Solve y = x2 +11 for x.
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Considering that the powers of 7 follow a pattern, it is found that the last two digits of are 43.
The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for , we have to look at the remainder of the division by 4:
- If the remainder is of 1, the last two digits are 07.
- If the remainder is of 2, the last two digits are 49.
- If the remainder is of 3, the last two digits are 43.
- If the remainder is of 0, the last two digits are 01.
In this problem, we have that n = 1867, and the remainder of the division of 1867 by 4 is of 3, hence the last two digits of are 43.
More can be learned about the powers of 7 pattern at brainly.com/question/10598663
18 students are 1/10 of the number of seventh graders 
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First choice.