A few weeks ago, I blogged about a calculator giveaway at Wild About Math. Since then, Sol has posted a submitted solution here (and here’s the direct link to the pdf solution by Nate Burchell).
Here’s the problem for those who didn’t see it:
One can create a triangle of consecutive positive integers as follows:
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
. . .
Each row, R, has R numbers. Each column, C, has infinitely many numbers. Rows and columns begin at 1. We define a function F(R,C) for row R and column C such that F(R,C) gives us a value in the triangle. Thus, F(1,1) = 1, F(2,1) = 2, and F(2,2) = 3. Note that F(R,C) is only defined when 1 < = C <= R.
Part 1: Come up with a formula that computes F(R,C) in terms of R and C for any positive values of R and C when 1 < = C <= R. Show your work.
Part 2: Come up with a formula or algorithm that, given a positive integer n, determines R and C.
I also solved the problem and submitted a solution but I didn’t win. Here’s my own solution.
Everyone should go enter this giveaway @ the Wild About Math blog. In order to qualify, though, you have to submit a solution to the problem Sol created for us. I just submitted mine!
A few weeks ago I posted this “powerful” problem:
Now, allow me to give you a major hint. Consider the simpler equation
What are the possible values of and ? Here are the possible combinations:
- and is anything
- and is any nonzero number
And here’s the tricky one that most people forget:
- and is even
You now have enough information to solve the original equation. I think you’ll be delighted with the solution!
I love this problem. I love it because it seems so complicated at first, just because we don’t teach students how to attack problems like this in Algebra class. There aren’t any “traditional” methods of attacking it, just a little mathematical reasoning/logic. Here it is:
And this is my new “super duper” problem which I post throughout the year on my board (I use a lot of the same problems each year). I first saw this problem at Messiah College where one of my professors shared it–either Dr. Phillippy or Dr. Brubaker, I can’t remember which.
So give it a try. It’s sure to delight you. My Precalculus class was sharp enough to solve it today in one period (albeit, while I was teaching about a completely different topic :-)).
If the cube has a volume of 64, what is the area of the green parallelogram? (Assume points I and J are midpoints.)
Go ahead, work it out. Then, go here for a more in depth discussion, including a video explanation. Also, see my very simple solution in the comments on that page. (My Precalculus students should especially take note!)
And, welcome, SAT Math Blog, to the internet! Thanks for pointing us to this great problem and creating the nice diagram above.
I posted a problem back in December that I never got back to answering. Sorry about that. The problem statement was:
Two jars contain an equal volume of soda. One contains Sprite, the other Coca Cola. You take a small amount of Coca Cola from the Coca Cola jar and add it to the Sprite jar. After uniformly mixing this concoction, you take a small amount out and put it back in the Coca Cola jar, restoring both jars to their original volumes. After having done this, is there more Coca Cola in the Sprite jar or more Sprite in the Coca Cola jar? Or, are they equally contaminated?
I have had the worked out solution for a while, just haven’t posted it until now. I’m relatively new with , but I’ve typed up the solution here, if you want all the gory details :-). And yes, Peekay, you got the right answer!