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

Question

Set T consists of 82 consecutive odd integers. If the sum of the integers is 3^8 - 165, what is the median of set T?

A) 78 
B) 79 
C) 80 
D) 81 
E) 82

BrentGMATPrepNow · Accepted Answer

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 = (3⁸ - 165)/82
= ( 3⁸ - 1  -  164 )/82    
= ( 3⁸ - 1 )/82 -  164 /82
= ( 3⁸ - 1 )/82 - 2

Now let's do some factoring    
=  (3⁴ + 1)(3⁴ - 1) /82 - 2
=  (81 + 1)(81 - 1) /82 - 2
=  (82)(80) /82 - 2
= 80 - 2
= 78

Answer: A

Cheers,
Brent

chetan2u · Answer

the series is consecutive ODD integers..
 so MEDIAN will be same as AVERAGE  
number of integers, so median= average = (41st+42nd)/2

here we are given sum and we know total number, so median =  \frac{3^8-165}{82} 
choice give us the answer..
 1) median will be a EVEN integer
2) it will be a multiple of 3, as numerator has 3 in it and denominator does not 
ONLY 78 matches

otherwise
 \frac{3^8-165}{82}=\frac{3(3^7-55)}{82}=\frac{3(2187-55)}{82}=78

A

vap311096 · Answer

Consecutive odd integers => (3^8)-165=82xAverage=82xMedian
(3^8) has 1 as the digit number, subtracted by 165 will end up having 6 as the digit number.
Now we need to find a median such that 82xMedian ends up having 6 as the digit number. 82x78 is the only valid choice.
Answer: A

RB15 · Answer

I am confused.

The sum of first n consecutive odd integers is of the form n^2. Now if we look at the sum in the given equation: it is ((3^4)^2) - 165 => (81^2 - something). So it is less than the sum of first 81 consecutive odd integers

(1) How can there be 82 terms when the sum is less than the sum of first 81 odd terms.

(2) also sum of any consecutive odd integers should be of the form x^2 - y^2, where x-y = n. so here 165 should also have been a perfect square.

Please help.

BrentGMATPrepNow · Answer

Your statement that  "The sum of first n consecutive odd integers is of the form n^2"  is true for the first n POSITIVE integers. 
However, the set of 82 consecutive odd integers in the question starts with odd integers that are LESS THAN zero.

Cheers,
Brent

mimajit · Answer

Set T consists of 82 consecutive odd integers. If the sum of the integers is 3⁸ - 165, what is the median of set T?

A) 78 
B) 79 
C) 80 
D) 81 
E) 82

This is how i solved ..... as brent mentioned

" There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."

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.

For which we can use( 3^8 - 165 )/ 82 = Mean or in this case also the median .

Solve to get 3^8 - 165 = 6396 /82 = 78 ans choice A

ScottTargetTestPrep · Answer

Solution:
 
In a set of consecutive odd integers, which are evenly spaced, the median of the set is also the average of the set. Therefore, if we let x be the median, we can create the equation:

82x = 3^8 - 165

82x = 3^8 - 1 - 164

82x = (3^4 - 1)(3^4 + 1) - 164

82x = (80)(82) - 2(82)

x = 80 - 2

x = 78

Answer: A

skmaqeel51 · Answer

Please correct the question - 3^8 is written as 38.

Bunuel · Answer

Fixed the typo. Thank you!

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

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

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