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The answer will be 5 as @x@ will represent all positive no less than x and we can find primes of negative numbers, so #x# = #(@x@)#

#x#=5 means there are five prime numbers less than @x@.The first five prime numbers are 2,3,5,7 and 11. so @x@=12 and #12#=5 Ans. B

Answer: E = 1

x >144 / @x@ = 12 / #x# = 5

@(@x@)@ = 3

#(@x@)# = 1

(E) 1

(A) 7

if #x# = 5; then @x@ has to be greater than or equal to 8 (the five positive primes less than 8 being 1, 2, 3, 5, 7)

hence # 8 # should be 5 i.e. no. of primes less than 8

correction (B)

“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@)#?”

We know #x# = 5, so five prime numbers are less than @x@. The first five prime numbers are 2, 3, 5, 7, and 11, followed by 13; this means that @x@ is either 12 or 13.

To find #(@x@)# we first have to find @(@x@)@: this is the number of positive perfect squares less than 12 or 13. The square numbers below 12 or 13 are 1, 4, and 9: so there are three of them. This means @(@x@)@ = 3.

Finally we need the number of primes less than 3; there is only one of these, the number 2.

Therefore, #(@x@)# = 1, choice E.

(E) 1

Is the answer (E) 1 ??

Edit : The answer is 1.

#x# = 5 (These are 2, 5, 7, 11, 13)

which implies that x=17

@x@ = 4 (These are 1, 4, 9, 16)

So , #(@x@)# = #(4)# = 1 (As 2 is the only prime no. less than 4)

Hence , the correct option is (E) .

5

#(x)# = 5

( these are 2, 5, 7, 11, 13 )

which implies that x = 17

@x@ = 4 (these are 1, 4, 9, 16 )

# (@x@) # = # (4) # = 1 (that no. is 2)

So the correct option is (e) .

Answer is 1.

Solution:

If #x# = 5, then @x@ = 12 or 13, because there are 5 primes less than 12 and 13.

Question becomes the value of #12# or #13#.

If x = 12 or 13, then @12@ or @13@ = 3, which are 1, 4 and 9 as the perfect squares.

If @12@ or @13@ = 3, then #12# or #13# = 1, which is 2 as the prime number.

Therefore if #x# = 5, then #(@x@)# = 1.

Ans E 1

Answer: E-1

#x# = 5 = number of primes less than @x@

so @x@ = 12 or @x@ = 13 -> (1)

#(@x@)# = number of primes less than @(@x@)@

= number of primes less than (@(12)@ or @(13)@) [from (1)]

= number of primes less than (number of perfect squares less than 12 or 13)

= number of primes less than (3)

= 1

(E)1

1

Ans: E

@x@ is 2( perfect squares – 1 and 4)

so #x# will be 1.

answer is E= 1..

Ans – 1

(E) 1

The answer is (E).

Given: #x# = 5

So, the number of primes less than @x@ = 5

The primes are 2, 3, 5, 7, 11, 13, …

Since the 5th prime is 11, @x@ is either 12 or 13.

Then, @(@x@)@ = the number of positive perfect squares less than @x@ = 3 (1, 4 and 9).

Therefore, #(@x@)# = The number of primes less than @(@x@)@ = 1 (Since there is only 1 prime less than 3)

If #x# = 5, the five prime numbers will be 2, 3, 5, 7, 11 and x is either 12 or 13.

Now y = @x@ = 3 (1, 4, 9)

Now #y# = 1 (2)

Hence, E

13

All the prime no. Less than and 37.