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3 years ago

Please help with this question

Mathematics

1 answer:

My name is Ann [436]3 years ago

4 0

In a symmetric histogram, you have the same number of points to the left and to the right of the median. An example of this is the distribution {1,2,3,4,5}. We have 3 as the median and there are two items below the median (1,2) and two items above the median (4,5).

If we place another number into this distribution, say the number 5, then we have {1,2,3,4,5,5} and we no longer have symmetry. We can fix this by adding in 1 to get {1,1,2,3,4,5,5} and now we have symmetry again. Think of it like having a weight scale. If you add a coin on one side, then you have to add the same weight to the other side to keep balance.

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Find an integer x such that 0<=x<527 and x^37===3 mod 527

Greeley [361]

Since $527=17\times31$ , we have that

$x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}$

By Fermat's little theorem, and the fact that $37=2(17)+3=1(31)+6$ , we know that

$x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}$
$x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}$

so we have

$\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}$

Consider the first case. By Fermat's little theorem, we know that

$x^{17}\equiv x^{16}x\equiv x\mod{17}$

so if we were to raise $x^5$ to the $n$ th power such that

$(x^5)^n\equiv x^{5n}\equiv x\mod{17}$

we would need to choose $n$ such that $5n\equiv1\mod{16}$ (because $16+1\equiv1\mod{16}$ ). We can find such an $n$ by applying the Euclidean algorithm:

$16=3(5)+1$
$\implies1=16-3(5)$
$\implies16-3(5)\equiv-3(5)\equiv1\mod{16}$

which makes $-3\equiv13\mod{16}$ the inverse of 5 modulo 16, and so $n=13$ .

Now,

$x^5\equiv3\mod{17}$
$\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}$

$3^1\equiv3\mod{17}$
$3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}$
$3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}$
$\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}$

Similarly, we can look for $m$ such that $7m\equiv1\mod{30}$ . Apply the Euclidean algorithm:

$30=4(7)+2$
$7=3(2)+1$
$\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)$
$\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}$

so that $m=13$ is also the inverse of 7 modulo 30.

And similarly,

$x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}$

Decomposing the power of 3 in a similar fashion, we have

$3^{13}\equiv(3^3)^4\times3\mod{31}$

$3\equiv3\mod{31}$
$3^3\equiv27\equiv-4\mod{31}$
$\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}$

So we have two linear congruences,

$\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}$

and because $\mathrm{gcd}\,(17,31)=1$ , we can use the Chinese remainder theorem to solve for $x$ .

Suppose $x=31+17$ . Then modulo 17, we have

$x\equiv31\equiv14\mod{17}$

but we want to obtain $x\equiv12\mod{17}$ . So let's assume $x=31y+17$ , so that modulo 17 this reduces to

$x\equiv31y+17\equiv14y\equiv1\mod{17}$

Using the Euclidean algorithm:

$17=1(14)+3$
$14=4(3)+2$
$3=1(2)+1$
$\implies1=3-2=5(3)-14=5(17)-6(14)$
$\implies-6(14)\equiv11(14)\equiv1\mod{17}$

we find that $y=11$ is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

$x\equiv31(11)(12)+17\equiv12\mod{17}$

To satisfy the second condition that $x\equiv24\mod{31}$ , taking $x$ modulo 31 gives

$x\equiv31(11)(12)+17\equiv17\mod{31}$

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find $z$ such that $17z\equiv1\mod{31}$ . Euclidean algorithm:

$31=1(17)+14$
$17=1(14)+3$

and so on - we've already done this. So $z=11$ is the inverse of 17 modulo 31. Now, we take

$x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}$

as required. This means the congruence $x^{37}\equiv3\mod{527}$ is satisfied by

$x=31(11)(12)+17(11)(24)=8580$

We want $0\le x$ , so just subtract as many multples of 527 from 8580 until this occurs.

$8580=16(527)+148\implies x=148$

3 0

3 years ago

Jackie has already baked 17 pies, and she can bake 1 pie with each additional cup of sugar she buys.

tia_tia [17]

Answer:

35 pies

Step-by-step explanation:

17 + 18 = 35 pies

18 cups of sugar = 18 pies

6 0

3 years ago

I need help on combining like terms, they make no sense whatsoever. Here are the questions: y + 4 + 3(y+2)..........z^2 + z+ 4z^

vodomira [7]

Y + 4 + 3(y+2)
y + 4 + 3y + 3*2
4y + 4 + 6
4y + 10

z^2 + z+ 4z^3 + 4z^2
5z^2 + z + 4z^3

1/4 (16 + 4p)
16/4 + 4p/4
4 + p

5 0

3 years ago

The distance between the two islands shown on the map is 210 miles. A ruler measures this distance on the map as 3.5 inches. How

AlexFokin [52]

The 'scale' of the map is a ratio (or a fraction).

It's (length on the map) / (distance in the real world) .

Different maps have different scales, but the scale is normally the same
everywhere on the same map. So no matter where it is on the map, or how 
long the distance is on the map, the ratio is always the same number on that
map.

On this particular map in this question, the ratio is (3.5 inches) / (210 miles) .
Any other measurement on the same map has the same ratio ... and what do
you have when you have equal ratios ?? That's right ! A proportion ! !

The other measurement has the ratio (1.75 inches) / (X miles) , and THAT
fraction is equal to the other one.

(1.75) / ( X ) = (3.5) / (210)

Cross-multiply in the proportion: (1.75 times 210) = (3.5 times X).

Can you find 'X' now ?

Hint: Divide both sides of that equation by 3.5 .

8 0

3 years ago

Read 2 more answers

What do you notice about these numbers? VeryEASY 4 YOU just EXPLANATION

BabaBlast [244]

The slope of AD is the reciprocal of the slope of DC. The slope of BC is the reciprocal of slope BA. The slope of AD is equal to the slope of BC (making those lines parallel). The slope of DC is equal to the slope of Ba (making those lines parallel). I hope this was what you were looking for!

8 0

3 years ago