Use modulo reduction of the base, if applicate, and the power decomposition calculations on congruences, to calculate the least non-negative residue (or equivalently the congruence class) of the folowing:

${8052}^{403}\text{mod}5$

ankarskogC
2020-11-08
Answered

Use modulo reduction of the base, if applicate, and the power decomposition calculations on congruences, to calculate the least non-negative residue (or equivalently the congruence class) of the folowing:

${8052}^{403}\text{mod}5$

You can still ask an expert for help

berggansS

Answered 2020-11-09
Author has **91** answers

Given that $8052}^{403$

This can be rewritten as,

$8052}^{403}\equiv {(8050+2)}^{403$

$\equiv {2}^{403}\text{mod}5$

So, the equation can be rewritten as,

${8052}^{403}\equiv {2}^{402}\left(2\right)\text{mod}5$

$\equiv {4}^{201}\left(2\right)\text{mod}5$

$={(5-1)}^{201}\left(2\right)\text{mod}5$

$\equiv -{1}^{201}\left(2\right)\text{mod}5$

$\equiv -2\text{mod}5$

$\equiv (5-2)\text{mod}5$

$\equiv 3\text{mod}5$

This can be rewritten as,

So, the equation can be rewritten as,

asked 2021-08-06

To write:

An inequality that compares m$\mathrm{\angle}CQD$ and m$\mathrm{\angle}AQB$

An inequality that compares m

asked 2022-07-20

Distance between (-5,13) and (4,7)?

asked 2022-07-18

A parametrization of a circle by arc length may be written as

$\gamma (t)=c+r\mathrm{cos}(t/r){e}_{1}+r\mathrm{sin}(t/r){e}_{2}.$

Suppose $\beta $ is an unit speed curve such that its curvature $\kappa $ satisfies $\kappa (0)>0$.

How to show there is one, and only one, circle which approximates $\beta $ in near $t=0$ in the sense

$\gamma (0)=\beta (0),{\gamma}^{{}^{\prime}}(0)={\beta}^{{}^{\prime}}(0)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\gamma}^{{}^{\u2033}}(0)={\beta}^{{}^{\u2033}}(0).$

I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

$\gamma (t)=c+r\mathrm{cos}(t/r){e}_{1}+r\mathrm{sin}(t/r){e}_{2}.$

Suppose $\beta $ is an unit speed curve such that its curvature $\kappa $ satisfies $\kappa (0)>0$.

How to show there is one, and only one, circle which approximates $\beta $ in near $t=0$ in the sense

$\gamma (0)=\beta (0),{\gamma}^{{}^{\prime}}(0)={\beta}^{{}^{\prime}}(0)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\gamma}^{{}^{\u2033}}(0)={\beta}^{{}^{\u2033}}(0).$

I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

asked 2021-05-14

Write the area A of a square as a function of its peremiter P.

asked 2022-07-09

Show that the curve defined by

$\gamma (s):=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Is on a circle with the radius $a$ and the center (0,0),also show that $\gamma (s)$ has been parameterized by its arc length.

I know that the parametric equation of $a$ circle with radius a and center $({x}_{0},{y}_{0})$ is :

$x={x}_{0}+a\mathrm{cos}\left(t\right)$

$y={y}_{0}+a\mathrm{sin}\left(t\right)$

If we denote the parametric equation of a circle with $\gamma (t)=({x}_{0}+a\mathrm{cos}\left(t\right),{y}_{0}+a\mathrm{sin}\left(t\right))$,then we have:

$\frac{d\gamma (t)}{dt}=(-a\mathrm{sin}\left(t\right),a\mathrm{cos}\left(t\right))$

$\Vert \frac{d\gamma (t)}{dt}\Vert =\sqrt{{a}^{2}}=a$

So the arc length is :

$s={\int}_{0}^{t}\Vert \frac{d\gamma (\tau )}{dt}\Vert d\tau =a{\int}_{0}^{t}\phantom{\rule{thickmathspace}{0ex}}d\tau $

Which implies $s=at$.

And if we parameterize the circle by its arc length then:

$\gamma (s)=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Which is the given parametrization.

But I have not shown that the center is at the origin.

$\gamma (s):=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Is on a circle with the radius $a$ and the center (0,0),also show that $\gamma (s)$ has been parameterized by its arc length.

I know that the parametric equation of $a$ circle with radius a and center $({x}_{0},{y}_{0})$ is :

$x={x}_{0}+a\mathrm{cos}\left(t\right)$

$y={y}_{0}+a\mathrm{sin}\left(t\right)$

If we denote the parametric equation of a circle with $\gamma (t)=({x}_{0}+a\mathrm{cos}\left(t\right),{y}_{0}+a\mathrm{sin}\left(t\right))$,then we have:

$\frac{d\gamma (t)}{dt}=(-a\mathrm{sin}\left(t\right),a\mathrm{cos}\left(t\right))$

$\Vert \frac{d\gamma (t)}{dt}\Vert =\sqrt{{a}^{2}}=a$

So the arc length is :

$s={\int}_{0}^{t}\Vert \frac{d\gamma (\tau )}{dt}\Vert d\tau =a{\int}_{0}^{t}\phantom{\rule{thickmathspace}{0ex}}d\tau $

Which implies $s=at$.

And if we parameterize the circle by its arc length then:

$\gamma (s)=(a\mathrm{cos}\left(\frac{s}{a}\right),a\mathrm{sin}\left(\frac{s}{a}\right))$

Which is the given parametrization.

But I have not shown that the center is at the origin.

asked 2021-08-14

Find $m\stackrel{\u2322}{AB}$ =

asked 2021-02-13

Are pairs of triangles similar?

Given information: