Part I work :
Review:
2010 AMC-8 # 25
What about if John can only take 1 or 2 stairs at a time and now there are 10 stairs ?
Make sure you know how to solve this as well.
2010 AMC-8 #23
At least 2 ways : 45-45-90 degree angle ratio or Pythagorean theory; this is not hard.
2011 AMC-8 #23
This one is tricky. Make sure you understand how to solve this.
What about this question? If you can use 2, 3, 4, 6, 9 to make a 3-digit number.
You can repeat the same digit, just that the same digit is not next to each other.
How many ways?
2012 AMC-8 #23 Ratio, proportion way is faster than the other method.
1999 AMC-8 #25 Use Sam's way; it's much faster.
However, review the geometric sequence here from AoPs.
Sequences from Mathcounts Mini
1998 AMC-8 #24
This is to find all the remainders at least once when triangular numbers are divided by 8.
Now the 1st, 2nd, 3rd, 6th and 7th rows are filled once in black, so we need to find the 4th, 5th an 8th row.
5 + 2 *8 = 21 (a triangular number), 4 + 8* 3 = 28 (a triangular number). Now, we only need to find one number
that is a triangular number and when divided by 8, the remainder is "0". This can be achieved first when n is 15.
\(\dfrac {n\left( n+1\right) } {2}=\dfrac {15*16} {2}=120\)
1998 AMC-8 #25 Explanation from my blog.
Part II work :
This week, work on the last 5 problems from AMC-8 year 1993, 1994, 1995, 1996, 1997.
Here is the link from AoPs.
Review:
2010 AMC-8 # 25
What about if John can only take 1 or 2 stairs at a time and now there are 10 stairs ?
Make sure you know how to solve this as well.
2010 AMC-8 #23
At least 2 ways : 45-45-90 degree angle ratio or Pythagorean theory; this is not hard.
2011 AMC-8 #23
This one is tricky. Make sure you understand how to solve this.
What about this question? If you can use 2, 3, 4, 6, 9 to make a 3-digit number.
You can repeat the same digit, just that the same digit is not next to each other.
How many ways?
2012 AMC-8 #23 Ratio, proportion way is faster than the other method.
1999 AMC-8 #25 Use Sam's way; it's much faster.
However, review the geometric sequence here from AoPs.
Sequences from Mathcounts Mini
1998 AMC-8 #24
This is to find all the remainders at least once when triangular numbers are divided by 8.
Now the 1st, 2nd, 3rd, 6th and 7th rows are filled once in black, so we need to find the 4th, 5th an 8th row.
5 + 2 *8 = 21 (a triangular number), 4 + 8* 3 = 28 (a triangular number). Now, we only need to find one number
that is a triangular number and when divided by 8, the remainder is "0". This can be achieved first when n is 15.
\(\dfrac {n\left( n+1\right) } {2}=\dfrac {15*16} {2}=120\)
1998 AMC-8 #25 Explanation from my blog.
Part II work :
This week, work on the last 5 problems from AMC-8 year 1993, 1994, 1995, 1996, 1997.
Here is the link from AoPs.