KnockOut

Off Topic/Spam. Thanks!

Okay.

Let A, B, C, D, and E stand for the five letters of the Mumblian alphabet.

Each word can be from 1-3 letters long, and letters can repeat.

Okay. Lets start with the words with 1 letter in them.

There are \(\boxed {5}\) possibilites of this.

Next up, the words with 2 letters in them.

There are \(5 * 5 = \boxed {25}\) possible 2 letter words.

Last, the words with 3 letters in them

There are \(5 * 5 * 5 = \boxed {125}\) possible 3 letter words.

There are a total of \(5 + 25 + 125 = \boxed{\boxed{155}}\) total words possible.

Nice!

The Greatest Common Factor

The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two numbers:

List the prime factors of each number.

Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1.

The Least Common Multiple

A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....

The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

I typed GCF(34,784) into my calculator, and got \(\boxed 2\) as my GCF.

I did the same for the LCM and got  \(\boxed{13328}\).

http://www.mathematicsmagazine.com/applications/GCF_LCM.htm

You can use this to find both of them if you do not have a calculator that can calculate GCF and LCM.

And... if you just want a written out solution...

Suppose the six cookies to be chosen are the stars, as we attempt to implement stars and bars.

We take two dividers, and place them between the cookies, such that the six cookies are split into 3 groups, where the groups are the number of chocolate chip, oatmeal, and peanut butter cookies, and each group can have 0.

First, assume that the two dividers cannot go in between the same two cookies.

By stars and bars, there are \(\binom{7}{2}\) ways to make the groups.

Finally, since the two dividers can be together, we must add those cases where the two dividers are in the same space between cookies. There are 7 spaces, and hence 7 cases.

Our final answer is\(21 + 7 = \boxed{28}\) assortments.

Precise solution right here

https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_21

This question is from an AMC 10A math competition. I like those...they're fun.

So the question is:

Solve the quartic equation- \(2x^4 + 14x^3 + 6x^2 + 14x + 2 = 0\)

The figure below can be helpful in understanding the solution.

https://i.stack.imgur.com/2SopB.png

We are given \(\angle ACD = \angle EAC= 90^{\circ}\), so \(AE\) is parallel to \(CD\), so \(\angle EAB = \angle BDC\).

We also have \(\angle EBA= \angle CBD = 90^{\circ}\), so that \(\angle AEB = \angle BCD\).

So angles of triangles \(ABE\) and \(BCE\) are the same, hence they are similar.