NotThatSmart

Let's note something important before we begin. 

All the points 5 units away from (2, 7) forms a circle with radius 5 and center of (2, 7). 

A circle with that equation would look something like

$(x-2)^2+(y-7)^2 = 5^2$

Combining that with the second part of the problem, we have a system with 

$(x-2)^2+(y-7)^2 = 25\\y=5x-28$

Now, we already have a value for y, and we subsitute that into the first equation to isolate x. 

We find that

$(x-2)^2+(5x-35)^2 = 25$

Expanding and bringing all terms to one side of the equation, we get a quadratic. We find that

$26x^2-354x+1204=0$

This gives us 2 x values that will satisfy thsi equation. We get that

$x=7,\:x=\frac{86}{13}$

Now, we plug these two values of x back into the equation to find y, so that we can complete our problem. 

We find that $y = 7, y= \frac{66}{13}$

So our final two abswers that satisfy all these conditions given above in the problem are

$(7, 7), (86/13, 66/13)$

Thanks! :)

We have a system of equations to solve. We have

$N \equiv 2 \pmod{3}, \\ N \equiv 2 \pmod{7}, \\ N \equiv 2 \pmod{10}.$

Now, let's note something really important. All 3 equations have a remainder of 2. 

Let's anakyze. 

N leaves a remainder of 2 when divided by 3, 7, and 10. 

Therefore, our answer is the LCM of 3, 7, and 10. 

This is because since they all leave the same remainder, the same number must be for all of them. 

Therefore, we have $LCM[3, 7, 10] = 210$

Adding 2, we have $210+2=212$

So our answer is 212. 

Thanks! :)

We need to solve the system of equations

$N \equiv 2 \pmod{5}, \\ N \equiv 2 \pmod{7}$

Well, let's analyze this equation. 

N leaves a remainder of 2 when divided by 5 and 7. 

This means that N is two greater than the LCM of 7 and 5. 

This is because since it leaves the same remainder for both numbers, then it must be the same number. 

Therefore it is the LCM we are looking for. 

We have

$LCM[5, 7] = 35$

Adding two, we have $35+2 = 37$

thus, our answer is 37. 

Thanks! :)

First, let's note that if a number has an odd number of factors, then it must be a perfect square. 

In the problem, it also must be divisible by 8. 

Thus first few that satsify this number is $16,64,144$

However, none of them work, as 16 and 64 have too few factors, and 144 has too many. 

The next number is 256. 

Note that 256 has exactly 9 factors of $1, 2, 4, 8, 16, 32, 64, 128,256$

Of these 9 factors, $2$ is the only prime divisor!

So out of the 9, only 1 is prime

So 1 is our answer.

Thanks! :)

First, let's note that if a number has an odd number of divisors, then it must be a perfect square. 

It also must be a multiple of 8.

So we need a perfect square divisble by 8. 

First, we have

$4*4=16$

However, 16 only has 5 factors, with $1,2,4,8,16$

Next, we have $8*8=64$. However, 64 only has 7 factors, with $1, 2, 4, 8, 16, 32, 64$

144 doesn't work, as it has way too much. 

However, we note that $16*16=256$

256 has exactly 9 divisors, of $1, 2, 4, 8, 16, 32, 64, 128,256$

Thus, the smallest possible number is 256. 

So 256 is our answer. 

Thanks! :)

There isn't really a model you can follow to solve this problem. 

However, there are multiple numbers that satisfy the condiitions given. 

First, we have

$244224 \mod 64 = 0$

Since the final result is 0, then 244224 is an answer. 

Also, let's note that another number, 

$444224 \mod 64 = 0$ also works. 

So 444224 satisfies the equation as well. 

Thus, the two answers we have are $444224, 244224$

Thanks! :)

We want two quadratics with the same constant and degree. So let's try converting the quadratic we are given. 

First, let's combine all like terms to the quadratic we are comparing to. We find that

$2x^2 - 5x + 6$

Let's note the constant is different. Thus, we must divide this by 6 so that we have the same constant as Ax^2+Bx + 1. We have

$(1/3)x^2 - (5/6) x + 1$

Thus, we know what A and B are. We simply plug those in and get

$A + B = (1/3) - (5/6) = (2/6) - (5/6) = -3/6 = -1/2$

So -1/2 is our final answer. 

Thanks! :)