sainaren wrote:
For every integer a>1, a* is defined as the least prime factor of a. Is b*<c*?
1. b is an odd integer greater than 1
2. c is an even integer
When they define a weird function like this one, it's not a bad idea to jot down a couple of examples before you start, just to make sure you understand what they're telling you.
For instance, here, we'd quickly calculate x* for a couple of different values of x.
If x is prime, then x* is just equal to x itself, since it doesn't have any smaller prime factors.
If x = 10, then x* = 2.
If x = 15, then x* = 3.
You might notice, at this point, that whenever x is even, x* should equal 2, because 2 is a prime factor of every even number and 2 is the smallest prime number. The words even and odd in the statements also give a clue that you should be thinking about how even and odd numbers behave differently from each other.
Statement 1: b is an odd integer greater than 1. We already tried the example of 15; if b = 15, then b* = 3. c* could be smaller than this (for example, if c = 2, then c* = 2, which is less than 3.) Or, c* could be bigger than this (for example, if c = 7, then c* = 7, which is more than 3.) Therefore, this statement is insufficient.
Statement 2: As discussed above, this one tells you that c* = 2.
The question asks whether b* is less than c*. Since 2 is the smallest prime number, 2 is the smallest value that b* could possibly have. b* definitely can't be
less than 2. (Remember that 1 isn't prime.)
Therefore, the answer to the question is 'no' - since c* = 2, b* definitely is NOT less than c*. This statement is sufficient.
The answer is B.