Saturday, August 27, 2011

Do the Primes go to Infinity and Beyond?!!

We've been doing a lot of work with patterns, or more formally called sequences or infinite series. Infinite series are a little different though, as they are guaranteed to go on forever without stopping. For instance, the sequence with formula n^2, or 1, 4, 9, 16..., is an infinite series. However, one like √-n + 3 is not infinite, as it will go √2, 1, 0, and then will be hitting complex numbers, which are not valid for these sequences.

Most patterns, it is pretty obvious, and would never be on a test, or even a valid question for a teacher to ask. However, what about the prime numbers. If you don't know, primes are numbers with only two factors, one and itself. So, 5 is prime, because its factors are 1 and 5. However, 6 is not prime, because it has more than two factors, namely 1, 2, 3, and 6. That means that 6 is composite, which is having three or more factors. Numbers such as one are known as universal, as they have only one factor.

Anyways, are prime numbers infinite? This is definitely a valid question, and answerable too! Can you create a list of all prime numbers on it? How about you try to. I'll bet you can't.

Say someone pops up and says,"I have made a list with all of the prime numbers that exist on it." The list would be much longer if someone did say that, but I made a small list below:


Okay. Let's multiply all of these numbers you've found together. 2 x 3 x 5 x 7 x 11 = 2310. Great. According to you, this is the product of all of the prime numbers out there. Try adding one. Now, we have 2311, which is a multiple of no prime numbers. Since every number is composed of primes, or is prime, this is not possible. That means the number is either prime, as 2311 happens to be, or could be a multiple of another prime that is not present on the list.

Is that all of the primes? No! We can do that process forever, and always find a prime that is missing. This is not a formula to generate prime numbers, as the first ten primes multiplied together is 6,469,693,230 which if you add one gives you 6,469,693,231, which you have no clue if it is prime or not! You guys can figure that one out. However, it is a cool proof that answers a question that definitely gets you thinking!

Saturday, August 20, 2011

Multiplying Giant Two-Digit Numbers Together with No Effort!!!

In the first week, we learned how to multiply two numbers in the teens together instantly, as if you memorized the answer. To the average person, this is the only way possible to get the answer so quickly! Now, you can take that up a notch, and multiply together the biggest two digit numbers. Say someone says 97 x 94. You can say, "That's easy! It's 9118!"

We will take it step by step here. For the first two digits, you take the bigger number and see how far away it is from 100. In this case, it is three away. Then, you subtract that from the other number. 94 - 3 = 91. There's your first two digits.

For the last two digits, take how far both numbers are from 100. 97 is 3 away and 94 is 6 away. The answer is precisely 3 x 6 = 18. Put them together and you have 9118.

Let's try another one, 95 x 89. 95 is 5 away from 100, and 89 - 5 = 84. Then, 89 is 11 away from 100, so 11 x 5 = 55. Then, we put them together to get 8455.

You are probably wondering why this works, for teens or nineties. Let's start with the teens. Pretend that the problem is (z + a)(z + b) with z being 10. We will leave it as z for the moment.

(z + a)(z + b) = z(z + a + b) + ab

If you factor it out, the z(z + a + b) becomes z^2 + za + zb.

(z + a)(z + b) = z^2 + za + zb + ab

If you FOIL out the (z + a)(z + b), you get:

z^2 + zb + za + ab = z^2 + za + zb + ab

This shows that they are equal. If you think about it, this is what we are doing. Take 17 x 16.

(10 + 7)(10 + 6) = 10(10 + 7 + 6) + (7)(6)

This is what you are actually doing. What about for 89 x 95.

(100 - 5)(100 - 11) = 100(100 - 5 - 11) + (-5)(-11)

I got commented about the fact that 20 wouldn't work for the teen method I described in the first post. Like 20 x 18 wouldn't work. However, this formula directs us to do it as so.

(10 + 10)(10 + 8) = 10(10 + 10 + 8) + (10)(8)

This will give us the 360 as promised. If you can do 2 x 1 multiplication problems in your head, you might want to try things like 32 x 37 with 30 as your z, or 68 x 66 with 70 as your z. If you get really good at that, you could even try doing 448 x 442 with 400 as your z, which makes you add 48 x 42, using 40 or 50 as your z. This would be difficult, but you would get 198016 as an answer.

Problem of the Week Solutions (from July):

b = 16
z = 1
n = 42
p = 116
odds = 58%

s = 53.1 degrees
t = 36.9 degrees
a = 1
b = 8
c = 0
h = -4
k = -16
area = 50.3 sq. cm

Friday, August 19, 2011

The Problem of the Week Day 5: Week of 8/14 - 8/20

Today, we will finish up August's problems. Since the easy problem usually has an order of operations problem and a probability problem, we will have both. Since we did deal with chess this week, the probability problem requires knowledge of chess. If you are not familiar with chess, please do the other problem, because I don't want you to get the wrong answer because of expertise in an area besides mathematics. However, the hard problem has only one way to find the answer, which is something I have not introduced yet.

Easy Problem:

Probability: If a chess player knows nothing about chess, and makes a completely random first move, what are the chances he will do d4 as his first move?

p = ___

Order of Operations: p = (25b + g - a - 3e)/3

p = ___

Hard Problem: Today, I will introduce a great tool in Algebra, the Quadratic Formula. The Quadratic Formula states that if ax^2 + bx + c = 0, then x = (-b ± √(b^2 - 4ac))/2a. So, if x^2 - 5x + 6 = 0, then to find x, you would do:

(-(-5) ± √(5^2 - 4(1)(6)))/2(1)
(5 ± √(25 - 24))/2
(5 ± 1)/2
(5 + 1)/2    OR    (5 - 1)/2
6/2      OR    4/2
3   OR   2

x = 3
x = 2

1) I made a couple of errors on Monday's problem. Please complete these simple calculations to have the correct z. We will call this number y.

y = 4z/5 + (4)(5) + 1

y = ___

2) If Ax = y, what does x equal? Use the explicit formula you created with the quadratic formula to achieve the answer.

Tip: There will be two possibilities. Choose the one that is reasonable. For instance, if you had 3 and -5, 3 would be the correct answer because you cannot cut a pizza with -5 straight lines.

x = ___

Also, let's call your false answer f.

f = ___

Thursday, August 18, 2011

The Problem of the Week Day 4: Week of 8/14 - 8/20

Today, the easy problem will move on and work on some geometry. For the hard problem, we will still be cracking away at the pizza problem!

Easy Problem: You probably know that to find the area of a circle, you square the radius and multiply it by π (3.14...). However, that is a little too easy. So, we will try finding the area of a quarter circle. In order to do that, we will do (πr^2)/4, or the radius squared times pi divided by four. Basically, you've found the area of the circle, and divided it by four.

The number of games it takes for b players will be called g.

What is the area of a quarter circle with radius g? Round to the nearest whole number this time.

a = ___

Hard Problem: You should remember the process behind solving a system from the last few months. If you don't, you will just create a variable with the same coefficients, and subtract the two equations. Continue this until you have isolated a variable. Then, you will substitute to find the rest.

Solve yesterday's system from the pizza problem. What is the explicit formula for this sequence?

Hint: It is quadratic (polynomial of degree two).

If you want, try finding the recursive formula. There is a systematic way to do it, which I'll bet you can figure out! You are solving a system! Remember, the recursive formula is An-1 + d with d being the difference.

Wednesday, August 17, 2011

The Problem of the Week Day 3: Week of 8/14 - 8/20

Today, we will finish up on the chess tournament problem. You should have already noticed a pattern.

Easy Problem: There are two types of formulas you can have in a sequence. One of which is the explicit formula, which is a formula based on the nth term. For instance, the explicit formula for the sequence 2, 4, 6, 8, 10, ... would be An = 2n. If you wanted the 6th tern, you would plug 6 in for n to get 2(6) = 12.

The other type of formula is the recursive formula, which is based on the previous term. In the even number sequence, the recursive formula would be An = An-1 + 2 because it is the previous term plus two.

1) What is the explicit formula for the chess tournament problem?

2) What is the recursive formula for the chess tournament problem?

3) If there were b players in the tournament, how many games would it take to find a winner fairly?

Hint: Use the explicit formula for number three!

Hard Problem: At CTY, we used various strategies to determine explicit formulas. However, I tend to lean towards the method I taught last month, with the systems. Just to remind you how to find the system, you must first find your common differences. For the sequence 2, 4, 6, 8, 10, ..., the differences are in the first row. Therefore, you are dealing with a first-degree, or linear, equation. So, we take our base equation for this: An = mn + b. Then, plug in values for n and An to create the system. For instance, you would first plug 1 in for n and 2 in for An, and have the first equation, m + b = 2. Then, you would create a second one, 2m + b = 4, to get our constants, m = 2 and b = 0. This makes the explicit formula An = 2n + 0 which becomes An = 2n.

1) Find common differences in the Pizza Problem.

2) Create a system of equations.

If you want to get ahead, try solving the system, or even try finding a recursive formula.

Tuesday, August 16, 2011

The Problem of the Week Day 2: Week of 8/14 - 8/20

Today, we will do start our main problems! We will mainly be generating values to work with today.

Easy Problem: If there is a chess tournament with n people in it, how many games must they play to get a winner?

For example, if there are seven players. One player will get a bye (sit out for the round and automatically move on) while the other six play. There will be three winners plus the bye makes four. Then, they will play two games to get the two finalists who will play for the win. So, in the first round, there were three games, then two, then one, giving you a six game tournament.

Find data for the values from 1 - 6.

n 1 2 3 4 5 6 7
An 6

Hard Problem: If you cut a pizza with n straight cuts, what is the maximum pieces of pizza you can get?

If you were to use one straight cut, you would just be able to split the pizza in half, which would make two pieces of pizza.

Find data from 2 - 5 or 6.

n 1 2 3 4 5 6 7
An 2

Hint: The pieces don't have to be equal in size, and shouldn't be equal.

Monday, August 15, 2011

The Problem of the Week Day 1: Week of 8/14 - 8/20

Since I just recently came back from CTY, the problems will mainly be a problem we discussed in class. The easy one will be a problem about a chess tournament, and the hard will be about slicing pizzas. However, we are going to start off with some triangles as always!

Easy Problem: In the last two problems, we worked with the famous Pythagorean Theorem, developed to find the missing side of a right triangle when given two. Just to remind you, the shortest side is labeled a, the medium labeled b, and the longest labeled c. The formula states that a^2 + b^2 = c^2. 

If you have a right triangle with a equalling 26.25 and with c equalling 43.75, what is the value of b? Remember to round to the nearest tenth!

b = ___

You won't need b's value until Wednesday, so hold onto it after you find it. 

Hard Problem: Last month, we used the button on the calculator that turns a sine into its angle. Let me review how it works. 

"You might not have it on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle."

If a right triangle's sides are lengths 27, 29.5, and 40, what is the measurement of the angle opposite to side a? This angle will be referred to later on as z. 

z = ___

You will not need this value until Friday. 

Additional Challenge: At camp, we also had a puzzle called region revenge. It was pretty difficult, but I'd like to share it with you. I will put up the answer in two months.

The problem is basically to create a vertex and see how many shaded areas are in the circle. So, for one vertex, there is one shaded region. Now, make two vertexes and connect them with a line. That makes two shaded areas. Then, you make three, and connect every vertex to one another, making a triangle. There, there would be four shaded areas. Keep doing this for five to seven circles, and look for differences. Can you find an explicit formula? It is not what you think it is.

Hint: The formula is a quartic equation (equation of degree four).

Saturday, August 13, 2011

Fibonacci Day: Adding the Evens

If you look closely, you'll find that today is in fact a Fibonacci day. 13 is the seventh Fibonacci number, so we're in for a treat!

Last month, we added up the Fibonacci numbers. This time, let's add the Fibonacci numbers in the even positions. We'll list some out.

1 = 1
1 + 3 = 4
1 + 3 + 8  = 12
1 + 3 + 8 + 21 = 33

See the pattern? Same thing as last time! Just the odd spots minus one!

1 = 1 = 2 - 1
1 + 3 = 4 = 5 - 1
1 + 3 + 8  = 12 = 13 - 1
1 + 3 + 8 + 21 = 33 = 34 - 1

Why does this work? Let's split each number into the previous two Fibonacci numbers.

1 + 3 + 8 + 21 = 33 = 34 - 1

1 + (1 + 2) + (3 + 5) + (8 + 13) = 34 - 1

Because of the associative property, we can eliminate the parentheses, giving us what we did last time, adding all the Fibonacci numbers. That would always give us a Fibonacci number minus one as well.

If you know something cool about Fibonacci numbers, please let me know!

Saturday, August 6, 2011

CTY Challenge: Calculating pi like in the olden days!

This is the last week of CTY camp. Of course, a few activities deal with π because, who doesn't love pi? We did two activities that were attempts on calculating pi, with very little to use. Of course, we did the old make a circle and divide the circumference by the diameter. However, we also did a variation on the Buffon's Needle experiment.

The problem goes like this: take a ruler and draw a bunch of parallel vertical lines that are two inches apart and a foot or two long. Then, take a two inch needle and toss it onto the grid and determine if it crosses the gridlines or not. What do you think the probability is that the needle will cross the line?

If you do the work, you should get a number around 63.66%. However, try plugging in these values into this equation:

2t/c =

t = total number of tosses
c = total amount of times the needle crosses the line

What do you get? Pi, exactly. Our whole class tried it, and with over 1000 tosses, our class average is 3.35. Pretty good, right?

Our instructor never went over the proof in class, though it involves geometry, calculus, and more advanced statistics. I think it has to do with the center point of the needle, and where it is in relation to the gridlines. If you have a proof, please post it!

Additional Puzzle: A census taker walks up to a house, and records the house number. Then, he knocks on the door and a man answers the door. The census taker asks if anybody else lives with him. The man responds that he lives with his three children. The census taker then asks for their ages. The man responds, "Their ages add up to the number on the door and their product is 36." The census taker then says, "I need one more clue. Is the youngest child a twin?" The man says that the youngest child is not a twin. What are the kids' ages, and what is the number on the door?

Hint: Find all the possible combinations of three numbers multiplied together to get 36 (including ones where the smaller numbers are equal).