BTGmoderatorDC wrote: ↑

Tue Jan 14, 2020 6:00 pm

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



OA B

Solution:

Statement One Only:

a, b, c, and d are consecutive integers where (a + b) and (c + d) are prime.

We see that b = a + 1, c = a + 2 and d = a + 3 since a, b, c, and d are consecutive integers. Therefore, a + b = 2a + 1 and c + d = 2a + 5. In other words c + d is 4 more than a + b. We are given that (a + b) and (c + d) are primes, however, there are many primes such that one is 4 more than the other. For example, 7 is 4 more than 3 and 17 is 4 more than 13. In the former case, a + b + c + d = 1 + 2 + 3 + 4 = 10, while in the latter, a + b + c + d = 6 + 7 + 8 + 9 = 30. Statement one alone is not sufficient.

Statement Two Only:

ac + ad + bc + bd = 21

Simplifying, we have:

a(c + d) + b(c + d) = 21

(a + b)(c + d) = 21

Since a, b, c, and d are positive integers, we see that a + b = 3 and c + d = 7 OR a + b = 7 and c + d = 3. Either way, a + b + c + d = 3 + 7 = 7 + 3 = 10. Statement two alone is sufficient.

Answer: B