2019 AMC 8 problems and solutions, for students, by students
A student's reflection on this year's test :
Mrs. Lin,
I did the AMC 8 yesterday, and it was actually quite easier than last year. I was reviewing my answers, and I believe I only got the last two wrong. I used stars and bars for the last one, but did 21C2 instead of 20C2. I could’ve done number 24, because geometry is really my best subject in math. I realized that I should’ve used mass points later on. It’s fine, though, because it’s still a good score. I think that many people could solve this test pretty well because in many of the last questions you could plug in the multiple choice answers and get the right answer. Also, a lot of it was just plain algebra. Question 20 was actually just an equation, which anybody who knows basic algebra can solve. I thought that I would never say this, but I honestly wish that it was harder, because I was hoping for some interesting problems. Those are the problems that get people’s gears turning; this year the problems were quite basic. I think many people will get really good scores on this test, which, along with a good thing, is also not so good because it brings down the credibility of the test.
Thanks,
some links that you can review those very basic, but extremely useful strategies on this
year's seemingly harder, but not really last two questions.
mass points learn together with triangles sharing the same vertex
dimensional change / scaling
balls and urns, stars and bars (lots of variations or twists on this one, so
you need to fully understand the concept so to use it well. Be patient !!!!!)
Showing posts with label AMC problems. Show all posts
Showing posts with label AMC problems. Show all posts
Thursday, November 21, 2019
Wednesday, August 20, 2014
Notes to Sunday Nights' Problem Solving Group Lessons
This week's work : Please review the last 8 hardest AMC-8 problems from 2005 to 2011 + finish the last 8 hardest AMC-8 problems from 2012 and 2013 if you haven't done that.
Some notes and questions.
Question #1 : How many ways can you climb up a ten-step staircase if you climb only one or two steps at a time ?
Solutions I :
Make a chart, starting with a smaller case.
one-step staircase -- 1 way, which is 1.
two-step staircase -- 2 ways, which is 1, 1 or 2.
three-step staircase -- 3 ways, which is 1 1 1, 1 2 or 2 1.
four-step staircase -- 5 ways, which is 1111, 211, 121, 112, 2222.
Notice the pattern - - 1 , 2, 3, 5, 8, 13, 21, 34, 55, 89, which is the answer; it also happens that it's
part of the Fibonacci numbers.
Why does it work that way ?
Well, if you climb the 3-step staircase, there are two cases :
You either take one step at first, and there are 2-step left, which leaves you two ways to climb the remaining staircase.
Or you take two step at first, and there are 1 step left, which leaves you one way to climb the remaining staircase.
If you climb the 4-step staircase, again, there are two similar cases :
You either take one step at first, and then there are 3-step left, which leaves you 3 ways to climb the remaining staircase.
Or you take two step at first, and there are 2 step left, which leaves you 2 way to climb the remaining staircase.
Thus, it's always the sum of the previous two terms for the next staircase steps.
This concept is called recursion.
Now try another question :
2010 AMC-8 #25 : Every day at school, Jo climbs a flight of 6 stairs. Joe can take the stairs, 1, 2 or 3 at a time. For example, Jo could climb, 3, then 1, then 2. In how many ways can Jo climb the stairs ?
Solution I :
List out all the possible ways.
111111 -- 1 way
21111 -- 5C1 = 5 ways to arrange the steps
2211 -- 4!/2! x 2! = 6 ways to arrange the steps
222-- 1 way
3111 -- 4C1 or 4 ways to arrange the steps
123 -- 3! or 6 ways
33 -- 1 way
Sum them up and the answer is 24.
Solution II :
Using recursion, starting with the smallest case.
1 step -- 1 way
2 steps-- 2 ways (11, or 2)
3 steps -- 4 ways (111, 21, 12, or 3)
4 steps -- 1 + 2 + 4 = 7 ways (why??)
5 steps -- 2 + 4 + 7 = 13 ways.
6 steps -- 4 + 7 + 13 = 24 ways, which is the answer.
Besides these, please review the following questions.
Answer key below.
Q #1 1988 AMC-8 #25 : A palindrome is a whole number that reads the same forwards as backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are 1: 01, 4: 44 and 12: 21. How many times during a 12-hour period will be palindromes?
Q#2 2004 Mathcounts sprint #21: If |-2a + 1| < 13, what is the sum of the distinct possible integer values of a?
Q#3 2004 Mathcounts sprint #30 : A particular right square-based pyramid has a volume of 63,960 cubic meters and a height of 30 meters. What is the number of meters in the length of the lateral height (AB) of the pyramid? Express your answer to the nearest whole number.
Answer key :
#1 : 57
#2 : 6
#3 : 50
Some notes and questions.
Question #1 : How many ways can you climb up a ten-step staircase if you climb only one or two steps at a time ?
Solutions I :
Make a chart, starting with a smaller case.
one-step staircase -- 1 way, which is 1.
two-step staircase -- 2 ways, which is 1, 1 or 2.
three-step staircase -- 3 ways, which is 1 1 1, 1 2 or 2 1.
four-step staircase -- 5 ways, which is 1111, 211, 121, 112, 2222.
Notice the pattern - - 1 , 2, 3, 5, 8, 13, 21, 34, 55, 89, which is the answer; it also happens that it's
part of the Fibonacci numbers.
Why does it work that way ?
Well, if you climb the 3-step staircase, there are two cases :
You either take one step at first, and there are 2-step left, which leaves you two ways to climb the remaining staircase.
Or you take two step at first, and there are 1 step left, which leaves you one way to climb the remaining staircase.
If you climb the 4-step staircase, again, there are two similar cases :
You either take one step at first, and then there are 3-step left, which leaves you 3 ways to climb the remaining staircase.
Or you take two step at first, and there are 2 step left, which leaves you 2 way to climb the remaining staircase.
Thus, it's always the sum of the previous two terms for the next staircase steps.
This concept is called recursion.
Now try another question :
2010 AMC-8 #25 : Every day at school, Jo climbs a flight of 6 stairs. Joe can take the stairs, 1, 2 or 3 at a time. For example, Jo could climb, 3, then 1, then 2. In how many ways can Jo climb the stairs ?
Solution I :
List out all the possible ways.
111111 -- 1 way
21111 -- 5C1 = 5 ways to arrange the steps
2211 -- 4!/2! x 2! = 6 ways to arrange the steps
222-- 1 way
3111 -- 4C1 or 4 ways to arrange the steps
123 -- 3! or 6 ways
33 -- 1 way
Sum them up and the answer is 24.
Solution II :
Using recursion, starting with the smallest case.
1 step -- 1 way
2 steps-- 2 ways (11, or 2)
3 steps -- 4 ways (111, 21, 12, or 3)
4 steps -- 1 + 2 + 4 = 7 ways (why??)
5 steps -- 2 + 4 + 7 = 13 ways.
6 steps -- 4 + 7 + 13 = 24 ways, which is the answer.
Besides these, please review the following questions.
Answer key below.
Q #1 1988 AMC-8 #25 : A palindrome is a whole number that reads the same forwards as backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are 1: 01, 4: 44 and 12: 21. How many times during a 12-hour period will be palindromes?
Q#2 2004 Mathcounts sprint #21: If |-2a + 1| < 13, what is the sum of the distinct possible integer values of a?
Q#3 2004 Mathcounts sprint #30 : A particular right square-based pyramid has a volume of 63,960 cubic meters and a height of 30 meters. What is the number of meters in the length of the lateral height (AB) of the pyramid? Express your answer to the nearest whole number.
Answer key :
#1 : 57
#2 : 6
#3 : 50
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