Here's a challenge question for students to try: Set T consists of 82 CONSECUTIVE ODD integers. If the sum of the integers is 3� - 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

3� - 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)

Set T consists of 82 CONSECUTIVE ODD integers. If the sum of the integers is 3� - 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 = (3� - 1)/82 = (3� - 1)(3� + 1)/82 [factored the difference of squares] = (81 - 1)(81 + 1)/82 = (80)(82)/82 = 80 = C Cheers, Brent

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 = (3� - 83) Half of this would be equivalent to 82 consecutive integers with new sum, S =(3� - 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