fluke wrote:
Average of elements*Number of Elements=Sum of elements
\(\frac{k+1}{2}*(\frac{k-1}{2}+1)=441\)
\(k=41\)
Ans: "B"
fluke's explanation is really nice. A variant of this rationale would be the following:
From the question stem we know that the set is an evenly spaced one, so the average of the elements of the set is the average of both extremes. In this case, the smallest term is 1 and the biggest term is \(k\). Therefore, the average is:
\(avg=\frac{1+k}{2}\)
On the other hand, we know the sum of all the terms of the set. We can determine the average dividing the sum by the number of elements in the set. If \(n\) is the number of elements in the set, then the average is:
\(avg=\frac{441}{n}\)
Then,
\(avg=\frac{1+k}{2}=\frac{441}{n}\)
\(n=\frac{441*2}{k+1}\)
At this point, you can easily realize that 441 is divisible by 3 and by 9 (sum of the digits is divisible by 9).
\(n=\frac{7*7*9*2}{k+1}\)
Now, using the answer choices, you have to find the value of \(k\) that makes \(n\) an integer.
(A) 47 -> 47 + 1 -> \(48 = 3*2*8\) -> No 8 in the numerator
(B) 41 -> 41 + 1 -> \(42 = 3*2*7\) ->
42 is a factor of the numerator(C) 37 -> 37 + 1 -> \(38 = 2*19\) -> No 19 in the numerator
(D) 33 -> 33 + 1 -> \(34 = 2*17\) -> No 17 in the numerator
(E) 29 -> 29 + 1 -> \(30 = 3*2*5\) -> No 5 in the numerator