How many positive 3-digit integers from have an odd number of divisors?

Transcription

1 Jan. 4, 2006

2 How many positive 3-digit integers from have an odd number of divisors?

3 Why is it true that only perfect squares have an odd number of divisors? If a number has an odd number of divisors, then all the exponents in its prime factorization must be even (since we get our number of divisors by adding 1 to all the exponents, then taking the product - the only way this will end up with an odd number is if all the exponents were even to begin with)

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5 How many circles in the plane have a radius of 1 unit and pass through at least two of the four points (2,1); (2,2); (3,1); (3,2)?

6 First thing you probably want to do here is sketch it. How can you count the possible circles? We can count the number of ways we can pick two of the points for our circle to go through: there are C(4,2) = 6. But, what if all of them weren t one apart? Our circles have diameter 2, so each pair of points can have a circle with radius 1 through them.

7 So what s our answer? 6? For each pair of points, there are two circles with radius 1 through that pass through both points. (There's one circle on either side.). We have 6 pairs of points, and 2 circles that go through each pair. That gives us 12 circles.

8 Summary There are C(4,2) = 6 pairs of points. You may also look at this as: We can choose the first point in 4 ways, the second in 3 ways, but this counts each pair of points twice. Therefore, there are 4x3/2 = 6 pairs of points. The key here is to find an organized way to do your counting. You'll often see these kind of problems on MATHCOUNTS - problems in which you can just count by hand, but you might miss a case or two. On these problems, try to find an organized way to do your counting; then you can be more confident that you haven't missed any cases.

9 The surface area of the cube shown is 216. What is the area of triangle ABC? This is supposed to be countdown.

10 The length of the altitude is s*sqrt(3)/2. The area is (base)(height)/2 = s^2 * sqrt(3) / 4, which in this problem is 18*sqrt(3).

11 Summary for 3D Problems 3D geometry problems - when 3D problems deal with lengths, areas, angles (as opposed to volumes), they are usually 2D problems in disguise. The way we usually attack them is to consider the appropriately chosen 2D parts of the problem. In this case, we considered first the faces of the cube to find the lengths AB,BC,AC. Then we considered just the triangle ABC. Each of these steps are 2D, not 3D, problems.

12 Start with the number , in which the pattern continues for 1000 total digits. From the left, we take every fourth digit in order to form a new number. We keep doing this process over and over until there are only 3 digits left. What is that last 3-digit number? Play with the problem. What happens when we go through the numbers the first time?

13 You get There are 250 numbers in the list. Hmm Then what? Do it again But how many numbers are in this chain? 250/4 = 62.5 There s only 62 numbers (round down) because there's no 252nd number from which we'd get a 63rd number of our new list.

14 Next round of cuts leave This has 15 numbers, so we are at our final cut to

15 Summary Some problems you just have to play with a little & look for a pattern you can follow to the end. Sometimes these patterns have simple explanations (and we'll see plenty of these), but sometimes they don't. (Or at least, you won't have time enough the completely figure out why the pattern works during the test.) When you see a method that will work and won't take you too much time, go ahead with it and solve the problem.

16 Countdown Round What is the area of the rhombus in the diagram?

17 The area of a rhombus can be solved by: ½ the product of the diagonals base * height Which is easier here?

18 Base * Height How do we find the base? Use triangle, and remember, all sides of a rhombus are equal.

19 Solution and Summary 12 * 24 = 288 When we go hunting for lengths, this is often a very effective strategy - draw a perpendicular in a strategic position. We do this to create right triangles - right triangles allow us to use the Pythagorean Theorem and to use the special right triangles we know a lot about ( & triangles). This can be especially useful with trapezoids.