SAT Integer Questions: Even and Odd Integers

SAT Integer Questions

Even and Odd Integers

SAT test makers like to ask questions about even and odd integers. Part (c) of this problem is similar to an actual SAT question (the multiple choice options are removed). Use parts (a) and (b) to help answer part (c). Then explore even and odd numbers in more depth with parts (d) and (e).

  1. Explain why multiplying any integer by 2 results in an even integer.
  2. If k is a positive integer, write an expression for the following.
    1. An even integer
    2. An odd integer
  3. If k is a positive integer, write an expression for en even integer that is twice the value of an odd integer.
  4. When the positive odd integer n is doubled, the result is three times m. Which of the following, if any, MUST be false?
    1. n is an even integer
    2. n is an odd integer
    3. n is not an integer
  5. Twice the sum of three consecutive odd integers is the same as the sum of three consecutive even integers. Find one possible set of such integers.

Solution

  1. Explain why multiplying any integer by 2 results in an even integer.
  2. An even integer is any integer that is a multiple of 2. Thus, multiplying any integer by 2 will guarantee that the result is an even integer.

  3. If k is a positive integer, write an expression for the following.
    1. An even integer
    2. An odd integer


    Use the result from part (a) to find expressions for even and odd integers.

    1. Since multiplying any integer by 2 will guarantee that the result is even, multiply k by 2 to find an expression for an even integer.
      Thus, an expression for an even integer is 2k.
    2. Start with 2k, which is now known to be an even integer. The question becomes: How do we make an even integer odd? Well, the next number after any even integer is odd. So adding 1 to an even integer will make it odd.
      Thus, an expression for an odd integer is 2k+1.
  4. If k is a positive integer, write an expression for an even integer that is twice the value of an odd integer.
  5. Start with the expression for an odd integer, 2k+1. We want an expression that is twice this value that is also even. The question is, if we simply multiply by 2, will the result be even? Well according to part (a), yes. Multiplying any integer by 2 results in an even integer.

    Thus, an expression for an even integer that is twice the value of an odd integer is 2(2k+1), or 4k+2.
  6. When a positive number n is doubled, the result is three times an even integer m. Which of the following, if any, MUST be false?
    1. n is an even integer
    2. n is an odd integer
    3. n is not an integer


    We will consider two different methods for solving this part. The first will involve plugging in numbers and the second will involve some more careful reasoning.
    We can rule one out one of the given statements if we can find an example for which the statement is true. So, for the first method, pick values for m until n is even, odd, or not an integer.

    1. First try m=4. Three times four is twelve, so what number when doubled is twelve? That number would be six, which is even. So we have an example in which n is an even integer.
      Therefore, the statement “n is an even integer” is sometimes true.
    2. Now try m=2. In this case, n=3. So here is an example in which n is an odd integer.
      Therefore, the statement “n is an odd integer” is sometimes true.
    3. You may have trouble finding an example when n is not an integer. But how can you be sure that n is never an integer? The second solution method below will be more conclusive.

    For the second method it will help to write an equation relating n and m. The problem states that doubling n is the same as tripling m, so the equation is as shown below.

        \[       2n=3m    \]

    Solving for n gives the following.

        \[       n=\frac{3}{2}m    \]

    Since m is even, replace it with the expression for an even integer that we’ve been using throughout: 2k.

        \[       n=\frac{3}{2}m    \]

        \[       n=\frac{3}{2}(2k)    \]

        \[       n=3k    \]

    The crucial point is that k, a positive integer, can be even or odd. If m is 4 then k is 2, but if m is 6 then k is 3. Therefore, the above equation is telling us that n is equal to an odd integer (3) times an integer that is either even or odd (k).
    Now we are ready to attack the three statements.

    1. Consider the case when k is even. Since an odd times an even is even, n=3k is even. So, n is an even integer whenever k is even.
      Therefore, the statement “n is an even integer” is sometimes true.
    2. Consider the case when k is odd. Since an odd times an odd is odd, n=3k is odd. So, n is an odd integer whenever k is odd.
      Therefore, the statement “n is an odd integer” is sometimes true.
    3. Answering this part is where all the work pays off for this second method. Since k is a positive integer, and since any integer times any other integer results in an integer, n=3k must be an integer. So, n is guaranteed to be an integer.
      Therefore, the statement “n is not an integer” must be false.
  7. Twice the sum of three consecutive odd integers is the same as the sum of three consecutive even integers. Find one possible set of such integers.
    Define n as the first of the three consecutive odd integers. The next odd integer is n+2, and the next is n+4. Similarly, define m as the first of the even integers. Then m+2 is the next even integer and m+4 is the next. The sums are shown below.

        \[       n+n+2+n+4    \]

        \[       m+m+2+m+4    \]

    The problem states that twice the sum of the odds is equal to the sum of the evens. Use the sums above to write this statement as an equation.

        \[       2(n+n+2+n+4)=m+m+2+m+4    \]

    It will be convenient to solve this for m.

        \[       2(3n+6)=3m+6    \]

        \[       6n+12=3m+6    \]

        \[       6n+6=3m    \]

        \[       2n+2=m    \]

    Luckily, the question only asks to find one of the possible sets. So all we need to do is pick a value for n and find the value for m. Recalling that n must be odd, we pick n=1. Then m=4 (which is even).

    A set of three consecutive odd integers is {1,3,5}.
    A set of three consecutive even integers is {4,6,8}.

About Jared R

Jared, founder of The Knowledge Roundtable, is passionate about the advancement of knowledge. He has a B.S. in astronomy and physics from UMass and an MBA in Advanced Financial Analytics, also from UMass. He has a day job as a Big Data Analyst in Boston. He has over 500 hours of tutoring experience in everything from algebra to writing. He taught our SAT prep group courses for two years in NH, and before that developed educational content for math, stats, and finance textbooks for two years. His teaching style is hands-on with a focus on problem-solving and critical thinking.

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