Time to solve Monday’s math challenge: What the #@#@! Here’s the original question, in case you forgot:

## 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

## Answer and Explanation

The first order of business is to crack #x# = 5. A good way is to work backwards,

trying to figure out what a certain value of x will equal. The thing is @x@ is

embedded in #x#. For instance, #100# would give us @100@ = 9 (1^2 = 1, 2^2 =

4…9^2 = 81). The number of primes, the #x# function, less than 9 gives us 4.

Therefore, when we plug in 100 for x, we end up getting 4.

The thing is we need some number that will lead to 5. Well, counting out primes that

takes us to 2, 3, 5, 7, 11. Therefore, what value of @x@ gives us 11? Anything greater

than 121 but less than 145—quite a range, but it doesn’t really matter which of

those numbers we choose. See, when we plug that number into @x@, it will

automatically give us 11.

The twist, however, is in translating the #(@x@)#. We wouldn’t just want to plug in

11 in place of @x@ and then count the primes less than 11, since embedded into

#x# is an extra @x@. Therefore, we have to first count the number of perfect

squares less than 11, which is 3. Then, we figure out how many primes are less than

3, and the answer is 1.

Answer: 1

In the statement : ” Therefore, what value of @x@ gives us 11? “. Shouldn’t there be 12 instead of 11. If we consider 11, we get no of primes as 4 instead of 5.

I agree, #11# = 4, #12# = 5 and #13# = 5