Set A consists of integers {3, -8, Y, 19, -6} and Set B consists of integers {K, -3, 0, 16, -5, 9}. Number L represents the median of Set A, number M represents the mode of set B, and number Z = LM. If Y is an integer greater than 21, for what value of K will Z be a divisor of 26? (A) -2 (B) -1 (C) 0 (D) 1 (E) 2 OA is C It took me 4 mins to solve this question. Please advice any faster method. Sachin

The intent of the problem seems to be that Z = L^M, as shown in red below: sachin_yadav wrote:Set A consists of integers {3, -8, Y, 19, -6} and Set B consists of integers {K, -3, 0, 16, -5, 9}. Number L represents the median of Set A, number M represents the mode of set B, and number Z = L^M. If Y is an integer greater than 21, for what value of K will Z be a divisor of 26? (A) -2 (B) -1 (C) 0 (D) 1 (E) 2 The MODE of a set is the value that appears the GREATEST NUMBER OF TIMES. For Set B to have a mode -- for one of the values of Set B to appear more than any other value in set B -- K must be equal to one of the 5 values already present: -3, 0, 16, -5, 9. Of the 5 answer choices, only C -- K=0 -- is included in the list above. The correct answer is C.

Mitch, Thank you. Quick one Regards Sachin

But then Z becomes 0, and how can 0 be a divisor?

sidelinesk wrote:But then Z becomes 0, and how can 0 be a divisor? I believe that the posted problem was transcribed incorrectly. The intent of the problem is not that Z = LM but that Z = L^M. If the mode of set B is 0, then M=0, implying that Z = L� = 1.