What is the right order for high school Math classes? Washington Post

This is a response to Valerie Strauss' blog:
http://voices.washingtonpost.com/answer-sheet/math/high-school-math-whats-the-rig.html

Thanks for this article on order of Math courses. I am not sure what order they should be in -- it may depend on how the topics are divided up. A2 is usually more rigorous than the others but it can depend on the school/class/state.


It would benefit students to learn and know the Math on the ACT/SAT/GED/ACCUPLACER (placement tests used by colleges). Many students have not seen the topics enough times, or have had the topics slivered (and are unused to multiple topics on the same exam), or have not developed the speed that will help them problem solve 20 questions in 25 minutes.

While the Common Core are under development, we already have these standards at the high school and college level.

Studying multiple choice items can improve metacognition due to compare/contrast and by studying "good wrong answers" (for example, exponent rules questions always have "good wrong answers"!!).

Students, teachers and parents can use the free SAT Question of the Day (and other free or reasonably priced resources) to better scores and knowledge and skills!
http://sat.collegeboard.com/practice/sat-question-of-the-day

Perhaps, we can bring academic and cognitive abilities up to the level of respect that athletics commands.

Robin Schwartz
Author, Build Math Confidence e-newsletter
http://www.mathconfidence.com/

Posted by: mathconfidence
January 14, 2011 12:37 PM

January 2011 Brain Teaser Solution

Q: What is the largest number of consecutive integers that will add up to 2011?

A:  A good way to think about this problem is to do what Polya said "Solve a simpler problem"
So first think about  -- What is the largest number of consecutive integers that will add up to 11?

The least number of numbers would be one -- 11
You could use two consecutive numbers -- 5 and 6.
If you think about negative integers, -4, 3, -2, -1 would cancel out 1, 2, 3, and 4
so - 4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 would give a total of 11 numbers

But we can do even better

-10 would cancel out with 10
-9 would cancel out with 9 and so on

-10, -9, -8 , -7, -6, -5, - 4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10 would make zero if you sum them.
so just now add an 11
-10 thru -1 is 10 integers
0 is 1 integer
1 thru 10 is 10 integers and
11 is 1 integer for a total of 22.
(22 = 2 x 11)

so the number of integers is always 2 times the number itself.

For 2011,
The lowest number would be -2010 (it would cancel out with +2010)
The next number would be -2009 (it would cancel out with +2009)
The next number would be -2008 (it would cancel out with +2008)
and so on...
until -2 cancels out with +2
and -1 cancels out with +1

and then there is 0

so there would be 2010 negative numbers -2010 through -1
2010 postiive numbers 1 through 2010
plus 0
and also 2011
2010 negative integerss + 2010 positive integers + 2 more (for 0 and 2011)
for a total of 4022
4022 is the answer.