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Strange Symbols in GRE Math Operations

Quite possibly the most intimidating problem on the GRE contains strange symbols: @, #, *, or a black circle often accompany these problems. Many recoil in horror thinking – I’ve never learned that before! (Or perhaps more aptly, what the @#?!)

But don’t despair – the symbols are completely arbitrary and are defined on the spot by the GRE. Here is an example:

#x# = x^2-1. What is the value of #3# – #2#?

Again, the pound sign surrounding the number has no mathematical meaning outside the problem. For the question, you simply want to follow the rules. Here, wherever we see a number between the pound sign, such as #3#, we want to refer back to #x# = x^2-1. The 3 essentially is taking the place of the x. So if #x# = x^2-1, then #3# = 3^2-1 = 8.

Now do the same for #2#: 2^2-1 = 3.

So #3# – #2# = 8-3 = 5

 

Now let’s try another one. This time, though, I am going to put a little spin on it.

n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

  1. 1^^
  2. 1^^ – 2^^
  3. 3^^
  4. 3^^ + 4^^
  5. 5^^ – 2^^

First off, note that ! is the factorial sign. It is not a strange symbol, but standard mathematical notation.You should quickly see that after n = 3, n^^ is going to yield a positive result. For example, 4! = 4*3*2*1=24, so 4^^ {=} 24 - 4^2 = 8. So when n is greater than or equal to 4, n^^ is greater than zero.

Be careful: 1^^ = 0, so it is not less than zero. Therefore, there are two values (2, 3) for which n^^ is less than zero.

When we look at the answers, 2 is not among them. Instead, the strange symbol ^^ has been reintroduced. Therefore you have to figure out which answer choice equals 2, the number of values of n that are less than zero.

The answer is (B), which gives us 1^^ = 0 minus 2^^ = -2, so 0 - (-2) = 2

Okay, that was a tough one. Let’s make the problem easier, while adding a layer of complexity – the embedded strange symbol.

 

If x is even, @x = 3x -3; if x is odd, @x = 2x - 2. What is the value of @(@(@5)?

  1. 21
  2. 40
  3. 63
  4. 117
  5. 140

Notice I’ve used the strange symbol three times. Don’t worry – just follow the operation (the technical name of this process). Taking the problem apart one step at a time, we get @5 = 8. @8 = 21, and @21 = 40. Just like that, B.

 

Takeaway

Don’t be freaked out by strange symbols on the GRE. The question will always clearly define the symbol for you. Carefully follow the steps to the correct answer.

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28 Responses to Strange Symbols in GRE Math Operations

  1. Karan August 2, 2014 at 4:24 pm #

    Chris, just a quick suggestion. The comments on Magoosh GRE blogs do not allow editing of once posted comments 🙁 You should look at Disqus as that is the best comments plugin available! This might be helpful to students who accidentally make mistakes while typing their replies.

    Keep the awesome tips coming 🙂

    • Rachel Wisuri
      Rachel August 4, 2014 at 10:59 am #

      Thanks for the suggestion, Karan. 🙂 We’ll look into it!

      • Karan August 4, 2014 at 11:24 am #

        Hey Rachel, Forgot to mention this in my previous post. Another great benefit of using disqus is that people can vote up useful comments, which makes the comments very easy to read for everyone!

        • Rachel Wisuri
          Rachel August 5, 2014 at 2:07 pm #

          That is indeed very useful and intriguing! 🙂

  2. Gajendra K July 25, 2014 at 10:01 am #

    ¤ what does this strange symbol mean ?

    f y ¤ x = y2x for all positive integers, then (3 ¤ 4) ¤ 2 =

    A. 38
    B. 312
    C. 316
    D. 324
    E. 332

    • Gajendra K July 25, 2014 at 10:03 am #

      its Y to the power 2 (x)
      options are
      A.3 power 8
      B. 3 power 12
      and so on..
      now please clarify how it can be solved

  3. Patty January 23, 2012 at 1:58 pm #

    Hi Chris,

    I understand the other problems, but am struggling big time with n^^ = , where n is a positive integer. For how many values of n is n^^ less than zero? First, how did you determine from the questions that “You should quickly see that after n = 4,” there are no numbers involved. Did you just guess at a number or are you looking at possible answers and plugging those numbers in? I will be honest, I am totally lost on this question and would love further clarification.

    I have another problem that I was given on my practice test booklet that I was struggling with as well which led me out to seek some answers on the internet.

    • Chris Lele
      Chris January 23, 2012 at 2:34 pm #

      Hi Patty,

      The part that read, “you should quickly see…” should read including n = 4, not after n = 4. I’ve made the necessary changes. Let me know if that helps. Also, read the response above Nevin’s comment above.

      If that doesn’t help, I’d be happy to clear things up.

      As for the practice problem, I’d love to answer it. Out of curiosity, does it include strange symbols? That way, it would be relevant to the post.

      Either way is fine though.

  4. Nevin January 22, 2012 at 3:37 am #

    Hi Chris,

    Would you please do this problem for me?

    Q: A pair of dice are so produced that when rolled simultaneously, exactly one of them always shows a prime number, what is the probability that the products of the numbers on the dice is even?

    (a)7/9
    (b)1/2
    (c)2/9
    (d)7/18
    (e)7/36

    • Ajey January 22, 2012 at 11:37 pm #

      When two dice are rolled, there are 6*6=36 possible outcomes. Now, the prime numbers among numbers 1,2,3,4,5,6 are 2,3 and 5. So on exactly one dice, a prime number turns up.

      If 2 turns up on one, the other dice can show 1,4 and 6. (Only one number is prime).
      If 3 turns up on one, the other can show 4 and 6 (remember the product should be even).
      If 5 turns up on one, the other can show 4 and 6.

      Hence we have the pairs – (2,1), (2,4), (2,6), (3,4), (3,6), (5,4), (5,6).
      That’s totally 7 pairs. And the same thing for the other dice as well. Hence that’s a total of 7+7= 14 favorable outcomes and 36 possible outcomes. 14/36 = 7/18
      Answer = (d)

      • Chris Lele
        Chris January 23, 2012 at 2:31 pm #

        Ajey,

        Great response!

        Thanks for the clear step by step approach.

      • Nevin January 23, 2012 at 5:24 pm #

        Thanks Ajey, GRE problems always seem so easy …..After we get to know how its done 😉

      • vijay August 20, 2012 at 4:13 am #

        But why aren’t we taking into account (1,2) (4,2) (6,2) (4,3) (6,3) (4,5) (6,5)????

        • Chris Lele
          Chris August 21, 2012 at 2:40 pm #

          Hi Vijay,

          Ajey does include these numbers in his calculation, “That’s totally 7 pairs. And the same thing for the other dice as well.” Meaning he reverses dice to account for all the possibilities you’ve listed.

          Hope that helps!

          • vijay August 22, 2012 at 1:19 am #

            Yup.. Thanks a lot!!

      • vijay August 20, 2012 at 4:15 am #

        also when you are making combination as (2,1) then why not as (3,1) and (5,1)????

    • Chris Lele
      Chris January 23, 2012 at 2:30 pm #

      Nevin,

      Check out Ajey’s great response!.

      One thing I would elaborate on is the last step. The reason the outcomes could apply to the other die, is we never stipulated which of the two dice landed on a prime.

      Because either of the two could be the prime (but just one at the same time), we multiply by 2.

      • Arpita August 25, 2012 at 6:09 pm #

        I am having issues figuring out the possible number of outcomes. If the question states ‘exactly one of them always shows a prime number’ doesn’t that mean that the dice is not fair, and the possible outcomes will be less than 36.
        Total outcomes I get is 3 for the die that always shows prime #s. (2, 3 or 5).
        And since ‘exactly one’ shows prime, that means both at the same time can’t have prime, thereby reducing the possibilities for the other die as well.

        Can you clarify? Thanks!

        • Karan August 2, 2014 at 4:20 pm #

          Hey Arpita,

          Notice how the question does not tell us which one of the dice is fixed. It just says “one of them”. Since we do not know whether the 1st dice was fixed or the 2nd one. We have to consider both cases while calculating the possible outcomes. Notice that if we consider both the cases:

          Case 1: Dice 1 is fixed and Dice 2 is fair, we have 3*6(=18) outcomes.
          Case 2: Dice 1 is fair and Dice 2 is fixed, we have 6*3(=18) outcomes.

          Therefore, total outcomes = outcomes of Case 1 + outcomes of Case 2 = 18 + 18 = 36.

          If you do consider that 1 dice is fixed and it will always produce a prime number, the final probability remains unchanged.

          Here’s how: If we know explicitly that the first dice always produces the prime values and the 2nd one is fair, then we can only get 7 pairs of favorable outcomes.

          Also note that since the first die always produces 3 numbers(2,3, & 5). The total number of outcomes is 3*6 = 18.

          Therefore, the final probability is, still 7/18.

          • Queque January 11, 2015 at 10:59 am #

            Karan why do you consider that one of the dices is fair while the remaining dice is not? I would say neither is fair because if the first dice is the one who shows a prime, then the second dice will automatically have to show a non-prime (either 1, 4 or 6), making the possible outcome in this case 3*3, not 3*6. Do you see what I mean?

            • Rasitha July 12, 2016 at 10:59 am #

              Exactly, I also had this problem in mind.

              Question is: A pair of dice are so produced that when rolled simultaneously, exactly one of them always shows a prime number, what is the probability that the products of the numbers on the dice is even?

              Therefore, the only time there can be a combination on the two dice where the product is odd is the following value pairs.
              (1,3), (1,5), (3,1), (5,1)

              This is because the two dice must always have one and only one prime (as is the case in the above pairs) & for the product to be odd, both the values of the dice must be odd.
              One of the odd values must be the prime number (3 or 5) and the other value must be an odd value which is NOT prime (only 1 can fit here)

              So considering the above, there can be total of 14 + 4 =18 total combinations possible from the throw of the dice.
              i.e.

              (2,1), (2,4), (2,6), (3,4), (3,6), (5,4), (5,6)
              (1,2), (4,2), (6,2), (4,3), (6,3), (4,5), (6,5)
              &
              (1,3), (1,5), (3,1), (5,1)

              Of them, 14 give even products.

              Therefore probability asked in the question is 14/18 = 7/9

              • Magoosh Test Prep Expert
                Magoosh Test Prep Expert July 13, 2016 at 7:10 am #

                It looks like you’re right, Rasitha. Since exactly one of the two will be a prime number, we do not have 3*6 = 18 possibilities (assuming the first dice lands on a prime number) but rather 3*3 = 9:

                Dice 1 = 2, 3, or 5
                Dice 2 = 1, 4, or 6

                Since, as you and the others have shown, there are 7 ways that the product of the values of the two dice is even, given the first dice is prime, P(even product) = 7/9.

                • Rasitha July 19, 2016 at 1:08 am #

                  Also, this answer depends on the assumption that all individual pairs as outcomes have the same probability. Just because we have N total outcomes and r total successes, probability does not equal r/N.

                  i.e. If because of the imbalance of the dice, getting for example (1,2) was 0.99 probable, then the above calculation is incorrect.

  5. Nevin January 21, 2012 at 2:48 am #

    Hi Chris,

    n^^ =n!-n^2 , where n is a positive integer. For how many values of n is n^^ less than zero?

    Lets consider ans choices B and C,

    B. 1^^ = 0 minus 2^^ = -2, so 0-(-2)=2, therefore the result is positive.

    C. 3! – 3^2 = 6 – 9 = -3, Therefore the result must be “C”.

    Why do you say that its “B”?

    • Chris Lele
      Chris January 23, 2012 at 2:23 pm #

      Hi Nevin,

      This question is a little tricky. Let’s take apart one step at a time.

      First, the question is asking for the number of values of n that are negative. Even though the values of n are negative, the number of values (i.e. 1, 2, 3, etc) is a positive number. Thus, there are two values of n (2,3) in which n^^ is negative.

      Secondly, the answer choices are expressed as ^^. So we have to find which one of the answer choices yield the number 2. Only (B) does so. As you pointed out 1^^ – 2^^ = 2.

      Hope that helps!

      • Nevin January 23, 2012 at 5:19 pm #

        Got it, Thank you… I was asinine 🙂

  6. tabi January 20, 2012 at 5:06 am #

    I am not satisfied with the following question and explanation….. can you please elaborate on it ?
    If x is even, @x = ; if x is odd, @x = . What is the value of @(@(@5)?

    21
    40
    63
    117
    140
    Notice I’ve used the strange symbol three times. Don’t worry – just follow the operation (the technical name of this process). Taking the problem apart one step at a time, we get @5 = 8. @8 = 21, and @21 = 40. Just like that, B.

    • Chris Lele
      Chris January 20, 2012 at 12:27 pm #

      Sure!

      @5 = 2(5) – 2, where 5 takes the place of x in the operation @x = 2x – 2.

      So @5 = 8.

      8 is an even number so we use the other operation: @8 = 3(8) – 3 = 21.

      21, an odd number, must use x@ = 2x – 2. So @21 = 2(21) – 2 = 40 (B).

      Hope that helps!


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