Let x be defined as the number of positive perfect squares less than

Question

Let @x@ be defined as the number of positive perfect squares less than x
Let #x# be defined as the number of primes less than @x@
If #x# = 5, what is the value of #(@x@)#?

(A) 7
(B) 5
(C) 4
(D) 2
(E) 1

chetan2u · Accepted Answer

Hi, 
 VeritasKarishma  has explained it very well. Just a point from my side on why you could be going wrong..

Firstly the question is not simple and easy one by any standards, and reading too much into x can get you wrong.

Step I :- #x#=5
So 5 primes less than @x@. Now five primes are 2,3,5,7 and 11, but the next prime is 13. Thus @x@=12 or 13 but NOT 14, as then primes would become 6.
Also solved above...... @x@=12 or 13

Step II :- @x@ as 12 makes #(@x@)# = #12#.
 Now FORGET what you did earlier to get #x#=5 . What you know now is that you are working with  a new value of x and that is x=12 ..

Step III :- #x# is the number of primes less than @x@
#12# is the number of primes less than @12@, but  we do not know what is @12@

Step IV :- @x@ is the number of positive perfect squares less than x.
So  @12@ is the number of positive perfect squares less than 12 . Thus values are 1, 4, and 9, that is 3 values.
Therefore, @12@=3.

Step V :- Back to step III
#x# is the number of primes less than @x@
 #12# is the number of primes less than @12@ or 3 . Possible value is only 2.
Hence #12#=1

E

pushpitkc · Answer

Given:    #x# = 5
#x# is defined as the number of primes less than @x@
where @x@ is the number of positive perfect squares less than x

If #x# = 5, then @x@ = 12 which has 5 primes(2,3,5,7,11) less than @x@
and @x@ is number of positive perfect squares less than 12, which is 3(1,4,9)

Now, we have been asked to find the value of #(@x@)# - which is the number of primes less than 3

Therefore, #(@x@)# or #3# =  1(Option E)  as only 2 is a prime lesser than 3.

rahul16singh28 · Answer

Hi  pushpitkc

As per the question statement  Xmin = 145, @x@ = 12 & #x# = 5.

Now, As per your explanation -

If #x# = 5, then @x@ = 12 which has 5 primes(2,3,5,7,11) less than @x@
and @x@ is number of positive perfect squares less than 12, which is 3(1,4,9)

@x@ is number of positive perfect squares less than x but in your above highlighted solution you are considering x itself as 12 for finding the number of @x@. But that is wrong because x will be fixed to minimum of 145.

Please correct me if my understanding is wrong.

ScottTargetTestPrep · Answer

Since #x# = 5, we see that there are 5 primes less than @x@. Since the 5th prime is 11 and the 6th prime is 13, we see that @x@ must be 12 or 13. Therefore, we have:

#(@x@)# = #12# or #13#

Since #12# is the number of primes less than @12@ and @12@ is the number of positive perfect squares less than 12, @12@ = 3 (since there are 3 perfect squares less than 12: 1, 4 and 9). Notice that since @13@ = 3 also, #13# is equal to the number of primes less than 3 as well.

So, #12# = #13# is the number of primes less than 3, and since there is only 1 prime (namely, 2) less than 3, the answer is 1.

Answer: E

preetamsaha · Answer

pushpitkc   VeritasKarishma   chetan2u   Bunuel  
If #x# = 5, then @x@ = 12 then @x@ indicates there are 12 no perfect squares less than x , why are we considering the no of perfect squares less than 12 ? I am not getting the problem . can u please explain it ?

KarishmaB · Answer

Focus on how the operators are defined:

Positive perfect squares = 1, 4, 9, 16, 25 etc
So @2@ = 1 (only 1 perfect square less than 2 i.e. 1)
@10@ = 3 (3 perfect squares less than 10 i.e. 1, 4 and 9)
@30@ = 5 (5 perfect squares less than 30 i.e. 1, 4, 9, 16 and 25)
etc

Primes = 2, 3, 5, 7, 11 etc
#2# = Number of primes less than @2@ = Number primes less than 1 = 0
#10# = Number of primes less than @10@ = Number primes less than 2 = 1 (Only 1 such prime i.e. 2)
#30# = Number of primes less than @30@ = Number primes less than 5 = 2 (2 such primes i.e. 2 and 3)
etc

Now let us look at the question:

Given #x# = 5
#x# = Number of primes less than @x@ = 5
First 5 primes are 2, 3, 5, 7, 11
So @x@ = 12 or 13. The moment @x@ becomes 14, we have 6 primes less than it.

Find #(@x@)#
#(@x@)# = #12# or #13#

#12# = Number of primes less than @12@ = Number primes less than 3 = 1
#13# = Number of primes less than @13@ = Number primes less than 3 = 1

Answer (E)

preetamsaha · Answer

VeritasKarishma   chetan2u  
I'm thankful to your posts, but still the following thinking is spinning into my head .

#x#=5 = { 2, 3, 5, 7, 11} = 5 = Number primes less than 12 = Number of primes less than @145@

so, #(@x@)#
#(@145@)#
=5

where am I wrong ?

chetan2u · Answer

Hi,
You are not understanding the wordings completely. Would request you to go through above solutions and then ask any query if you have. I am sure if you go over the solutions given slowly, you will understand.

KarishmaB · Answer

Not correct. You haven't understood the defn of #x#. #x# is the number of primes less than @x@, not the number of primes less than x.

#x# = 5 = Number of primes less than @x@ = Number of primes less than 12

So
@x@ = 12
x = 145 or 146 or 147 till 169

#(@145@)# = #(12)# = Number of primes less than @12@ (not number of primes less than 12)
@12@ = Number of perfect squares less than 12 = 3

#12# = Number of primes less than 3 = 1

KarishmaB · Answer

Also note that you don't need to get the values of x here. Getting @x@ is enough to proceed with what is asked. Check my solution.

BrushMyQuant · Answer

#x# = 5 
=> Number of prime numbers less than @x@ = 5
5 prime numbers are 2,3, 5,7,11 and next prime number = 13

So, we will have two cases as x cannot be 14, then we will have 13 as the 6th prime number < 14  
 
   -   Case 1:  @x@ = 12
  
=> #(@x@)# = #(12)# = #12#
=> Number of prime numbers less than @12@
@12@ = Number of positive perfect squares less than 12
= 3 (1, 4 and 9)

=> Number of prime numbers less than 3 = 1 (i.e. 2) 
  -   Case 2:  @x@ = 13
 
 => #(@x@)# = #(13)# = #13#
=> Number of prime numbers less than @13@
@12@ = Number of positive perfect squares less than 13
= 3 (1, 4 and 9)

=> Number of prime numbers less than 3 = 1 (i.e. 2)

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

onlymalapink · Answer

we are essentially solving the funtion twice why?

Let x be defined as the number of positive perfect squares less than

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So, we will have two cases as x cannot be 14, then we will have 13 as the 6th prime number < 14
-Case 1: @x@ = 12 => #(@x@)# = #(12)# = #12# => Number of prime numbers less than @12@ @12@ = Number of positive perfect squares less than 12 = 3 (1, 4 and 9) => Number of prime numbers less than 3 = 1 (i.e. 2)	-Case 2: @x@ = 13 => #(@x@)# = #(13)# = #13# => Number of prime numbers less than @13@ @12@ = Number of positive perfect squares less than 13 = 3 (1, 4 and 9) => Number of prime numbers less than 3 = 1 (i.e. 2)

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