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\((1 - 3i)(2 - i)(2 + i + 1 + 3i)\\ =(1 - 3i)(2 - i)(3+4i)\\ =(2-i-6i-3)(3+4i)\\ =(-1-7i)(3+4i)\\ =(-3-4i-21i+28)\)

\(=25-25i\\ \color{blue}=25(1-i)\)

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Let $IJKLMN$ be a hexagon with side lengths $IJ = LM = 3,$ $JK = MN = 3,$ and $KL = NI = 3$. Also, all the interior angles of the hexagon are equal. Find the area of hexagon $IJKLMN$.    

The sides are equal and the angles are equal.    

That means the hexagon is a regular hexagon.    

                                                                                               3 sqrt(3)    

The area of a regular hexagon in terms of its sides is   A  =  ––––––––  x  a²  where "a" is a side.    

                                                                                                     2    

                                                                                       A  =  ( 3 x sqrt(3) x 3² ) / 2       

                                                                                       A  =  ( 27 sqrt(3) ) / 2    

Is that good enough or do you need a number.     

                                                                                       A  =  ( 27 x 1.732 ) / 2     

                                                                                       A  =  23.383    

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A terminal zero is a $0$ that appears at the end of a number. For example, the number $3,800$ has two terminal zeros.   
How many terminal zeroes does $40 \cdot 6 \cdot 75 \cdot 12$ have?    

          40 • 6 • 75 • 12 = 216,000        so, three terminal zeros    

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Find the $4000$th digit following the decimal point in the expansion of $\frac{1}{117}$.    

The 4000^th digit is 5.    

1 / 117 in decimal notation has a repeating phrase. 

1 / 117 = 0.008547     

Let's count 'em up:         0.008547(6^th digit)   008547(12^th digit)   008547(18^th digit)   008547(24^th digit)  • • •       

                                          008547(3990^th digit)   008547(3996^th digit)   0085(4000^th digit)      

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It turns out that there are 12 possible bases b.

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?    

The number 4 has two perfect square divisors, namely 1 and 4.    

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You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $5$ pieces of candy per scoop to the vat, and another machine that can remove exactly $3$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?    

Start the 1st machine and let it cycle 2 times ...  2 x 5 = 10 pieces added    

Start the 2nd machine and let it cycle 3 times ...  3 x 3 =   9 pieces removed    

                                                                                          1 piece remains in vat    

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How many three-digit numbers are equal to five times the sum of their digits?    

100a + 10b + c  =  5 • (a + b + c)    

100a + 10b + c  =  5a + 5b + 5c     

  95a + 5b – 4c  =  0     

                  95a  =  4c – 5b    

Stipulating that "a" cannot be zero; because, if so, the number would be a two-digit number.    

If a = 1 then   95 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9    

If a = 2 then 190 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9    

If a = 3 then 285 is too large, etc., and so on for the rest of the digits that "a" might equal.    

I think this little demonstration shows that there is no solution.    

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We know that P(a) - P(b) is divisible by a - b.  This means if a - b is not a prime number, then P(a) - P(b) is not a prime number.  So by modular arithmetic, we can find a positive integer n such that P(n) is composite.

If P(a_1) and P(a_2) are composite, then P(a_1) - P(a_2) will be divisible by some prime number.  Similarly, if P(a_1) - P(a_2) is divisible by some prime number, and P(a_1) or P(a_2) is composite, then the other one is also composite.

We can then take a_1 = n.  So there exists a positive integer a_2 such that P(a_2) is composite.  But since P(a_2) is composite, we can apply the result above, to find a positive integer a_3 such that P(a_3) is composite.  In this way, we can generate an infinite number of positive integers n such that P(n) is composite.

By the prime number theorem, the nth prime is like n*log n.  This is not a polynomial, so there must be an n such that p(n) is composite.

Let P(x) = x^2.  Then P(1) = 1, p(2) = 4, p(3) = 9, so p(n) is always a perfect square.  You can do this for any polynomial that factors, like P(x) = x^3 and P(x) = x(x + 1).

Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.   

          x^2 + kx - 9x + 25 - 9    ==>>    x² + (kx – 9x) + 16   

          I combined like terms and arranged the expression in standard form  ax² + bx + c    

          Note that (kx – 9x) can be written as (k – 9) • x ... and (k – 9) will be our "b" value    

          To be a square, the "b" coefficient in the middle will be twice the square root of "c"    

          The sqrt(16) is + 4, therefore "b" will be either + 8 or – 8    

          For "b" = + 8    then (k – 9) = + 8    which means k = + 17    

          For "b" = – 8    then (k – 9) = – 8    which means k =   + 1    

                                          the sum of both values of k is    + 18    

          The problem doesn't ask for the binomial(s) but they're (x + 4)² and (x – 4)²        

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The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$
If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?    

The tens digit will always be 2.  Take any number that ends with 5 and square it.  The square will end with 25.  Always.    

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Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 28/9 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?    

(x) + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)  =  (28 / 9)(x+6)    

                                                                 7x + 21  =  (28x + 168) / 9    

                                                          9 • (7x + 21)  =  (28x + 168)    

                                                              63x + 189  =  28x + 168    

                                                                        35x  =  – 21    

                                                                            x  =  – 21 / 35    

                                                      reduces to     x  =  – 3 / 5    

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Ten positive integers (not necessarily distinct) are arranged in a circle.  An integer is colored red if it is equal to the average of its two neighbors.  What is the largest possible number of red numbers?    

"not necessarily distinct" is the key to the answer.  All ten integers can be the same.         

Thus, every integer is the average of the two on either side.  Total red numbers, all ten.        

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Here is the answer: The largest possible value of c is 100.

The distance between the circumcenters of Triangle ABC and Triangle ADC is 6*sqrt(3).

The answer is 4810.

The answer is 25.

The answer is 14.