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10 Jul 2019, 00:59
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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
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
10 Jul 2019, 01:10
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Bunuel
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
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
Y is an integer greater than 21 => Median of A = 3 since A ={-8,-6,3,19,Y} = L => L=3
M is mode of set B. Since set B does not have any repetitive numbers M = K where K = -3 or 0 or 16 or -5 or 9
Z = L^M = 3^K and is a divisor for 26 = 2*13
For 3^K to be a divisor of 2*13, K=0 since 3 is not a factor of 26.
IMO C
10 Jul 2019, 09:01
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Bunuel
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
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
very good question
A = {3, -8, Y, 19, -6} re arrange set we get L median is 3
and mode of B is not know so let M=K
26 = 2*13
now Z=L^M
L=3 and M has to be 0 since 26 is not divisible by 3
IMO C ; 0
