RedDragonl

The answer is phi(1200) = 600.

The answer is phi(191) = 172.

The smallest number that works is 16.

The number we're looking at is interesting because it can be expressed as a difference of squares:

1,005,010,010,005,001 = (1005^2) - (1^2)

We can use a property of difference of squares to our advantage: a^2 - b^2 = (a + b)(a - b). In this case:

(1005^2) - (1^2) = (1005 + 1)(1005 - 1) = 1006 * 1004

Now, 1006 is even (because the last digit is 6), so it's not prime. We can focus on 1004.

Looking closely at 1004, we see that the last two digits are 04, which is a multiple of 4. Therefore, 1004 is also even and not prime.

However, 1004 can be further factored: 1004 = 2 * 2 * 251. Here, 251 is odd, and a quick primality check reveals it is a prime number.

Therefore, the largest prime factor of 1,005,010,010,005,001 is 251.

The answer to the problem is sqrt(17).

For

x+y+z=17, 

xy+xz+yz=94,

xyz=168

you can look at all the factors of 168

The answer is 3/4.

a^3 - 1/a^3 = 27.

Since B is on segment AC, we can divide triangle ABC into two right triangles: right triangle ABI and right triangle BCI.

Find the lengths of AI and CI:

We know AB = 6 and BC = 10. Since the semi-circle is constructed on AB, diameter AB is twice the radius of the semi-circle, so the radius of the semi-circle on AB is 6 / 2 = 3. This radius (3) is also the height of right triangle ABI with hypotenuse AB (6) and base AI. Using the Pythagorean Theorem on right triangle ABI:

AI^2 + 3^2 = 6^2 AI^2 = 27 AI = √27 = 3√3

Similarly, the radius of the semi-circle on BC is 10 / 2 = 5. This is the height of right triangle BCI with hypotenuse BC (10) and base CI. Using the Pythagorean Theorem:

CI^2 + 5^2 = 10^2 CI^2 = 75 CI = √75 = 5√3

Radius of the Tangent Circle:

Since the circle is tangent to all three semi-circles, the center of this circle must be at the intersection point of the angle bisectors of triangle ABC. In a right triangle, the angle bisector of the right angle coincides with the median drawn to the hypotenuse, which divides the hypotenuse into two segments with a 1:1 ratio. Therefore, point I, the intersection of AI and CI, is the center of the tangent circle.

We now have a right triangle with legs AI = 3√3 and CI = 5√3, and the hypotenuse is the diameter of the tangent circle (which is twice the radius). Using the Pythagorean Theorem:

(diameter of tangent circle)^2 = (3√3)^2 + (5√3)^2 (diameter of tangent circle)^2 = 48 diameter of tangent circle = √48 = 4√3

Since the diameter is twice the radius, the radius of the tangent circle is:

radius = diameter / 2 = (4√3) / 2 = 2√3

Therefore, the radius of the circle tangent to all three semi-circles is 2√3.

To solve the given system of equations

\[
\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4,
\]

we first rewrite each equation in terms of its cross-multiplication:

1. \( \frac{xy}{x + y} = 1 \)

\[
xy = x + y
\]

\[
xy - x - y = 0 \quad \Rightarrow \quad xy - x - y + 1 = 1
\]

\[
(x-1)(y-1) = 2
\]

2. \( \frac{xz}{x + z} = 2 \)

\[
xz = 2(x + z)
\]

\[
xz - 2x - 2z = 0 \quad \Rightarrow \quad xz - 2x - 2z + 4 = 4
\]

\[
(x-2)(z-2) = 8
\]

3. \( \frac{yz}{y + z} = 4 \)

\[
yz = 4(y + z)
\]

\[
yz - 4y - 4z = 0 \quad \Rightarrow \quad yz - 4y - 4z + 16 = 16
\]

\[
(y-4)(z-4) = 20
\]

We now have three transformed equations:

\[
(x-1)(y-1) = 2
\]

\[
(x-2)(z-2) = 8
\]

\[
(y-4)(z-4) = 20
\]

We solve these equations step by step. Start with the first equation:

\[
(x-1)(y-1) = 2
\]

Express \( y \) in terms of \( x \):

\[
y = \frac{2}{x-1} + 1
\]

Next, substitute \( y \) into the second equation:

\[
(x-2)(z-2) = 8
\]

Express \( z \) in terms of \( x \):

\[
z = \frac{8}{x-2} + 2
\]

Substitute both \( y \) and \( z \) into the third equation:

\[
(y-4)(z-4) = 20
\]

Substituting \( y = \frac{2}{x-1} + 1 \) and \( z = \frac{8}{x-2} + 2 \):

\[
\left( \frac{2}{x-1} + 1 - 4 \right) \left( \frac{8}{x-2} + 2 - 4 \right) = 20
\]

Simplify \( y - 4 \) and \( z - 4 \):

\[
y - 4 = \frac{2}{x-1} - 3
\]

\[
z - 4 = \frac{8}{x-2} - 2
\]

So, the equation becomes:

\[
\left( \frac{2 - 3(x-1)}{x-1} \right) \left( \frac{8 - 2(x-2)}{x-2} \right) = 20
\]

Simplify the terms inside the parentheses:

\[
\left( \frac{2 - 3x + 3}{x-1} \right) \left( \frac{8 - 2x + 4}{x-2} \right) = 20
\]

\[
\left( \frac{5 - 3x}{x-1} \right) \left( \frac{12 - 2x}{x-2} \right) = 20
\]

Solve for \( x \). By testing rational values, we find:

Let \( x = 3 \):

\[
y = \frac{2}{3-1} + 1 = 2
\]

\[
z = \frac{8}{3-2} + 2 = 10
\]

Therefore, \(z = 10\).

The number of solutions is 3218.

The answer is 22/9.

The answer is 65/72.

1. The answer is 24 degrees.

2. The number of possible locations is 4.

3. x = 56 degrees.