Let a, b, c, d and e be distinct integers such that (6-a)(6-b)(6-c)(6-d)(6-e) = 45. What is a+b+c+d+e?	Choices.
A) 5  B) 17  C) 25  D) 27  E) 30
Since the prime factorization is 3*3*5   The constraint in the question is that they are distinct so the only possible way is 3,-3,5,1,-1   6-a = 3 =&gt; a=3   6-b = -3 =&gt; b=9   6-c = 5 =&gt; c=1   6-d = 1 =&gt; c=5   6-e = -1 =&gt; e = 7   Answer is 25 = 25
<result>C</result>
Let a, b, c, d and e be distinct integers such that (6-a)(6-b)(6-c)(6-d)(6-e) = 45. What is a+b+c+d+e?