Answers and Solutions


    1. \(5,~10,~15,~20,~25\)
    2. \(7,~14,~21,~28,~35\)
    3. \(12,~24,~36,~48,~60\)
    4. \(16,~32,~48,~64,~80\)
    1. All numbers that are a multiple of \(6\) only go into the left region.  These numbers are: \(12,~ 30,~ 48,~ 84\)
      All numbers that are a multiple of \(9\) only go into the right region.  These numbers are: \(27,~45,~ 81\)
      All numbers that are a multiple of \(6\) and \(9\) go into the centre region.  These numbers are: \(18,~ 54,~ 90\)
      Any number that is not a multiple of \(6\) or a multiple of \(9\) does not get placed in the diagram.  We can place these numbers outside of the diagram.  These numbers are: \(38, ~52\)
    2. The numbers in the overlapping region represent multiples of \(6\) and \(9\).  In other words, this region is where we will find the common multiples of \(6\) and \(9\).
  3. We would start by listing some multiples of \(35\).  From this list of multiples, we would find the smallest number that is also a multiple of \(30\).
    1. Any number that is a multiple of both \(3\) and \(5\) must be divisible by both \(3\) and \(5\).  We can start by considering the multiples of \(5\).
      \(5,~ 10, ~15, ~20, ~25,~ 30, ~35, ~40, ~45,~ 50,~ 55, ...\)
      From here we can circle the numbers which are also a multiple of \(3\).
      \(5, ~10,~\) \(15,\) \(~20,~ 25,~ \) \(30, \) \(~35, ~40,~\) \(45,\) \(~50, ~55, ...\)
      We have found the first three common multiples.  If you keep going in this way, the first six common multiples of \(3\) and \(5\) are:
      \(15, ~30, ~45, ~60, ~75, ~90\)
    2. We notice that the common multiples of \(3\) and \(5\) are all multiples of \(15\), which is the lowest common multiple of \(3\) and \(5\).
    3. The second common multiple is two times the first common multiple.
    4. If the first common multiple is \(252\), then the second common multiple is \(252\times2=504\).
  5. To answer this question, we must find the first common multiple of \(6\) and \(10\).  Let's start by considering some multiples of \(10\).\[10,~20,~30,~40,~50, \ldots\] Next, we determine the smallest number that is also a multiple of \(6\).  This number is \(30\).
    Therefore, the drummer is expected to play both on the \(30\)th beat.
    1. If \(32\) is a common multiple of two numbers then it follows that it must be a multiple of each number.  For example,  \(4\) and \(8\) have a common multiple of \(32\).
    2. Yes, in fact there are many possible answers.  We start by listing all of the numbers of which \(32\) is a multiple: \(1, ~2,~ 4,~ 8, ~16, ~32\).  The possible answers can be made from pairing any two different numbers from this list.
    3. All of the possible pairs come from pairing two different numbers from the list of numbers of which \(32\) is a multiple: \(1, 2, 4, 8, 16, 32\)

      \(1,~2\)

      \(1,~4\)

      \(1,~8\)

      \(1,~16\)

      \(1,~32\)

      \(2,~4\)

      \(2,~8\)

      \(2,~16\)

      \(2,~32\)

      \(4,~8\)

      \(4,~16\)

      \(4,~32\)

      \(8,~16\)

      \(8,~32\)

      \(16,~32\)

  7. To answer this question, we must find the smallest common multiple of \(6\) and \(8\) that is greater than \(135\).  Let's start by considering some multiples of \(8\).\[8, 16, 24, \ldots, 120, 128, 136, 144, 152, \ldots\] Next, we need to find the smallest value above \(135\) that is also a multiple of \(6\).  This number is \(144\). To find the number of packages of each, we must divide \(144\) by \(6\) and \(8\) respectively to conclude that you should buy \(24\) packages of hamburgers and \(18\) packages of buns.
    1. True
    2. False. For example, \(\)\(4\) is a multiple of \(2\) and \(9\) is a multiple of \(3\), but \(4+9=13\) is not a multiple of \(6\).
  9. The mystery number is \(0.55\).