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In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?    

In order not to have two yellows next to each other,    

the yellow color must occupy squares 1, 3, and 5.   

That leaves squares 2 and 4 to fill, and two colors    

to do it with. There are only four ways.   

Blue Blue     Red Red     Blue Red     Red Blue    

So, the answer is four ways.    

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I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row?    

7 • 6 • 5 • 4 • 3 • 2 • 1  =  5040    

Seven books total, so you have 7 choices for the first book.    

For each of those choices, you have 6 choices for the second book.   

There are 5 choices left for the third book in the row.    

et cetera until all of the books are shelved.    

The expression for the arrangements can be shown as 7! (called seven factorial).   

Another way to express the number of arrangements is ₇P₇ (called seven permute seven).    

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The number $100$ has four perfect square divisors, namely $1,$ $4,$ $25,$ and $100.$    
What is the smallest positive integer that has exactly $2$ perfect square divisors?    

That would be 4.  Its two perfect square divisors are 1 and 4. 

We ignore 2 as a divisor of 4 because it isn't a perfect square,    

just as the example given in the problem ignored 10 and 20, etc.    

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For a positive integer $n$, $\phi(n)$ denotes the number of positive integers less than or equal to $n$ that are relatively prime to $n$.   
What is $\phi(191)$?    

191 is a prime number.  Therefore, it is relatively prime to every integer less than itself.    

You don't count 1, so there are 189 integers (i.e., 2 through 190) relatively prime to 191.    

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Now, let's me explain my tactic. Hang with me, it's quite tedious. 

Now first off, let's set a multivariable function to deal with this question, 

First, let's let \(f(a,b,c) = (ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3\). (This will come into handy later)

This following step is not necessary, as you can just easily follow through, but expanding everything, which is quite tedious, we get

\(3a^3b^2c+3a^2b^3c+3a^3bc^2+3a^2bc^3+3ab^3c^2+3ab^2c^3+6a^2b^2c^2\)

Now, let's note that every term is divisble by 3abc. Factoring this out, we get

\(f(a, b, c,)=3abc(b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2)\)

The following steps are complicated, but I do believe this was explained in an AOPS course, which is where this problem came from.

(I know because I had the same problem while taking the course myself, it was a writing problem)

Let's note the following ideas. 

If we let \(c=-a\), we find that \(f(a,b,-a) =0\), meaning that a+c is a factor of \(f(a,b,c)\) 

If we let \(-a = b\), we also find that \(f(a, -a, c) = 0\). This also means that a+b is a factor. 

If we let \(b = -c\),  \(f(a, -c, c) = 0\). This also means that b+c is a factor. 

This also means that \((a+b)(b+c)(c+a)\) should be in the final factorization. 

Wait a sec! Notice something really quickly!

\((a+b)(b+c)(c+a) = b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2\), which means that we can replace what we had in parethensis in the original facorization with \((a+b)(b+c)(c+a)\)

Thus, our final factorization is \(3abc(a+b)(b+c)(c+a)\)

Thus, our final answer is \(3abc(a+b)(b+c)(c+a)\)

Thanks! :)

Let's graph (yay!)

What's interesting about this is this is on a coordinate plane, so we can make quite a few deductions from it. 

First off, we have that

\(CD = BD = 10 \\ ED = ED\)

Also though, note that ED and BC are just the x and y axis, thus they are perpendicular lines. 

Which means that we have \(\angle CDE = \angle BDE\)

By SAS Congruence, we know that triangle CDE is congruent to triangle BDE.

Thus, we can conclude that corresponding sides \(CE = BE\)

Finally, we just simply right that

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

Thus, 10.35 SHOULD be correct, not sure, trig is never my thing :(

Thanks! :)

We know that AD bisects BC. Thus, we can set variables to get some common ground here. 

Let's let \(BD  = 3 - x\) and let \(CD =  x\)

There's a couple things we can do, but let's the Angle bisector thereom to get

\(BD / AB = CD / AC\)

We already set some variables, so subsituting in what we know, we get

\((3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

Alright, so now we know what CD is. 

Since the area of ABC can be found with CD and AB, we simply have

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

Thus, 10/3 is our final answer. 

Thanks! :)

*1900

The number of diagonals in a certain regular polygon is equal to $2$ times the number of sides. How many sides does this polygon have?    

The number of diagonals in a regular polygon with "n" sides is calculated using the formula: n(n-3) / 2.    

             from the formula                                d  = n • (n – 3) / 2    

             given                                                  d  =  2 • n    

                                                                      2n  =  n • (n – 3) / 2   

                                                                      2n  =  (n² – 3n) / 2     

                                                                      4n  =  n² – 3n    

                                                                        n²  =  7n    

                                                                         n  =  7    

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The answer on that page (by CPhill) is wrong however: 

Ball splits             Ways to distribute the balls

7  1                                  2

6  2                                  2

5  3                                  2

4  4                                  1

2 * 2 * 2 * 1  =  8 ways

but you cannot multiply as these are separate cases; you add them to get 7.

Then the answer would be 75.

There is also another way (stars and bars):

Choose 3 boxes to be empty: \({5 \choose 3}=10\)

Use stars and bars allowing empty boxes to calculate (or just count like CPhill)

CPhills method if he was here:

Ball splits

0 8

1 7

2 6

3 5

4 4

5 3

6 2

7 1

8 0

to get 9 different cases (stars and bars would be \({n+k-1 \choose k-1}={8+2-1\choose 2-1}={9\choose 1}=9\) ways)

Multiply by 10 to get 90 ways. However, this overcounts (0 0 0 8 0 is counted four times, and so are all other permutations) so we subtract 5*3=15 from 90 to get our final answer, 75.

Find the number of bases $b \ge 2$ such that $100_b - 1_b$ is prime.    

The third position to the left is the base raised to the second power position.    

So this problem could be viewed as "For which bases "b" can (b² – 1) be prime?"    

The only base I could find for which this works is 2 ... i.e., 2² – 1 = 3.    

Therefore, the answer to the problem is:  there is only one.    

_.   

The answer is f(17) = -21.

630 divided by 42 = 15     the other two integers, when multiplied = 15    so  either they are   3 and 5   or   1 and 15