How many integers??

Our set contains consecutive numbers starting from -33, and the sum of its elements is 34. So, our set must contains the following subsets

• A = {all numbers from -33 to -1}
• B = {0}
• C ={all numbers from +1 to +33}, these compensate the sum of the negative numbers from -33 to -1
• D {34}, which ensure that the total sum of all the elements is 34

Therefore our set has |A| + |B| + |C| + |D| = 33 + 1 + 33 + 1 =68 elements

Sn = n/2[2a + (n – 1)d

a = first number = –33 and n = total numbers and d = common difference = 1

n[– 66 + (n – 1)] = 68 solving n² – 67n – 68 = 0

(n – 68)(n + 1) = 0 then n = 67 taking +ve value

Note that |-33|=34-1 We can see that a symmetric sequence around 0 i.e (-33, -32, 31, 30... 31, 32, 33) yields a sum of 0. And simply adding the next term, 34 yields a sum of 34. i.e [(-33)+(-32)+(-31)...+(32)+(33)]+(34)=0+34=34 So the sequence is from -33 to 34. Number of terms is 2 33, plus we need to count the zero and the 34 which gives 2 33 + 2=66+2=68

More rigorously, this is an A.P of common difference one. We have the A.P sum formula S=n(a+l)/2, a and l being the first and last terms respectively. The idea behind this is to find the average term that represents the A.P and multiply it by the number of terms. To illustrate this, consider the A.P 3,4,5,6,7 Adding, we get 3+4+5+6+7=7+5+6+7=12+6+7=18+7=25 But we have the average term (3+7)/2=10/2=5, which when multiplied by 5 yields the same sum 25.

This is because 3 and 7 can be paired, as can 4 and 6. Take 2 from the 7 and give it to the 3 and we get two 5s. Similarly with 4 and 6.

Going back to the problem,

l is the last term, which is a+(n-1)d=a+n-1 [since d=1 for consecutive integers] Giving n(a+(a+n-1))/2=S=34

So n(-33-33+n-1)=68

n^2-67n=68 Solving the quadratic yields n=-1 or n=68. Since number of terms cannot be negative, n=68.

Sorry for the excessively detailed solution, it is my first attempt at writing one and I had hoped to provide some insight with regards to different ways to approaching problems.