Triangular Day... times nine plus one

Today is another triangular day! It is July 28th, and 28 is the seventh triangular number.

Before we start, let's remind ourselves of the explicit formula (you plug in n and receive the nth term in the sequence) for triangular numbers. Considering we will be proving patterns, it will definitely come in handy.

T(n) = n(n + 1)/2

We'll get back to that later. In the meanwhile, let's look at this pattern:

Add one to any triangular number multiplied by nine and you have a new triangular number.

This seems a little crazy; is this sequence really that cool? Let's check a few values.

1) 1 x 9 + 1 = 10
2) 3 x 9 + 1 = 28
3) 6 x 9 + 1 = 55
4) 10 x 9 + 1 = 91
5) 15 x 9 + 1 = 136

It is working. Those are all triangular numbers. Even that 136 is in fact sixteen times seventeen over two. But why?

First off, let's look for a pattern in the outcomes.

The first triangular number gives us the fourth
The second gives us the seventh
The third gives us the tenth
The fourth gives us the thirteenth
The fifth gives us the sixteenth

Basically, we are just adding three. To make it more mathematical (or complicated), the formula for the output of this pattern while putting in the nth triangular number is:

(3n + 1)(3n + 2)/2

That is a little confusing, but we aren't looking for the nth triangular number. We are looking for the 3n + 1st triangular number, the output of the pattern, which can be solved by this formula.

However, we are really doing 9 times T(n) plus one, which would look like this:

9n(n + 1)/2 + 1

We can simplify that to:

[9n(n + 1) + 2]/2
(9n^2 + 9n + 2)/2

If we also simplify the formula above, we get:

(3n + 1)(3n + 2)/2
(9n^2 + 6n + 3n + 2)/2
(9n^2 + 9n + 2)/2

They are both equal! And there is your proof.

I normally post the algebraic proof of things like this mainly because it is difficult to put shapes in a blog post. However, triangular numbers also have some very elegant geometric proofs that actually are like puzzles; fitting nine triangles together into another triangle with just one dot missing, or any other pattern you like.

Answer: On June 30th, I posted a problem that we learned at CTY. Let me write it down for you again.

By pure random guessing, what is the probability that you will get this answer correct.
  a. 50%
  b. 25%
  c. 0%
  d. 50%

This is a little confusing to follow, but let's think about it. With four choices, there is a 25% chance that you will get it correct. Therefore, it is obviously b: 25%.

However, what are the odds that you choose a or d? 50%, correct? So if one of those two were an answer, you would have a 50% chance of choosing it meaning those are correct as well.

Yet, this gives you a 75% chance of getting the answer correct. The probability that you will choose 75% is 0%, considering that it is not an option. This makes the correct answer 0%, since there is no way you would actually get the correct answer.

Round and round we go. There is a case for every single answer on the board, meaning all of the answers are correct. Personally, I think c has the best, most in depth case. However, which ever answer you thought when I gave the problem is correct. Good job!

Triangular Day: Huh?

Last year, we had a tradition where every Saturday whose date was a Fibonacci number, the post would be about Fibonacci numbers. Since my blog has been up for a year, I decided that we should change it up a little bit and look at a difference sequence of numbers.

I ended up choosing the triangular numbers: a sequence that is also simple to understand and has some really fascinating things about it. First off, what are they? Here are the first several.

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210

What is special about this sequence? Let's try to figure out a formula for this sequence where you plug in a number and it gives you the number in the sequence in that position. That is known as an explicit formula. We can find that by looking for common differences.

1     3     6     10     15     21
  \   /  \   /   \   /   \   /    \   /
   2      3      4      5       6
     \    /  \    /  \    /   \    /
       1      1      1       1

We found a common difference. You would then create a system to solve for a, b, and c: the variables in the quadratic equation (since it is second differences, the equation puts a number to the second power) that would be the formula. If you do that, you get the formula:

1/2n^2 + 1/2n

That looks a little ugly, but we can make it look nicer by going:

n(n + 1)/2

Okay, that's a cool formula. But that isn't so special, and this sequence obviously is or it wouldn't have it's own name.

Think about the name for a second; the triangular numbers. Let's think about triangles. If you were to make a triangular array (a triangle made up of dots) with one dot per side, it would look like this:

• = 1 dot

What about an array with 2 dots per side?

  •     = 3 dots
•   •

What about an array with 3 dots per side?

    •
  •   •    = 6 dots
•   •   •

How about 4 dots per side?

      •
    •   •
  •   •   •     = 10 dots
•   •   •   •

See the pattern? The number of dots in the triangular arrays make up the numbers of the sequence. Something you could use this formula for is to think if the back row of a bowling lane had 7 pins rather than 4, than how many pins is a strike/spare? Turns out, just plug it into the formula:

7(7 + 1)/2
7(8)/2
56/2
28

There would in fact be 28 pins in this lane.

One last pattern; look at the differences between all of the numbers in the sequence. You have:

2, 3, 4, 5, 6, 7, ...

For the seventh number, you would find it by adding seven to the sixth number. This formula is known as the recursive formula, which is written as:

T(n) = T(n-1) + n

There are so many more cool and easy to understand patterns involving this simple set of numbers that are just as surprising as the Fibonacci numbers. And lucky us, next week is a triangular day too!

June Problem of the Week answers:

Easy:

h = 2
a = 1
z = 4
n = 8
t = 1024
l = 16

Hard:

f = 53.1
g = 6
h = 10
a = 1
b = -21
c = 104
x1 = 8
x2 = 13
n = 6725
d = 82
z = 5.4

If you haven't already, make sure to do July's problem of the week.

Problem of the Week Day 5: Week of 7/16/12 - 7/20/12

Today is the final day of the problem of the week! Remember to email me your answers at Ethan@EthanMath.com. Don't forget that you must substitute in your answers from previous days. The answer will never contain a variable in it.

Easy: As usual, I like to finish off the week with a little geometry. Pretend we have a circle. It's circumference is below:

C = n√(π)

From that fact, determine the area A of the circle.

Hint: The formulas for circumference and area are below. The letter r stands for the radius, which is found in both formulas.

C = 2πr
A = πr^2

Hard: I have two shapes: a square and a circle. Their areas are below:

A(square) = s^2/2
A(circle) = πx

Figure out which shape is bigger. Then, find the perimeter of that shape.

P =

Hint: The circle's formulas are in the easy problem. The square's are (with s standing for side):

A = s^2
P = 4s

It may make it more complicated, but you can figure out which shape is bigger without multiplying the π and the x together. If you want to figure it out that way, determine the diameter of the circle and the diagonal length of the square (which is simple using the Pythagorean Theorem). Simple logic from there will determine their size.

Problem of the Week Day 4: Week of 7/16/12 - 7/20/12

Today is day 4 of the problem of the week. Good luck!

Easy: Take the following sequence:

6, m, n, e, h, ...

You may notice a letter in there that you haven't seen yet: n. I want you to determine the value of n.

Hint: Look for common differences.

Hard: Today's part will take a little longer than usual, but shouldn't be too bad.

First, figure out the following equations:

a = 10z/t
b = -t(9s - 7t - 10)/z
c = (t ÷ z ÷ s)^-8

a =
b =
c =

Remember the order of operations (which are listed on yesterday's post).

You might be wondering why I chose a, b, and c. It is because you then have to solve for x in this equation:

0 = ax^2 + bx + c.

x =

Hint: You can use any of the methods to solve quadratic equations you wish. The most straight forward is the quadratic formula which is below.

x = [-b ± √(b^2 - 4ac)]/2a

Problem of the Week Day 3: Week of 7/18/12 - 7/22/12

Today is day three of the problem of the week. Good luck!

Easy: For today, I want you to determine what m equals just using the two previous day's answers h and e.

m = 2(h - e)^[(h - e)/2] ÷ (h - e)^[(h + e)/(h - 3)]
m =

Hint: Remember to use the order of operations:

1. Parentheses/Brackets
2. Exponents/Radicals
3. Multiplication/Division
4. Addition/Subtraction

Hard: Take the following sequence:

t, z, z^2/t, ...

If you let this sequence go on all the way to infinity terms and added them ALL up, what would be the sum s?

s =

Hint: The answer is not infinity.

Problem of the Week Day 2: Week of 7/16/12 - 7/20/12

Today is day two of the problem of the week. Good luck!

Easy: Pretend you have a right triangle with sides 18, e, and h with h being the longest side. Determine what h equals.

h =

Hint: Use the Pythagorean Theorem; a^2 + b^2 = c^2

Hard: Pretend you have a right triangle with angles 90, t, and 90-t. You also have a hypotenuse of 1/5√(10t). Determine the area z of this triangle.

z =

Hint: The area formula for a triangle is (base x height)/2

Problem of the Week Day 1: Week of 7/16/12 - 7/20/12

Today is the first day of July's Problem of the Week. Rather than giving a thorough explanation of each problem, I will just give a hint. The hint may be the formulas you need or something to help you achieve the answer.

Remember to write down the answer you receive after each day so that you can plug it in for the next day's problem.

Easy: Normally, I begin the week with triangles, but I thought that I would change it up this time and start with a puzzle.

If a goose and a half can lay an egg and a half in a day and a half, then how many eggs e can a half a gross of geese lay in a half a day?

e =

Hint: a gross is a dozen dozens.

Hard: Again, I am not starting the week with triangles, but a puzzle instead. This puzzle is much harder.

You are stuck out in the wilderness and the only thing that you can possibly eat that is anywhere near you is a plant. The plant is poisonous, but it takes exactly 15 minutes over a flame to kill the poison. However, it takes exactly 15 minutes over the flame to activate another poison in the plant. So, if you cook it for exactly 15 minutes, it is safe to eat, but just a little more or a little less and you will die from the poison.

You don't have any timers, watches, phones, or clocks. All you have is two ropes that take exactly an hour to burn when the end is lit. However, it may not burn equally all the way through (a quarter of the rope probably won't burn in exactly 15 minutes).

Using just these two ropes and the fire, you can cook the plant for exactly 15 minutes. Once you figure out how (which is the hard part), tell me how long it will take to actually get your food.

t =

Hint: first, figure out the problem if the plant took exactly 30 minutes to become harmless.

CTY: My Envelope Has More Than Your Envelope

This is my last week at CTY. In class, we looked at this problem:

Someone hands you an envelope and asks you to look at the contents (let's say it is some dollar amount). They then say that they are holding a second envelope that has either double of yours or half of yours. You must decide whether to switch envelopes or keep the one you currently have.

To solve this mathematically, you would use the expected value formula we used last week. Say your envelope has X dollars. You can either:

EV (Stay) = X
EV (Switch) = 1/2(1/2X) + 1/2(2X) = 5/4X

In other words, you will make 25% more money on average by switching. However, that is not realistic in these situations.

You must examine the problem from a logical standpoint. If you have a very generous person offering this deal and showing an envelope with $10, it might be worth the gamble to switch. If a more conservative person offered the same deal, you would be better off staying.  I found this problem cool because once again, math failed to give a reliable answer.

Bonus: Here is another problem to solve. I will give the answer in a month.

There are a group of monks who all vowed not to communicate to one another in any way (speaking, codes, sings, etc.). Every morning, the monks all gather in a circle and the head monk speaks to them.  One morning, the head monk said that there were sinners among them. He waved his hand and a mark appeared on all of the foreheads of the sinners (everyone knew who had marks, but could not see if they had one). He asked for anyone who knew they were a sinner to leave.

The second morning, the head monk announced that there were still sinners left and put the marks back. Once again, he asked for anyone who knew they were a sinner to leave.

The third morning, the head monk announced that there were still sinners left and put the marks back. For a third time, he asked for anyone who knew they were a sinner to leave.

The fourth morning, the head monk announced that all of the sinners had left.  How did the sinners know to leave?

Answer: A month ago, I posted the locker problem where student one opens every locker, student two closes every even locker, student three opens/closes every third locker, and so on up through 1000 students. The question was which lockers remain open.

The answers are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961

There is a pattern among these 31 numbers: they are all square. The reason for this is that every non-square number doesn't just have factors, but they have pairs of factors. For instance, six has factors:

1 x 6
6 x 1
3 x 2
2 x 3

The six and the one can be written twice, giving it two factors. Therefore, six will get opened by 1, closed by 2, opened by 3, and closed by 6. However, a number like 9 has factors:

1 x 9
9 x 1
3 x 3

Here, the three cannot be written twice since it is paired with itself. So, locker nine will get opened by 1, closed by 3, and opened by 9. The same logic applies to all of the lockers.

CTY: Are Game Theory and Economics really that similar?

This is my second week at CTY studying Game Theory and Economics. For today, I decided to show a combination between the two.  Let's say two companies are producing Q goods overall, q1 from the first company and q2 from the second. The equation for their relation of production is: Pq1 = 200q1 - 2Qq1 Let's see what happens if we simplify.  Pq1 = -2q1^2 + (200 - 2q2)q1 This is a quadratic equation, which we have looked at several times. Since the leading coefficient is negative, there is a point that is the highest point on the graph which is the vertex.  To solve for the x-coordinate (q1's production), you would be doing -b/2a with a being the first coefficient and b being the second.  If you do this, you get: -(200 - 2q2)/2(-2) 2q2 - 200/-4 q2 - 100/-2 100 - q2/2 Basically, this number is the best your company can do with this equation. For q2, the same logic applies.  100 - q1/2 To find q1's actual number, we assume that q2 will play rationally and follow this formula. That means we can substitute this formula in for q2 in the q1 problem. This gives us: q1 = 100 - q2/2 q1 = 100 - (100 - q1/2)/2 4q1 = 200 - 100 + q1 3q1 = 100 q1 = 33.3... In other words, q1 and q2 should both always do 33.3... to get the best outcome for them. The game theory is their choosing of the strategy and the economics is its application. I found it cool how this relation is so clear in this problem even though they seem so different.  Bonus: Here is another fun problem that we learned last week.  There is a kingdom ruled by two tyrants: the king and the dragon. They are normally in good terms with each other until now. The townspeople are split between the two, and they decide that the only fair way to decide a leader is to fight to the death.  The dragon suggests a fire breathing contest and the king suggests a juggling contest, but that wasn't fair. The king then proposed this idea: There are ten numbered wells in our kingdom. Each well has a poison in the water that can only be treated by the poison in a higher numbered well. If the poison isn't treated in an hour, they are dead. One day, we will both serve each other a glass of water, and then go our separate ways. After an hour, we will see who is alive.  The dragon does not know where well #10 is and the king does, and the king is to smart to not catch a following dragon. However, the dragon accepts the king's challenge, and after the hour, the king is somehow dead and the dragon is left to rule the kingdom.  How did the dragon do it? As usual, I will post the answer in a month.