GYanggg

My mistake, I did not realize there was a less than, I thought it was just an equal sign. 

Let n be the integer such that $0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31}.$ What is $\left(2^n\right)^3 - 2 \pmod{31}$? Express your answer as an integer from 0 to 30, inclusive.

$3n\equiv1\pmod{31}\\ 3n\cdot21\equiv1\cdot21\pmod{31}\\ 63n\equiv21\pmod{31}\\ 62n+n\equiv21\pmod{31}\\ \boxed{n\equiv21\pmod{31}}$

If $0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31},$ n = 21

You can finish this on your own. 

What is the sum of all positive integer solutions less than or equal to 20 to the congruence $13(3x-2)\equiv 26\pmod 8$?

Note: you usually can not divide in linear congruences, but since 13 is relatively prime to 8, you can. 

$13(3x-2)\equiv 26\pmod 8\\ 3x-2\equiv2\pmod8\\ 3x\equiv4\pmod8\\ 3x\cdot3\equiv4\cdot3\pmod8\\ 9x\equiv12\pmod8\\ 8x+x\equiv12\pmod8\\ x\equiv12\pmod8$

I hope this helped,

Gavin

What is the residue of $9^{2010}$, modulo 17?

We can see the pattern:

$9^1\equiv9\pmod{17}\\ 9^2\equiv13\pmod{17}\\ 9^3\equiv15\pmod{17}\\ 9^4\equiv16\pmod{17}\\ 9^5\equiv8\pmod{17}\\ 9^6\equiv4\pmod{17}\\ 9^7\equiv2\pmod{17}\\ 9^8\equiv1\pmod{17}\\$

The residues then repeat after that. 

Therefore, since $2010÷8=251\ R\ 2$, the residue is 13. 

I hope this helped,

Gavin

Note: this is solution is incorrect, since I missed the less than, I thought it was just an equal sign.

We first need to determine the range of for $a$:

$1^3=1,\\ 2^3=8,\\ 3^3=27,\\ 4^3=64.\\ \Rightarrow{a=2}$

$a=2, \text{since a is even}\\ \Rightarrow2^3+b^2+c=50\\ b^2+c=42$

Repeating our process, b can be 2, 4, or 6

There is a fixed value for c for each value of b. 

Therefore, there are 3 triples that satisfy $a^3+b^2+c=50$

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The prime factorization of 18 is $2\cdot3^2$

Case 1: {1, 2, 3, 3}

If these digits are used, there are 4!/2! combinations. 

Case 2: {1, 1, 3, 6}

If these digits are used, there are 4!/2! combinations. 

Case 3: {1, 1, 2, 9}

If these digits are used, there are 4!/2! combinations. 

There are a total of 4!/2! + 4!/2! + 4!/2! = 36 positive four digit integers. 

However, excluding permutations with ones in the unit digits, we get 36 - 15 = 21

As far as I know this is the best way. 

I hope this helped,

Gavin

1.

From the definition of logarithm, we obtain:

$\frac{1}{log_b{a}}=\log_a{b}$

The equation can therefore be written in the form:

$\log_n2+\log_n3+\cdots\log_n100+=\log_n(2\cdot3\cdots100)$

from which the desired conclusion immediately follows. 

2. 

https://www.desmos.com/calculator/vp8etonnna 

First, if $\sin{x}=\log{x},$  then $x\le10$ (inasmuch as $\sin{x}\le1$. 

Since $2\cdot2\pi>10$, the interval on the x-axis between x = 0 and x = 10 contains one complete period of the sine curve plug part of a second period. The graph of $\log{x}$ intersects the first wave of the sine curve at precisely one point. Futher, since $2\Pi+\frac{\Pi}{2}<10$, than at the point $x=\frac{5\Pi}{2}$, we have $\sin{x=1}>\log{x}$, which means that the graph of log x intersects the first half of the second positive wave of sin x. Since, at $x=10, \log{x=1}>\sin{x},$ the graph intersect this second wave another time. Therefore we conclude that the equation $\sin{x}=\log{x},$ has exactly three roots.

Wow! I haven't done these in SO long! Thanks for bringing me back to the more beautiful realm of mathematics. 

I hope this helped,

Gavin

Hey Rollingblade!

When N is divided by 10, the remainder is a, 

When N is divided by 13, the remainder is b.

From these two statements, we can setup the modular arithmetic systems:  

$\begin{align*} x &\equiv 1 \pmod{10}, \\ x &\equiv 0 \pmod{13}. \end{align*}$  

The solution is: $X\equiv91\pmod{130}$

Then we solve:

$\begin{align*} y &\equiv 0 \pmod{10}, \\ y &\equiv 1 \pmod{13}. \end{align*}$ 

The solution is: $y\equiv40\pmod{130}$

Therefore, the answer is $N\equiv91a+40b\pmod{130}$

I hope this helped,

Gavin

Since the equation is already factored, we just need to consider the three cases:

To solve this problem, we need to understand that the product of zero and any number is equal to zero. 

Case 1: 3x = 0 

3x = 0 

x = 0 

Case 2: x + 4 =0

x + 4 = 0 

x= -4 

Case 3: x - 6

x - 6 = 0 

x = 6 

The three solutions are 0, -4, and 6. 

I hope this helped,

Gavin

$\sqrt{2x-5}+4=x\\ \sqrt{2x-5}+4-4=x-4\\ \sqrt{2x-5}=x-4\\ (\sqrt{2x-5})^2=(x-4)^2\\ 2x-5=x^2-8x+16\\ x^2-10x+21=0\\ (x-7)(x-3)=0\\ x_1=7, x_2 = 3$

We plug the solutions to verify:

$\sqrt{2\cdot7-5}+4=7\\ \sqrt9+4=7\\ 3+4=7\\$ 

Seven works!

$\sqrt{2\cdot3-5}+4=3\\ \sqrt1+4=3\\ 5\ne3\\$  

Three doesn’t work. 

Therefore, the answer you seek is 7!

I hope this helped,

Gavin

In radians, π is also equal to 180 degrees.

510 degrees is also 2π + 150 degrees, 

150 degrees is also 5/6 π.

So the answer is 2 5/6π

Or 17/6 π

I hope this helped,

Gavin