Question

What are the last two digits of 7^1867? ​

Answer 1

Considering that the powers of 7 follow a pattern, it is found that the last two digits of $7^{1867}$ are 43.

What is the powers of 7 pattern?

The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for $7^n$, 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 $7^{1867}$ are 43.

More can be learned about the powers of 7 pattern at brainly.com/question/10598663