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Let's set up for a really cool function. 

First, note that \(191=191*1\)

Now, we use Euler's Totient Function. We have

\(\phi{191} = 191/ ( 191) * (191-1) = 190\)

The answer is 190

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! :)

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! :)

Let's set up for a really cool function. 

First, note that

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2\)

Now, we use Euler's Totient Function. We have

\(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320\)

The answer is 320

Thanks! :)

*700th answer

1=1             5 = 5            9 = 3^2             13 =13 

2=2             6 = 2 * 3      10 = 2*5            14= 2*7

3=3             7 = 7            11=11                15 = 3*5

4=2^2         8 = 2^3         12 = 2^2 * 3      16 = 2^4

LCM = 2^4 * 3^2 * 5 * 7 * 11 * 13  = 720720 = M

No. of Divisors  =  5*3*2*2*2*2 = 240

(n + 1)^2 = n^2 + 2n + 1

            n

n + 2  [ n^2 + 2n + 1 ] 

            n^2 + 2n

            ___________

                             1

Note that this will never produce an integer because we will always have a remainder of  1 / (n + 2)

5*6*7*8 + 1  =  1681

630 / 42  = 15

Either   3 and 5    or     1 and  15

x^2 + 12x - 4 = 0

Let the roots be a,b

ab  = -4

2ab = -8

a + b = -12       square both sides

a^2 + 2ab + b^2 =144

a^2 - 8 + b^2  = 144

a^2 + b^2 = 152

x + y = 3 + k

y = -x + (3 + k)

The slope of the line = -1

The slope of a tangent line at any point on the circle  = -x/y

Set the slopes equal

-x/y = -1

-x= -y

x = y

So....subbing into both equations

x + x = 3 + k  →  2x = 3 + k  →  x = (3 + k)/2   (1)

x^2 + x^2 = k  →  2x^2 = k   (2)

Sub (1)  into (2)

2 [ (3 + k)/2]^2 = k           simplify

k^2 + 6k + 9  = 2k

k^2 + 4k + 9 = 0

k cannot be a real number because the discriminant =  -20

Had there not been any limitations; we could've done the following:

First, we can start with some basic rearrangements.

\(2x^2 - 10x + 13 = -6x^2 - 18x - 15\),

\(8x^2 + 8x + 28 = 0\), 

\(x^2 + x + \frac{7}{2} = 0 \)

.
We turn this into a monic quadratic to simplify our working out. Since the roots of the quadratic are \(a+bi, a-bi\)

, we can use Vieta's formula to see that:

\((a+bi)+(a-bi) = 2a = -1 \rightarrow a = -\frac{1}{2}\)
\((a+bi)\cdot(a-bi) = a^2 + b^2 = \frac{7}{2}\)

We can substitute our value for a and get \(a^2 + b^2 = \frac{7}{2} \rightarrow (-\frac{1}{2})^2 + b^2 = \frac{7}{2} \rightarrow b^2 = \frac{13}{4} \therefore b = \frac{\pm\sqrt{13}}{2}.\)

Which means \(a\cdot b = -\frac{1}{2}\cdot\frac{\pm\sqrt{13}}{2} = \frac{\pm\sqrt{13}}{4}.\)

\(\dfrac{x+4}{x+5} = \dfrac{x-3}{2} + \dfrac{x + 7}{5}\\ \dfrac{10(x+4)}{10(x+5)}=\dfrac{(x+5)(x-3)+(x+5)(x+7)}{10(x+5)}\\ 10x+40=x^2+2x-15+x^2+12x+35\\ 2x^2+4x-20=0\\ x^2-2x-10=0\\ x_{1,2}=1\pm \sqrt{1+10}\\ \color{blue}x\in \{-2.3166, 4.3166\}\)

 !

The value for a can't be positive for the solutions for that equation. Since the roots are literally \(\frac{-1\pm\sqrt{13}i}{2}\). b > 0 but not a.

Last question of the day for me. 

First, let's simplify the equation. Bringing all terms to on side and combining all like terms, we get

\(x^2 - mx + 14 = 0 \)

Now, we simply search for factors of 14, both negative and positive. We have

Factors of 14                                                m

(1 , 14)                   (x + 1) (x + 14)              -14

(-1, -14)                  (x - 1)  (x - 14)               14

(2,7)                       (x + 2) ( x + 7)                -9

(-2, -7)                   (x - 2) ( x - 7)                   9

4 should be the answer. 

Thanks! :)

250 in base 10 is FA in base 16.    

_.

Combining like terms and moving all terms to one side, we get

\(x^{2}+12x-4=0\)

Using the quadratic equation, we get

\(x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}\)

Finally simplifying, we get

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Thanks! :)

First, let's simplify the equation so that all terms are on one side.

Combining all like terms and moving them all to one side, we get the equation

\(x^2 + (17 - m)x + 4 = 0 \)

Now, since it has two distinct roots, the descriminant of the quadratic must be greater than 0.

Thus, we have the equation

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16 \)

We get two roots from this equation. 

Let's calculate both of them. For the first one, we have

\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

For the second root, we have

\( 17 - m < -4 \\ 21 < m \\ m > 21\)

Thus, in interval notation, we have \(m = ( -\infty, 13) U ( 21, \infty) \)

Thanks! :)

Ok, so we start off with the equation

\(ab -3a + 4b = 131 \)

I know this is random, but let'st ake the product of the coefficients on a,b.....add this product to  both sides

So this gets us

\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)

This eventually achieves the equation

\((a + 4) ( b - 3) = 119\)

Now, the factors of 119 are \(1,   7 ,   17    , 119\)

Since we want the minimum, let's use 7 and 17. We have

\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)

Thus, our final answer is 3. 

Thanks! :)

Let's create a system of equations to solve this question. 

First, we know that the two points must be on the equation \(x^2 + y^2 = 25\)

We also know that x must be 4. 

Thus, plugging in x=4 into the first equation, we can find y values. We have

\(4^2+y^2=25\\ y^2=25-16\\ y=\pm\sqrt9\\ y=3, y=-3\)

Thus, the two points A and B are \((4,3), (4,-3)\)

Since they share an x value, the difference between the y values is the distance. We have

\(3-(-3)=6\)

Thus, 6 is our final answer. 

Thanks! :)

We can use Euler's Totient Function. 

Let's find all the distinct prime factors of 2835 first. 

The prime factors of 2835 are 3, 5, and 7

Now, we simplfy do \(2835(1-1/3)(1-1/5)(1-1/7)\)

Simplfying this, we have

\(2835 * \frac{2}{3}*\frac{4}{5}*\frac{6}{7} =\frac{2835\cdot \:2\cdot \:4\cdot \:6}{1\cdot \:3\cdot \:5\cdot \:7}=1296\)

So 1296 is our answer. 

I'm not sure if I did this correctly...

Thanks! :)

Let's use a whole number of equations to solve this problem. 

First off, we know that \((a+b)^2=a^2+2ab+b^2\)

We ALSO know that \((a+b)^2=4^2=16\)

Thus, we have the equation \(a^2+2ab+b^2=16\)

We are given a^2+b^2 in the problem, so plugging that in, we get

\(6+2ab=16\)

We can now find ab. This will come in handy later. 

\(2ab=10\\ ab=5\)

Alright, let's move on to what we are TRYING to find. 

Let's note that we can split a^3+b^3 into 

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

WAIT! We already know the values of every number! Plugging in 4, 6, and 5, we get

We have

\((4)(6-5)=4\)

So 4 SHOULD be the answer. 

I might have a mistake during the calculations...not sure. 

Thanks! :)

(b) Let's use a different tactic here. 

We want the fact that 4x subtracted by the remainder is divisble by 75. 

We can have the equation \(75y=4x-71\), where y is another integer. 

Isolating x from this equation, we get

\(x=\frac{75y+71}{4}\)

Now, any value of y we plug in should work. We just need an integer, and we're done. 

When we plug in \(y=3\), we find that

\(x=\frac{75(3)+71}{4}=\frac{296}{4}=74\)

Thus, 74 is our answer. Just fit it into the interval! Lol

Thanks! :)

(a) Let's note a pattern in the equation. 

\(x=1; 346 \equiv 346\pmod{347}\\ x=2; 346*2 \equiv 345\pmod{347}\\ x=3; 346*3 \equiv 344 \pmod{347}\\ x=4; 346*4 \equiv 343 \pmod{347}\\ etc. \)

Thus, we have 

\(347-x=129\) as the pattern for this congruence. 

thus, solving for x, we find that \(x=218\)

Testing if this is indeed true, we find that 

\(x=218; 346*218 \equiv 129 \pmod{347}\)

So 218 is the answer. 

Thanks! :)

Let's note that 1/17 is a repeating decimal. 

\(1/17 = 0.\overline{ 0588235294117647}\)

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)