Explanation:

The wording might be considered a little ambiguous. I will find the smallest number which can be written as the sum of 1111 and of 1010 consecutive integers. If the first of the 10 is a, then the sum is 10a+45. If the first of the 11 is b, then the sum is 11b+55. Therefore,

10a = 11b+10, and we can conclude two things: 10|b, 11|a−1.

The values of a and b which produce the smallest sum are 1 and 0 respectively and the common sum is 55, which by default is the sum of consecutive integers.

Hence, option 2 is the correct answer.