## 82 consecutive odd integers

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### 82 consecutive odd integers

by [email protected] » Sun Nov 02, 2014 12:06 pm
Here's a challenge question for students to try:
Set T consists of 82 CONSECUTIVE ODD integers. If the sum of the integers is 3â�¸ - 1, what is the median of set T?

A) 41
B) 42
C) 80
D) 82
E) 164
Aside: Since students will invariably ask about the difficulty level, I'd place it around 650/700.

Cheers,
Brent

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by nipunranjan » Mon Nov 03, 2014 6:28 am
3â�¸ - 1=( 3^4 + 1)( 3^2 + 1)(3+1)(3-1)=82*10*4*2

Median will be equal to mean in this case as the number of terms in the set are even.

Median=Mean= (Sum of all nos)/(No of terms)=82*10*4*2/82=80

Ans =(C)

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by [email protected] » Mon Nov 03, 2014 6:49 am
Set T consists of 82 CONSECUTIVE ODD integers. If the sum of the integers is 3â�¸ - 1, what is the median of set T?

A) 41
B) 42
C) 80
D) 82
E) 164
Nice work, nipunranjan!
Here's my full solution (with a minor shortcut at the end)

There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
Since the consecutive odd integers are equally spaced, the mean = median.
So, let's find the mean.

The mean = (SUM of all 82 integers)/82
= (3â�¸ - 1)/82
= (3â�´ - 1)(3â�´ + 1)/82 [factored the difference of squares]
= (81 - 1)(81 + 1)/82
= (80)(82)/82
= 80
= C

Cheers,
Brent

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by Mathsbuddy » Mon Nov 03, 2014 8:30 am
Immediate inspection tells us that the answer cannot be 41 because it must be even (because the median will lie half-way between 2 odd values). This reduces the problem to 4 possible answers.

The solution below contains a small error, but the principle takes me to the right answer.
If we subtract 1 from each integer, we would get 82 consecutive even numbers with sum Z = (3â�¸ - 83)
Half of this would be equivalent to 82 consecutive integers with new sum, S =(3â�¸ - 83)/2 = 3239
If our set started with 1 and ended with N, then S would be N(N+1)/2 = 82(82+1)/2 = 3403 > 3239
3403-3239 = 164 = 82*2
So let's try starting with 0 and ending with (N-1): S = 81*82/2 = 3321
So let's try starting with -1 and ending with (N-2): S = 80*81/2 = 3240 (extremely close to 3239)
The median would be the 41.5th value, i.e. 41.5 -2 = 39.5
Double this and add 1 to return to our original set of consecutive ODD numbers: M = 80

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