* I will post* seven new problems every weekend, one for each day of the week. The idea is that you work on your own during the week and I will post answers the following Saturday. For those of you who practiced with me last year, the process is similar.

* For instructions* on submitting weekly answers, click on the “Logistics” tab.

*posted problems, click on the “Archive” tab.*

**For previously****, click on the “News” tab.**

*For previous prizes*** Week of Nov 6 – Nov 12: **This is the last week before ACM 8 takes place on Nov 15. We will do another timed practice this week. Again,

*Like last week, I will also ask you to submit your solutions, not just the multiple choice answers, to three of the harder problems at the end of the test. This will give you a chance to work on the harder problems if you don’t get to finish the entire test in 40 minutes. The best solution(s) will be posted and acknowledged.*

**40 minutes, 1 sitting,****no calculator, no team work, no web help, and also no penalty for GUESSING.**Click Googleform to submit solutions to the 3 written problems by Nov 12.

* Nov 6:* AMC8 test 2011, https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems. Please take the test before looking at the three problems below.

* Nov 7-8:* (Problem 22) What is the

**tens**digit of ?

* Nov 9-10: * (Problem 23) How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

* Nov 11-12: *(Problem 24) In how many ways can 10001 be written as the sum of two primes?

** Week of Oct 30 – Nov 5: **We will do another timed practice this week. Again,

*Like last week, I will also ask you to submit your solutions, not just the multiple choice answers, to three of the harder problems at the end of the test. This will give you a chance to work on the harder problems if you don’t get to finish the entire test in 40 minutes. The best solution(s) will be posted and acknowledged.*

**40 minutes, 1 sitting,****no calculator, no team work, no web help, and also no penalty for GUESSING.**~~Click Googleform to submit solutions to the 3 written problems by Nov 5.~~ See solutions: https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Answer_Key

* Oct 30*: AMC8 test 2012, http://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems. Please take the test before looking at the three problems below.

* Oct 31- Nov 1: *(Problem 18) What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?

**[Om’s solution]** The smallest prime number greater than 50 is 53. This can’t be the answer though, because it is prime. This means you have to multiply it by a number, and that number will have to be a prime number greater than 50. It can’t be 53 though, because then the answer would be a square. So you multiply 53 by the next largest prime, 59, to get 3127.

* Nov 2 – 3: *(Problem 23) An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?

**[Stephen’s solution] **Let’s call the side length of the triangle s. Since the triangle and and the hexagon have the same perimeter the side length of the hexagon is s/2. Then we can divide the hexagon into 6 equilateral triangles. Those triangles each has an area quarter to the big triangle. 4*(1/4)=1. 1*6=6

* Nov 4-5: *(Problem 25) A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length , and the other of length . What is the value of ?

**[Stephen’s solution]**

a^2+b^2=4 (due to pythagorean)

a+b=sqrt(5) (side length of square)

(a+b)^2-2ab=4 (re-write first equation)

5-2ab=4 (apply second equation)

2ab=1 (simplify)

ab=1/2 (simplify)

**[Om’s solution] **We know that one square has an area of 4, and the larger one has an area of 5. Subtracting 4 from 5 shows that the area of the 4 triangles(which are the areas inside the large square and outside the small square) is 1. This means that the area of one triangle is 1/4, or 0.25. A is the width and b is the height, so A*B*1/2 is the area, which is 1/4. You multiply by two to get A*B, which is 1/2. We could try to find A and B individually, but the problem asks to find A*B, which we have. The answer is 1/2.