MaxWong

If x^3 = 1, then x^3 - 1 = 0.

Factorizing gives (x - 1)(x^2 + x + 1) = 0.

If the root x is non-real and its cube is 1, then x^2 + x + 1 = 0.

Comparing coefficients gives a = 1, b = 1.

We use a table to summarize the cases. 

Out of 36 cases, 22 cases satisfy the condition.

Probability = 22/36 = 11/18

Rearranging terms,

x = y^2 - 8y + 13.

Completing the square gives

x = (y - 4)^2 - 3.

Therefore, the vertex is (-3, 4).

By triangle inequality, sum of any two sides must be greater than the other side.

If the third side is 5, then 9 + 5 < 15, which contradicts triangle inequality.

The answer is 5.

For rational functions, asymptotes occur when the denominator is 0.

Then x = 1 and x = 2 are roots of x^2 + ax + b = 0.

Using some precalculus, a = -3 and b = 2.

The value of a + b is obvious at this point.

First term is 1/3.

Common ratio is (1/6)/(1/3) = 1/2.

Using the formula sum = a/(1 - r), sum = \(\dfrac{\frac13}{1 - \frac12} = \dfrac23\)

A dilation with scale factor k applies a scale factor of k^2 on the area, so A/[XYZ] = (-3/4)^2 = 9/16.

We calculate the trailing zeroes of each of 100!, 200!, 300!, and 400!.

Note that we have this nice formula for calculating the trailing zeroes of n! : \(\#\text{ trailing zeros} = \displaystyle\sum_{k = 1}^\infty \left\lfloor\dfrac{n}{5^k}\right\rfloor\).

Then \(100!\) has \(\left\lfloor\dfrac{100}5\right\rfloor + \left\lfloor\dfrac{100}{5^2}\right\rfloor = 24\) trailing zeros.

Similarly, 200! has 49 trailing zeros, 300! has 74 trailing zeros, and 400! has 99 trailing zeros.

Then, (100!)(200!)(300!)(400!) has 24 + 49 + 74 + 99 = 246 trailing zeros.

Simplifying the left hand side gives 32x^3 = 2.

DIviding by 32 on both sides and taking cube root gives the answer.

Answer: \(x = \dfrac1{2\sqrt[3]2}\)

Hint: If you project the vector \(\overrightarrow{AV}\) onto the line, take the projection as \(\overrightarrow{p}\), \(\overrightarrow{p}\) is actually \(\overrightarrow{OP} - \overrightarrow{OA}\).