nikhilgmat31 wrote:If a, b, c, and d are integers where 0 < a < b < c < d, what is the value of a + b + c +d?
(1) a, b, c, and d are consecutive integers where (a + b) and (c + d) are prime.
(2) ac + ad + bc + bd = 21
Statement 1:
Case 1: (a+b) + (c+d) = (1+2) + (3+4) = 3 + 7 = 10.
Case 2: (a+b) + (c+d) = (3+4) + (5+6) = 7 + 11 = 18.
Since a+b+c+d can be different values, INSUFFICIENT.
Statement 2:
ac + ad + bc + bd = 21
a(c+d) + b(c+d) = 21
(a+b)(c+d) = 21.
Implication:
a+b and c+d are factor pairs of 21:
1*21
3*7.
Since a, b, c and d are integers -- and 0<a<b<c<d implies that a+b < c+d -- only one case is possible:
a+b = 3 and c+d = 7, with the result that a+b+c+d = 3+7 = 10.
SUFFICIENT.
The correct answer is
B.
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