You can also do this without making a common demoninator. If the numbers were increased to i.e something like 15234/15235 and 15676/15677, finding the common denominator would be reduntant and be extremely time confusing. Instead, we can notice that both the fractions are 1/n away from 1. 1/5 is clearly greater than 1/7, as spliting the same thing into 5 things compared to 7 things, is clearly greater. Thus, we are subtracting 1/5 from 4/5, SOMETHING THAT is greater than subtracting 1/7, so therefore 6/7 is greater .
Sol for ASHME Q20 1992:
Since it is an n-pointed star, we have a 2n-polygon. As we know, the sum of angles in a polygon formula is (n-2)(180).
Thus, the sum of angles in our n-pointed star is 180(2n-2). If the angle at each vertex A(n) is some number A(i) (Note that the () represent subscripts) and the angle at each vertex B(n) equals B(i), we have the following equation:
n*A+n*B=180(2n-2). Note that the acute angle at B(i) is 360-B, so therefore A(i) is simply 360-B-10=350-B Hence, A+B=350. Therefore we can set up the following equation:
After we solve for n, we get that n=36!
I have qualified for USJMO twice, and USAMO once. Currently I am in 12th grade. I would say that AoPS certainly helps, but I never really used Alcumus, as the questions there are quite easy. Currently, there is around 1 more month left 'till the amc10a, and if you want to make AIME, you should start spamming tests as of now. I would say 5 amc10 tests MINIMUM a day should prepare you fine. However, you should make sure you understand the concepts to solving the problems. Furthermore, I would say you should get some AoPS books, the introduction to alegebra being the most important. If you want to make USJMO and above, I would say you'd need the precalculus book, which has much information in it.
Now, often times there is time pressure, along with the difficulty of the problems. I would suggest "ftw" on AoPS to help increase your speed in solving and skimming the problems. Note that amc10 questions 1-10 should take each maximum of 2 mins per question. Questions 10-20 take some time, and usually for me, require around 4-5 min to figure out 17-20. I suggest taking some of the newest tests, to get an idea of the difficulty of the current tests. If you can score a 120 and above consistently on amc10 tests, i would say you are fine to make AIME.
Some important concepts for the AMC tests include but are not limited to:
Basic trig (Sum of sins, cosines, law of sines, etc)
Veita's formula (extended)
Geometric and Arithemetic series
Difference and Sum of squares
Floor and Ceiling
Advanced Factorial Concepts
Combination & Purmutation
Hockey Stick Identity
Mass point geometry
Stars and Bars
Descartes Circle theorem
Roots of Unity
That is all of I can think of now. These concepts are extremely useful in doing well in the AMC tests. Good luck! :_)
Notice our equations 2 and 3 are equivalent.
Now we can rewrite our 2nd equation as x(z-1)=0
First we note that x cannot =0, as x,y, and z are all positive. Thus, z=1.
We can plug this in, giving us x^2-1=15. Hence, x=4 (x cannot be -4)
Therefore, x=4, y=4, and z=1. Thus, our answer is 4*4*1= 16.
Hello! I like the letters of the problem :-).
Notice that "USAMO" is one of the last numbers in the five letter words. If we let A=1, M=2, O=3, S=4, and U=5, we can write the numbers in term of "12345." Clearly, USAMO is only of the last letters in the sequence, as it starts with U. Note that "USAMO" is equal to 54123. Now, we can simply count backwords! The last number is 54321, followed by 54312, 54231, 54213,54132, and 54123. Thus, we can clearly see that "USAMO" is 5 away from the last number. There are 5! ways to order the numbers. The 5th to the last is 120-5= 115.
First note that the first equation is clearly a circle. We can complete the square on the first equation to get (x+2)^2+(y+3)^2=25 This is a circle with a center (-2,-3) and radius 5. The second equation can be expressed as y=4x/5 +13/5. After doing a quick graph, we can clearly see that there is 2 intersections.
First, we can fix the spot for where the teachers go. Then, we can simply plug in places for where the students can go. Note that there is 4 spaces for the students to sit. Thus, there is 4! ways to seat the students. In addition, there are 4! ways to order the teachers. Thus, there is 24*24 ways to order them. HOWEVER, the question states that you are ordering them in a circular table, meaning that you must divide by 8 to account for the number of rotations and reflections. Thus, our answer is 24*24/8, giving us 72.