Bunuel
For distinct positive integers x and y, where x < y, the function FP(x, y) returns the smallest prime number between x and y, exclusive, or the text string ‘NULL’ if no such number is found. If FP(a, b) +FP(c, d) = FP(e, f), where a, b, c, d, e and f are distinct positive integers, what is the value of ca ?
(1) FP(g, h) = a, where g and h are distinct positive integers
(2) c is less than the minimum possible value of the function FP(x,y).
Wow . Such a good ques . Kudos to this ques !! Took me >3 min to deduce the logic .
Ans is
E IMO .
FP(a, b) +FP(c, d) = FP(e, f) , a, b, c, d, e and f are distinct positive integers ..
Prime + Prime = Prime
we know that addition of 2 prime can only yield prime number if one of them is 2 .
so either a or c must be 1 .
We need to find value of
ac .
(1) FP(g, h) = a, where g and h are distinct positive integers
a must be a prime number .
So now c must be 1 . But a can be any prime 3 or 2 .
FP(2, 4) +FP(1, 3) = FP(4, 6) ===> 3 + 2 = 5
or
FP(3, 6) +FP(1, 9) = FP(5,8 ) ===> 5 + 2 = 7
ac can be 2 or 3 .
So statement 1 is not sufficient . (2) c is less than the minimum possible value of the function FP(x,y).
minimum possible value of the function FP(x,y) is 2 .
Now
C < 2 and C > 0
so C = 1 .
But a can be anything .
So statement 2 is not sufficient . Now Combining both
we have c=1 and a can be still 2 or 3 .
So statement 1 and 2 , combing is not sufficient .
Ans
E