Answers and Solutions

1. 1. \((-5) \times 4 = -20\)
   2. \((-12) \times (-3) = 36\)
   3. \(7 \times (-8) = -56\)
   4. 
      \( \begin{align*} (-6) \times (+3) \times 2 &= (-18) \times 2 \\ &= -36 \end{align*} \)
   5. 
      \( \begin{align*} 8 \times \Big( (-9) + 3 \Big) &= 8 \times (-6) \\ &= -48 \end{align*} \)
2. 1. The missing integer is \(6\).
      Since \((-) \times (+) = (-)\), we know the missing integer is positive.
      We also know that \(4 \times 6 = 24\).
      Putting this all together, the missing integer is \(6\), and we have that \((-4) \times 6 = -24\).
   2. The missing integer is \(-8\).
      Since \((-) \times (+) = (-)\), we know the missing integer is negative.
      We also know that \(8 \times 7 = 56\).
      Putting this all together, the missing integer is \(-8\), and we have that \((-8) \times 7 = -56\).
   3. The missing integer is \(-9\).
      Since \((-) \times (-) = (+)\), we know the missing integer is negative.
      We also know that \(9 \times 6 = 54\).
      Putting this all together, the missing integer is \(-9\), and we have that \((-9) \times (-6) = 54\).
3. We start by considering the factor pairs of \(20\), which are:\[1,~20 \qquad\qquad 2,~10 \qquad\qquad 4,~5\]In order for two integers to multiply to \(-20\), one integer must be positive and the other negative. Here are the possible pairs:\[1,-20 \quad\quad\quad\quad 2,-10 \quad\quad\quad\quad  4,-5\]\[-1,20 \quad\quad\quad\quad -2,10 \quad\quad\quad\quad  -4,5\] Notice that there is more than one possible answer.
4. The product of two negative integers is always positive and the sum of two negative integers is always negative. As a result, the product will always be greater than the sum.
   We can use these two examples to help convince ourselves that the product of the negative integers is always going to be greater than the sum. 
   \( \begin{align*} \text{Integers:} & \quad -4, -3 \\ \text{Product:} & \quad 12 \\ \text{Sum:} & \quad -7 \end{align*} \)                  \( \begin{align*} \text{Integers:} & \quad -8, -5 \\ \text{Product:} & \quad 40 \\ \text{Sum:} & \quad -13 \end{align*} \)
   Notice, in our examples the product of two negative integers is positive and the sum of two negative integers is negative.  This result is true for any pair of negative integers, because
   \((-) \times (-) = (+)\qquad\)and \(\qquad(-) + (-) = (-)\)
   Since a positive integer is always greater than a negative integer, the product of two negative integers will always be greater than their sum.
5. There are two possible solutions that meet the conditions.
   
6. 1. \(-8 \) and \(3\); since \((-8) + 3 = -5\) and \((-8) \times 3 = -24\)
   2. \(-3\) and \(-4\); since \((-3) + (-4) = -7\) and \((-3)\times (-4) = 12\)
   3. \( -1 \) and \(5\); since \((-1) + 5 = 4\) and \((-1) \times 5 = -5\)
   4. \(2\) and \(-5\); since  \(2 - (-5) = 7\) and \(2 \times (-5) = -10\)
7. 1. \(38 \times 80 + (60-38) \times (-40)\)
      Evaluating the expression, we get
      \( \begin{align*} 38 \times 80 + (60-38) \times (-40) &= 38 \times 80 + 22 \times (-40) \\ &= 3040 + 22 \times (-40) \\ &= 3040 + (-880) \\ &= 2160 \end{align*} \)
      If the hotel has \(38\) rooms occupied, then their net profit is \($2160\).
   2. \(18 \times 80 + (60-18) \times (-40)\)
      Evaluating the expression, we get
      \( \begin{align*} 18 \times 80 + (60-18) \times (-40) &= 18 \times 80 + 42 \times (-40) \\ &= 1440 + 42 \times (-40) \\ &= 1440 + (-1680) \\ &= -240 \end{align*} \)
      If the hotel has \(18\) rooms occupied, then their net profit is \(-$240\), which means the hotel loses \($240\).
   3. \((60-10) \times 80 + 10 \times (-40)\)
      Evaluating the expression, we get
      \( \begin{align*} (60-10) \times 80 + 10 \times (-40) &= 50 \times 80 + 10 \times (-40) \\ &= 4000 + 10 \times (-40) \\ &= 4000 + (-400) \\ &= 3600 \end{align*} \)
      If the hotel has \(10\) rooms vacant, then their net profit is \($3600\).
   4. Solution 1
      From parts a) and b), we know o have a net profit of \(0\), the hotel must have greater than \(18\) rooms occupied and less than \(38\) rooms occupied.  Furthermore, from part b), this number will be closer to \(18\). We can find the exact number of rooms through trial and error.

      Number of Rooms Occupied	Net Profit
      \(19\)	\(-$120\)
      \(20\)	\($0\)

      Therefore, to have a net profit of \(0\), the hotel must have \(20\) rooms occupied.
      
      Solution 2
      Note that \(1 \times 80 + 2 \times (- 40) = 0\). So if exactly \(1\) out of every three rooms is occupied, the net profit will be \($0\). Since there are \(60\) rooms in the hotel. \(\dfrac{60}{3} = 20\) rooms must be occupied to have a net profit of exactly \($0\).

8. 1. The maximum score is a result of all \(5\) questions being answered correctly.
      \( \begin{align*} \text{Total Score} &= 3 \times 5 + (-2) \times 0 + 0 \times 0 \\ &= 15 \end{align*} \)
      The maximum score is \(15\).
   2. The minimum score is a result of all \(5\) questions being answered incorrectly.
      \( \begin{align*} \text{Total Score} &= 3 \times 0 + (-2) \times 5 + 0 \times 0 \\ &= -10 \end{align*} \)
      The minimum score is \(-10\).
   3. A student scoring \(1\) question correct, \(0\) questions incorrect, and \(4\) questions not answered produces a score of \(3\).
   4. A student scoring \(1\) question correct, \(3\) questions incorrect, and \(1\) question not answered produces a score of \(-3\).
9. 1. \(0\).
   2. First let's examine the two expressions. The number \((-n)\) is the opposite of \(n\) and \((-n)^{2} = (-n)\times (-n)\).  In the order of operations, exponents come before any other operations and so, in \(-n^{2}\), we calculate the power before taking the opposite.  This means \(-n^{2} =-(n^{2})\) represents the opposite of the integer \(n^{2}\).  
      
      Now if  \(n = 0\) then \((-0) = 0\).  Therefore, \((-0)^{2} = 0^{2} = 0\) and \(-(0^{2}) = -0 = 0\) and so the two numbers are equal.  If \(n\) is not equal to zero, then we can show that \((-n)^{2}\) is positive and \(-(n^{2})\) is negative and so the two numbers cannot be equal.  Below we justify this by considering when \(n\) is positive, and then when \(n\) is negative:
      • If \(n\) is positive then \((-n)\) is negative. Since a negative number multiplied by a negative number results in a positive number, we determine that  \((-n)^2 = (-n) \times (-n)\) must be positive.  Since \(n\) is positive, \(n^{2} = n\times n\) is also positive.  Since the opposite of a positive number is a negative number, \(-(n^{2})\) is negative. 
      • If \(n\) is negative then \((-n)\) is positive (since the opposite of a negative number is a positive number). Since a positive number multiplied by a positive number results in a positive number, we determine that  \((-n)^2 = (-n) \times (-n)\) is positive. Since \(n\) is negative \(n^{2} = n \times n\) is a product of two negative numbers and so it is a positive number.  This means that its opposite, \(-(n^{2})\), must be negative.
         
10. 1. 1. \(\lvert 2 \times 3 \rvert= \lvert 6 \rvert = 6\)
       2. \(\lvert (-4) \times 8 \rvert = \lvert -32 \rvert = 32\)
       3. \(\lvert5 \times (-6) \rvert =\lvert -30 \rvert= 30\)
       4. \(\lvert (-4) \times (-7) \rvert = \vert 28 \rvert= 28\)
    2. From our previous work in part a), we notice that it appears the absolute value of a number is never negative, even though one or more integers being multiplied inside the absolute value signs are negative. 
       We know that \(\lvert 1 \times 13\rvert = 13\). Since we can make the integers \(1\) and/or \(13\) negative without changing the result, we have
       \(\lvert 1 \times 13 \rvert = 13 \qquad \lvert (-1) \times 13 \rvert = 13 \qquad \lvert 1 \times (-13) \rvert = 13 \qquad \lvert (-1) \times (-13) \rvert = 13\)
       We also know that multiplication is commutative, and so we can change the order to produce
       \(\lvert 13 \times 1 \rvert = 13 \qquad \lvert 13 \times (-1) \rvert = 13 \qquad \lvert (-13) \times 1 \rvert = 13 \qquad \lvert (-13) \times (-1) \rvert = 13\)
       Therefore, the integers satisfying \(\lvert m \times n \rvert=13\) are\[1,13 \qquad (-1),13 \qquad 1,(-13) \qquad (-1),(-13)\]\[13,1 \qquad 13,(-1) \qquad (-13),1 \qquad (-13),(-1)\]
    3. First, we consider positive integers that have a product of \(4\) or less. They are
       \(\begin{align*} 1 \times 1 &\quad& 1 \times 2 &\quad& 1 \times 3 & \quad & 1 \times 4 \\ 2 \times 2 &\quad& 2 \times 1 &\quad& 3 \times 1 & \quad & 4 \times 1 \\ \end{align*}\)
       When we find the absolute value of any of the above products, our previous work indicates that either integer in the product can be positive or negative. Since there are \(8\) different products, each with \(4\) different combinations of positive/negative values from part b), there are a total of \(32\) pairs of non-zero integers that satisfy \(\lvert m \times n \rvert \leq 4\).