Let's multiply the polynomials and see what we get. 

We have \((x^3)(x^2 + tx)\)

Distributing in the x^3, we find that we get the expression \(x^5+tx^4\)

there is actually no x^2 term, meaning t can pretty much be anything. 

As long as t doesn't contain a x^-2 term, then it isn't possible there is an x^2 term. 

Thanks! :)

First, let's note that
\(3 + 4 + 5 =  3 *4 \)

\(13 + 14 + 15 = 3 * 14\)

This pattern will keep going. Thus, we just have

\(3 [ 4  +  14 + 24  +  ....... + 994 ] \) 

No. of terms =  100

Thus, our final answer is

\(3 [  994 + 4 ] [ 100 / 2 ]   =  149,700\)

Thanks! :)

Ok, let me give this a shot....

The Law of Cosines state that \(cos PMQ = - cos PMR \)

\(PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)\)

Subsituting in the values we already know from the problem. we get

\( 8^2 = 7.5^2 + PM^2 + 2(7.5 * PM)cos(PMR) \\ 9^2 = 7.5^2 + PM^2 - 2(7.5 * PM) cos (PMR)\)

Now, adding these two, we get

\(8^2 + 9^2  = 2 * 7.5^2 + 2PM^2 \)
\(145= 112.5+ 2PM^2 \\ 32.5/ 2 = PM^2 \\ 16.25 = PM^2 \\ PM=\sqrt{16.25}\approx 4.03112887415\)

Is this correct, CPhill? I'm not sure. Can u verify?

Thanks! :)

WAIT A SECOND! Let's note something real quick. 

First, let's look at the first equation. We want only two equations, we isolate p in terms of r. 

We get that

\(p=3+2q\)

Subsituting this value of p into the second and third equation, we get the system with r and q, we get

\(q-2r=-2+q\\ 3+2q+r=9+3+2q\)

Now, focusing on the first equation of the new system, isolating r, we find that

\(r=1\)

Plugging in this value for r into the second equation, we have that

\(3+2q+1=2q+12\\2q+4=2q+12\\-8=0\)

THIS IS DEFINITELY NOT TRUE. 

In conclusion, we know that this system has NO SOLUTIONS. 

I think this is right...

Thanks! :)

Here's a similar prob that I answered earlier :

https://web2.0calc.com/questions/geometry_2526

See if you can solve it now....

Oh, your right! Sorry about that. I was coming up with numbers...sorry.

How about

PQ = 8, PR = 9, QR = 15

Nice catch...should have caught that!

Note, NTS, that PQR can't be a triangle because

PQ + PR  = QR

But the triangle inequality says that

PQ + PR  must be > QR

Glad I can help!

Hope you understood my explenation!

Nice work! :)

~NTS

First, let's simplify the equation so that we have a quadratic equal to 0. We have

\( x^2 – mx + 24 = 10    \\ x^2 – mx + 14 = 0  \)

The coefficient of x is the sum of the factors of 14   

The integer factors of 14 are (2, 7) and (–2, –7)   

                                         and (1, 14) and (–1, –14)   

So m could equal 9 or –9 or 15 or –15   

Therefore, m could have 4 possible values    

Thanks! :)

Let P  have the coordinates  ( x , -x + 6)

PA^2  = ( x - 10)^2 + (- x + 6 - -12)^2  =  (x -10)^2 + ( -x + 18)^2

PO^2 = ( x - 2)^2 + ( -x + 6 - 8)^2 = (x -2^2 + ( -x - 2)^2

We want   PA^2 = PO^2....so....

(x -10)^2 + ( -x + 18)^2 = ( x -2)^2 + (-x -2)^2

x^2 -20x + 100 + x^2 - 36x + 324  =  x^2 -4x + 4 + x^2 +4x + 4

-56x + 424  =  8

-56x = -416

x = 416/56 = 52/7

y = - 52/7 + 6 =  -10/7 

P = (52/7, -10/7)

Thanx, Not That Smart !!   I check in once in a while !   ,,,,,

(n+2) / n    *   ( n+3)/ (n+1)  *  (n+4) / (n+2) = 10      <==== Given 

Simplifies to 

1/n * (n+3) / (n+1) * (n+4) = 10 

then:

(n^2 + 7n + 12 )/ (n^2 +n) = 10        doing the fraction division results in:

9n^2 + 3n-12 = 0 

Use Quadratic Formula with    a = 9    b= 3    and c = -12     to find  n = 1      (or - 4/3  <=== does not work) 

So the first fraction would be 3/1 

Thanks again. I think I can fill in the gaps!

 Hello EP! It's been a while! 

Good job! :)

Ok, first off, let's split the one equation into a system of equations so that each term is equal to another. We get

\(x + 1/y = y + 1/z\\ x + 1/y = z + 1/x\\ y + 1/z = z + 1/x\)

I won't really show the steps, although I will if needed. Let me know if you want to. 

\((x, y, z) = (-\frac{1}{z-1}, \frac{z-1}{z}, z)\\ (x, y,z) = (-\frac{1}{z+1}, -\frac{z+1}{z}, z)\)

We want to find the value of xyz since we need the product. 

Thus, multiplying these together, we get

\( (-\frac{1}{z-1})(\frac{z-1}{z})z=-\frac{1}{z-1}\left(z-1\right)=-1\)

\( (-\frac{1}{z+1} )(-\frac{z+1}{z}) z=\frac{1}{z}z=1\)

So good job! You were correct about 1. The other value is -1. 

I didn't go into too much detail, but let me know if you need more assistance. 

Thanks! :)

If John has 13 students he will not be able to form a rectangle using 'whole' students 

  Factors of 15 :    1   3  5  15

    he could make a 3 x 5 rectangle 

14  factors:      1 2 7 14     a 2 x 7 rectangle 

13  factors:     1 13        only a 1 x  13   rectangle (straight line ) 

stop posting your AoPs Intro to Counting and Probability class with Charles Buehrle week 2 homework. just use the message board or do it yourself. its not that hard

'

See attached image:

Thank you! I just really want the methods.

Hi, AUnVerifiedTaxPayer! I will give this problem a shot!

So a lattice point is essentially any point on the graph. (1, 1), (2, 2) ,(3, 3) are ALL latice points!

However, we need to find ones on the line connecting  (3, 17)  and (81, 131)

To do this, let's first find the slope of the line. We have

\(\frac{131-17}{81-3}=\frac{114}{78} = \frac{19}{13}\), so you are correct about the slope. 

Now, one way you can do this is by just counting. Draw a graph, and use the slope to go up 19 and to the right 13 starting from (3, 17), eventually getting the next point. We count \((3,17),(16,36),(29,55),(42,74),(55,93),(68,112),(81,131)\)

However, I will now do this another way. 

Let's do this a simpler way rather than using the slope. 

The slope​ indicates that for every increase of 13 units in the x-direction, the line moves 19 units in the y-direction.

The number of steps in the x-direction is given by the equation \(\text{Number of steps}= \frac{\text{total change in x}}{{\text{step size in x}}}\)

Now, from the slope, we know that the step size is 13. 

Since we go from 3 to 81, the total change in x is 78. 

Thus, we have 78/13 = 6.

Now, we are not done yet. We haven't included the start point, so adding 1, we have 6+1 = 7. 

So 7 is our answer. I hope I did it correctly!

Thanks! :)

I don't think it's possible for this to occur. 

In the problem, it said 

 each scientist must sit next to a mathematician

However, take a look at the graph. 

I don't think it's possible for that sentence to be true. 

Thus, this problem is invalid. 

Thanks! :)

Let's focus on the two smallest numbers first. 15 and 24. 

Since N leaves a remainder, we can elmiinate the factors.

Thus, N cannot be \(3, 5, 2, 4, 6, 8, 12\)

This leaves 7, 9, 10, 11, 13, and 14.   Now, let's test out all the other numbers

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

mmm...I don't think there is a solution

Thanks! :)

The first 5 primes are

\(101, 103, 107, 109,113 \)

113 is our answer. 

Thanks! :)

Well, let's notice something interesting about this problem. 

The number two has only two residues. 0 and 1. 

If the number is odd, then the residue is 1. If the number is even, then the residue is 0. 

We just have to figure out whether or not \(n \equiv 43 \pmod{165}\)is even or odd. 

We can easily figure this out though. 

Let's note that since n is 43 mod 165, we can write n as

\(165x+43 =n\) where x is a random integer. 

In this case, when x is even, then n is odd. 

If x is odd, then n is even. 

so in turn, both residues are completely possible for the final result. 

Thus, our answer is \(0,1\)

Thanks! :)