P* is defined as the number of positive even integers less than P, if

7* = 6*

7* = 3 ( 2,4,6)
6* = 3 (2,3,5)

The answer choices are in the form <integer>* and not what that integer would evaluate to.

I think 5* should be {0,2,4}, since 0 is considered and even positive integer, hence it evaluates to 3
10* would be {2,3,5,7}
so (3+4)* = (7)*

Since 5 is odd, 5* represents the number of even integers less than 5. Since the even integers less than 5 are 4 and 2, 5* = 2.

Since 10 is even, 10* represents the number of primes less than 10. Since the primes less than 10 are 7, 5, 3, and 2, 10* = 4.

So, (5* + 10*)* = (2 + 4)* = 6* = 3, since 6 is even and the primes less than 6 are 5, 3, and 2.

We need to find an answer choice that is equal to 3.

We see that 7* = 3, since there are 3 even numbers less than 7, namely, 6, 4, and 2.

Answer: C
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Hi Bunuel,

For an odd P, P* is defined as the number of positive even integers less than P
For an even P, P* is defined as the number of prime integers less than P.

Since we are asked the value of (5* + 10*)*

5* = 2,4(2)
10* = 2,3,5,7(4)

Therefore we have been asked to find (2+4)* = 6*

But I do not find that answer option. Am i doing something wrong?

Hi avaneeshvyas

0 is neither positive nor negative.

Consider this: if 0 is positive, then on multiplying it with a negative number, say -1, it should become negative because positive*negative=negative. But 0 multiplied by any number is 0.

For this solution 5* ={2,4}= 2 values
10*={2,3,5,7}=4 values
hence (5*+10*)*=(2+4)*=6*
again 6*={2,3,5}= 3 values
and 7*={2,4,6}= 3 values

Hence 6*=7*

So option C

hi ..
6* means prime number LESS than 6 so 2,3,5 - ans 3
in choices 7* means EVEN integers LESS than 7 so 2,4,6 again answer 3
EDIT - SOMEONE has already clarified above...
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I would like others to pitch in with their thoughts on this since for GMAT, as far as I know, 0 is an even number. Positive/negative/otherwise I am not very sure.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.


Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Just to Add to Bunuel 's point for 0 being even number despite 0 being divisible by Odd number but still it is not an Odd no.

Any even number is of the form 2k (that is a multiple of 2) and Odd no is of the form 2k+1, where k is an integer.

so 2k=0 => k=0 (an integer). Hence 0 is even

but if 0 were Odd, then 2k+1=0 => k=-1/2 =not an integer. Hence 0 is not odd

Also Odd ± Odd = Even, so if 0 were Odd, then

0+0=0, so 0 becomes even despite being an Odd as per our earlier assumption. Hence the assumption is invalid

Therefore, 0 is even as Even ± Even=Even

so we can say 0+0=0=Even perfectly fine assumption

In conclusion 0 is a unique no that is Even and neither positive nor negative.

I approached this question like this. Please correct me if I am wrong:

(5* + 10*)* = [5*(1+2)]* = [5* x 3]* = (15*)*

Therefore, P is odd and so, no. of even integers less than 15 are 2, 4, 6, 8, 10, 12, 14 i.e 7 integers

hence, we get: (7)* (C)

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