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?
a) 1^^
b) 1^^ – 2^^
c) 3^^
d) 3^^ + 4^^
e) 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)?
a) 21
b) 40
c) 63
d) 117
e) 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.
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.
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?
a) 1^^
b) 1^^ – 2^^
c) 3^^
d) 3^^ + 4^^
e) 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)?
a) 21
b) 40
c) 63
d) 117
e) 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.
Final Word
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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