GYanggg

\(1+2+3+\cdots+198+199+200=?\)

a. 

First number: 1, 

Last number 200. 

\(1+200=\boxed{201}\)

b. 

Second number: 2, 

Second to last number: 199.

\(2+199=\boxed{201}\)

c. 

Third number: 3, 

Third to last number: 198.

\(3+198=\boxed{201}\)

d. 

When you pair up all the numbers as suggested above, each pair adds to 201. 

Since there are 2 numbers in each pair, and there are 200 numbers, there are 200 ÷ 2 = 100 pairs. 

e. 

Since each pair sums to 201, and there are 100 pairs, \(1+2+3+\cdots+198+199+200=201\cdot100=\boxed{20100}\)

The shaded figure shown in the image is a parallelogram. 

The area of a parallelogram is base x height. 

In this image, the base is 12 and the height is 6. 

6 * 12 = 72, this is the area. 

I hope this helped,

Gavin.

I oppose questioners having answerers "private message" them the answers. This is because of three reasons:

1. 

How are we supposed to know whether or not someone has already answered the question, or are you just expecting 20 people to "private message" you the answer. It's a waste of time for all the effort in composing the answer to satisfy your uncanny request. 

2. 

"Private messaging" you do answer prevents other users from viewing the solution. This way, other people cannot learn, just to satisfy your uncanny request.

3. 

To my knowledge, you do not receive points from sending private messages. Not only you are preventing others from learning, and causing confusion for the answerers, you are completely screwing up the point system. 

1. Since the balls are indistinguishable, we just need to keep track of how many balls in each box. 

The possible arrangements are:

\((4, 0, 0); (3, 1, 0); (2, 2, 0); (2, 1, 1) \Rightarrow \boxed{4}\)

2. We start by splitting the numbers into three cases:

The one digit numbers don't have any zeroes. 

The two digit numbers use two zeroes: 10 and 20. 

There are \(3^2 = 9\) three-digit numbers starting with 1, and 9 starting with 2. For each leading digit, a zero appears in each digit in \(9\div3 = 3\) of the numbers, so each has a total of \(3 + 3 = 6\) zeroes. Thus, the 3-digit numbers contain \(2\cdot6 = 12 \) zeroes. 

3. Here is the 3 by 3 grid:

For the first column, we have 3! = 6 arrangements of red, green, and blue. 

For the second column, we have only \(\binom21\) arrangements of red, green, and blue. 

For the last column, we have only 1 arrangements of red, green, and blue. 

For a total of 6 * 2 = 12 arrangements. 

I hope this helped,

Gavin.

From 20 to 69, they are a total of 50 integers. 

To choose 5 of them, we do: \(\binom{50}{5}\)

Now, we need to find out the number of successful outcomes. 

We need an integer between 20 and 29 

Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. 

There are 10 numbers in each category, and therefore \(10^5\) ways of doing this. 

Your final answer is: \(\boxed{\frac{2500}{52969}}\)

I hope this helped,

Gavin

Yup!

I totally agree with you, that's why I put together this list of all the patterns hidden inside the Pascal's  triange. 

I hope you enjoyed.

Hey CPhill!

I don't think your answer accounts for this part of the question: 

After it is rolled, 5 of the 6 faces of the cube are visible, with the face on the bottom unseen. What is the probability that exactly one of the five faces is painted.

In your case there isn't the rolling process, hence the "5 of the 6 faces of the cube are visible, with the face on the bottom unseen" part was neglected. 

I'm still trying to find out why you also got the same answer. 

I might have missed something about your solution and I'm the one at fault, and you solution is actually correct. 

We first need to determine how the 27 unit cubes are painted. 

Cases:

Case 1: We select a cube with 0 painted sides

It is not possible to see a painted side, since none are painted. 

Case 2: We select a cube with 1 painted sides

The roll is successful as long as an unpainted side is on the bottom. 

Case 3: We select a cube with 2 painted sides

The roll is successful as long as a painted side is on the bottom. 

Case 4: We select a cube with 3 painted sides

It is not possible to see only one painted side, since only on can be hidden, but the other two cannot. Therefore, we will see more than one painted side. 

Probabilty:

We only need to consider cases 1 and 2. 

Case 2:

6 sides to a cube, 5 unpainted and 1 painted. 

We need an unpainted side on the bottom, 5/6. 

Case 3: 

We need a painted side of the bottom, 2/6. 

\(\frac{12}{27}\cdot\frac26+\frac{6}{27}\cdot\frac56=\boxed{\frac13}\)

I hope this helped,

Gavin

Hey Rollingblade!

There are myriads of patterns to the Pascal's triangle. Here is the basic Pascal's triangle. To begin, let's just look at the numbers is each row. 

We can already see that the sum of the numbers in each row is a power of 2. 

\(1=2^0\\ 1+1=2^1\\ 1+2+1=2^2\\ 1+3+3+1=2^3\\ ...\)

Futhermore, just looking at the digits, they are powers of 11!

\(1=11^0\\ 11=11^1\\ 121=11^2\\ 1331=11^3\\ 14641=11^4\\\)

However, when we get to the next row, we need to advance the ten's digit. 

\(15101051\Rightarrow161051=11^5\)

Then when we look down the diagonal's, from the top, we see:

- The first row is all ones: 1, 1, 1, ...

- The second row is counting numbers: 1, 2, 3, ...

- The third row are triangular numbers: 1, 3, 6, ...

- The fourth row are tetrahedral numbers: 1, 4, 10, ... 

From the diagram above, we see the Fibonacci Numbers as the sum of each row. 

Some more patterns: 

-The numbers in the triangle can be represented by combination. 

-If you shade all the odd numbers, you see a beautiful fractal. 

These are just some patterns, you can see all of them on this amazing website: https://www.cut-the-knot.org/arithmetic/combinatorics/PascalTriangleProperties.shtml. 

The Pascal's triangle was one of the most beautiful mathematical creations ever made. There are so many arrangments and identities derived from a simple pattern. 

I hope this helped,

Gavin. 

\(\sqrt{\frac{1}{45}+\frac{1}{36}}\\ =\sqrt{\frac{4}{180}+\frac{5}{180}}\\ =\sqrt{\frac{9}{180}}\\ =\sqrt{\frac1{20}}\\ =\sqrt{0.05} \)

I'm pretty sure you can do the latter on your calculator. 

I hope this helped,

Gavin.