\!-1\!:\ \ \dfrac{1}{\color{#c00}3}\equiv {\color{#c00}n\equiv -7}\!\! \overset{\times\ 3\!\!}\iff 1\equiv -21\iff 3n\!-\!1\mid 22$
Remark $ $ This is a special Easy Inverse ...
- 267k
0
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Show some polynomial is irreducible over the field of 7 elements.
Assume $F$ is a ring ( commutative, with $1$) and $K$ is an extension of $F$ in which $x^4 + x^3 + x^2 + x + 1$ has a root $a$. Then one checks that we have the following decomposition in $K[x]$:
$$x^...
- 53.2k
2
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With polynomial long division, why are the un-applied divisor terms still multiplied and subtracted?
The idea behind the inductive step of the polynomial division algorithm for $\,G\div F\,$ is as follows. Write $\, G = ax^{k+j}+g,\,$ $\,F = x^k + f \,$ as its (highest degree) lead_term plus lower ...
- 267k
5
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With polynomial long division, why are the un-applied divisor terms still multiplied and subtracted?
The point of the first step in the long division of polynomials is to figure out how to reduce the degree of the polynomial you are dividing.
In your example that means writing
$$
x^2+x+8 = a(x+2) + ...
- 89.4k
2
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How to get the smallest $n$ that $n^n$ is divisible by $m$.
Let $\,m=p_1^{r_1}p_2^{r_2}\ldots p_t^{r_t}$ be the factorization of the positive integer $\,m\geqslant2\,.$
We can obtain the smallest positive integer $\,n\,$ such that $\,n^n$ is divisible by $\,m\,...
- 11.4k
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Algorithm for allocating whole gold bars amongst thieves by % share of the booty
This is known as an "apportionment problem".
See Mathematics of apportionment of representation in a legislative body, How to round numbers fairly, https://stackoverflow.com/q/35931885/...
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