Set T consists of 82 consecutive odd integers. If the sum of

[Brent@GMATPrepNow]

Set T consists of 82 consecutive odd integers. If the sum of the integers is 3� - 165, what is the median of set T?

A) 78
B) 80
C) 82
D) 84
E) 86

Answer: A
Source: www.gmatprepnow.com
Estimated difficulty level: 650-700

Cheers,
Brent

[deloitte247]

The series is consecutive odd integers
The median= average ; (41st+42nd)/2
We are given sum & we know total number
$$median=\frac{\left(3^8-165\right)}{82}$$
Note that
-Median will be EVEN integer
-It will be a multiple of 3 because numerator has 3 in it and denominator does not.
$$median=\frac{\left(3^8-165\right)}{82}$$
$$median=\frac{3\left(3^7-55\right)}{82}$$
$$median=\frac{3\left(2187-55\right)}{82}=78$$
Hence, the correct answer is A

[Brent@GMATPrepNow]

Set T consists of 82 consecutive odd integers. If the sum of the integers is 3� - 165, what is the median of set T?

A) 78
B) 80
C) 82
D) 84
E) 86

Answer: A
Source: www.gmatprepnow.com
Estimated difficulty level: 650-700

Cheers,
Brent

Here's another approach:

There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}

Since set T consists of 82 consecutive odd integers (e.g., 5, 7, 9, 11, etc), we can see that these values are equally spaced.
So, the mean and median of set T are equal.

So, to find the answer, we need only find the mean of set T.

Median = mean = (3� - 165)/82
= (3� - 1 - 164)/82 [you'll see why I did this shortly]
= (3� - 1)/82 - 164/82
= (3� - 1)/82 - 2 [PERFECT!! 164/82 simplifies to 2]

Now let's do some factoring [since 3� - 1 is a DIFFERENCE OF SQUARES]
= (3� + 1)(3� - 1)/82 - 2
= (81 + 1)(81 - 1)/82 - 2
= (82)(80)/82 - 2
= 80 - 2
= 78

Answer: A

Cheers,
Brent

[GMATGuruNY]

Set T consists of 82 consecutive odd integers. If the sum of the integers is 3� - 165, what is the median of set T?

A) 78
B) 80
C) 82
D) 84
E) 86

For any set of consecutive integers:
SUM = (COUNT)(MEDIAN).

In the problem above:
SUM = 3� - 165 = (3�3�) - 165 = (81*81) - 165 = (integer with a units digit of 1) - (integer with a units digit of 5) = integer with a units digit of 6.
COUNT = 82 integers.

Since the sum has a units digit of 6, multiplying the count (82) by the correct answer (the median) must yield an integer with a units digit of 6.
Only A is viable:
78*82 = integer with a units digit of 6.

The correct answer is A.