OrcSlop

In total, we must make 10 steps, 2 up, and 8 to the right.  

In order to do this, we must choose  any 2 out of the 10 moves to go up.  

Examples are R R U R R R R U R R

The number of ways to do this is 10 choose 2, \({10 \choose 2}\), or \(\fbox{45}\) as the final answer.  

You should note that because we can never travel down, anything below C (the entire 3rd row) is irrelevant in computing the answer.  

22 green marbles.  

To solve, let x be the number of marbles added.  

\(\frac{5+x}{12+x}>\frac{79}{100}\)

 simplifying, we see that 

\(x>21\frac{1}{3}\), 

so x is at least 22.  

If we put this back into the original problem, there would be 27 green marbles and 34 marbles total.  

27/34 is about 0.79411, which is slightly above 79%.

Visualizing, you can see that for the angle to the left of the top angle to be to the right of the top angle, we have to rotate the polygon 144 degrees around A_1.  A way to think about it is to have the side to the left of A_1 be moved to the right of A_1.  In order to do this, we must rotate 144 degrees, shown in the diagram.  

Thus, the answer is 144 degrees.

n=3

calculating a few terms, we have f(2)=4/3

f(3)=9/7

f(4)=16/13

f(5)=25/21

We see that each term becomes larger than the last (easily proved and intuitive)

Thus, we know that the product will keep on increasing as n increases.  

letting n=3, we have the product equal to 12/7, or about 1.714

n=4, the product is about equal to 2.110.  

From this, we know the largest value is n=3.  

It really helps to graph on this question.  

Essentially, a function is its own inverse if it can be reflected over the line and still contain the exact same values.  

If we graph f(x) = -3x + 8 if x <= 2, we have something that passes through (2, 2), and (0, 8).

Thus, k(x) must be the reflection of the line -3x+8 over the line y=x, meaning that it must pass through (2, 2) and (8, 0)

The line that passes through these points is k(x)= -1/3x + 8/3.

If k(x) equals this, then the graph is symmetric about y=x, and thus, is its own inverse.  

176.  (in base 8)

or 126 base 10

33394400 license plates.  

If it is composed of 3 letters, then 3 numbers, there is 26^3 * 10^3 ways to create a license plate.  

If the plate starts with 3 numbers however, there are 9 choices for the first number (1 through 9), and 10 choices for the other 2.  There are 26^3 ways to choose 3 letters after that, meaning that there are 9* 10^2 * 26^3 ways to arrange a license plate this way.  

In total, there are 26^3 * 10^3 +9 * 10^2 * 26^3, or 33394400

10.5, or 21/2

5.

say X divides both 11n+3 and 2n+1, and is the greatest common divisor.

then we have the following system of equations

11n+3=ax

2n+1=bx 

where a, b, and x are all positive integers.  

then, multiplying 2n+1=bx by 3 and subtracting it from 11n+3=ax (elimination for a system of equations), 

we have ax-3bx=5n

x(a-3b)=5n

n=x(a-3b)/5

using this value for n, we substitute it into the first equation 11n+3=ax

\(\frac{11x(a-3b)}{5}+3=ax \)

we simbplify this equation to get \(x(2a-11b)=5\)

However, because all x, a, and b are positive integers, we have either x=1, or x=5.  The greatest value of x is 5.  

we can note that when n=2, both 11n+3 and 2n+1 are divisible by 5 (25 and 5 respectively).  

I don't have a definative anwswer, but this was how I attempted this problem.  

The first thing to notice is that it is impossible to derive 2z from any term other than \((1-z)^{b_1}\) due to how the binomials expand.  

Moreover, for any \(b_1\) other than 2, it is impossible to derive -2z (getting coefficients from binomial theorem). 

In particular, when \(b_1\)= 2, then the (1-z) term expands to \(1-2z+z^2\).  

Moreover, when we expand \((1-z^{32})^{b_{32}}\), the only terms that matter are the 1, and \(b_{32}C1-z^{32}\), or \(b_{32}z^{32}\).  (we again expand using the binomial theorem)

All the other terms contain z to a power greater than 32, so they are irrelevant once we expand and discard the terms.  Moreover, the 1 simply "maintains" all the terms in the expansion of things before the term containing \(b_{32}\), so it keeps the \(1-2z\) preserved.  

The \(b_{32}z^{32}\) only matters when we multiply by the term "1" in the expansion of \((1-z)^{b_1}(1-z^2)^{b_2}(1-z^3)^{b_3}(1-z^4)^{b_4}...(1-z^{31})^{b_{31}}\)

Because of this fact, we know that the exponent of (1-z^2) must be 1 in order to make the coefficient of the z^2 term 0 (the z^2 term is not altered in any way after we multiply past (1-z^2)).  

Afterward, using a similar strategy (and admittedly, wolfram alpha to hopefully find a pattern within the exponents), I found that the coefficient of the term with z^3 is 2

z^1 is 2

z^2 is 1

z^3 is 2

z^4 is 3

z^5 is 6

z^6 is 9

z^7 is 18

z^8 is 36

and this is where the calculator could not compute the expression any further.  

I can't find a pattern for the numbers 2, 1, 2, 3, 6, 9, 18, 36, and so on, but hopefully you can continue this approach.  \(b_{32}\) is the coefficient of the \(z^{32}\) term in the expansion

\((1-z)^{b_1}(1-z^2)^{b_2}(1-z^3)^{b_3}(1-z^4)^{b_4}...(1-z^{31})^{b_{31}}\)

, but I have no clue on how to get there.  The reason why this is the case is because we have to cancel out the z^32 term, in accordance with the situation descibed in the problem.  If we substitute the values for the b's we found earlier into the expression, it becomes 

\((1-z)^{2}(1-z^2)^1(1-z^3)^{2}(1-z^4)^{3}(1-z^5)^{6}(1-z^6)^{9}(1-z^7)^{18}...(1-z^{31})^{b_{31}}\)

Im sure there are some types of combinotorical ways to use the binomial theorem in order to calculate the rest of the terms, but this is my best shot at this question.  

Sorry I couldn't find a more specific answer, but tell me what the answer is if you manage to get it, this problem is pretty interesting, especially for a concise, efficient solution.