Step-by-step explanation:
To find the last two digits of \(7^{1867}\), we can use modular arithmetic.
First, we notice that for any integer \(n\), \(7^n\) will end in the same digit as \(7^{\text{mod } 4}\), because the last digit of \(7^n\) repeats in cycles of 4.
So, let's find \(7^{\text{mod } 4}\):
\[7^1 = 7\]
\[7^2 = 49\]
\[7^3 = 343\]
\[7^4 = 2401\]
Thus, \(7^{\text{mod } 4} = 1\).
Next, we know that \(7^{1867} = (7^{4})^{466} \times 7^3\).
We already know that \(7^{4}\) ends with 01. So, we just need to find the last two digits of \(7^3\):
\[7^3 = 343\]
Therefore, the last two digits of \(7^{1867}\) are 43.
