Mathc1

You can evaluate it as the difference of two sums (relating problems to similar problems).

One sum is 0+2+4+6+8+...+86.

Hint: Build right triangles using the radii to points of tangency. Check AoPs and mindyourdecisions (YouTube) for examples.

If our number is divisible by 4, then the last two dights are divisible by 4. In this case, we only have 2 options for the last digit: 24 and 64.

What caught my eye in this question is the 2 and the -2.

Hypatia's and Euclid's numbers have a sum of 2. However, Plotemy cubes -2 twenty times. We can notice a pattern: \((-2)^3=-8,(-8)^3 = -512.\) and so on (when dealing with big numbers, look for patterns or try examples/smaller numbers). Answer this: Is there a cancellation that we can use?

Consider the total (the denominator). For the total (without considering restrictions) there are \(10!/10= (10*9*8...2*1)/10=9!.\) (due to rotational fixations). This is because there are 10 ways that we could rotate the table (so we overcount by a factor of 10).

Numerator case: You should be able to find the ways that the women sit next to each other by considering where to fix the womens' spots then using constructive counting. It may be easier if you use variables. Hint: \(W_1,M_1\).

As I am thinking about it, I notice that for any b(i), the angle for b is 180 minus the reflex angle. So, nA+180n-n(times the reflex angles) is some quantity. But, I am not sure how to procced.

How is nA+nB equal to 180(2n-2)? Angles b(1),...,b(n) look like exterior angles.

I have done the whole intro series, but when I do questions on the AMC, I get stuck around question 18. I find the difficult questions hard to navigate. Anyways, on Alcumus, I usually get stuck on questions that are level 25, and waste lots of time to get the answer wrong. Can you please explain (but not solve) how to relate the angles on ashme question 20 (1992).

Consider the possibilities, and overcounting.