\section*{Problem 4 - 041114}

What are the last two digits of the number \(3^{999}-2^{999}\) (in the decimal system)?
}

We first prove the following lemma:

If \(a_{i}\) and \(b_{i}\) are the second to last and the last digit of the natural number \(z_{i}\) in the decimal system, \(i=1,2\), then the second to last and the last digit of \(z_{1} \cdot z_{2}\) coincide with the corresponding digit of \(\left(10 a_{1}+b_{1}\right)\left(10 a_{2}+b_{2}\right)\).

Proof:

By the assumptions, there exist two natural numbers \(c_{1}\) and \(c_{2}\) such that

\[
z_{i}=100 c_{i}+10 a_{i}+b_{i}, \quad i=1,2
\]

This implies

\[
z_{1} \cdot z_{2}=100\left[100 c_{1} c_{2}+\left(10 a_{1}+b_{1}\right) c_{2}+\left(10 a_{2}+b_{2}\right) c_{1}\right]+\left(10 a_{1}+b_{1}\right)\left(10 a_{2}+b_{2}\right)
\]

from which the claim follows immediately.

If we calculate the last two digits of the power \(3^{n}\) for \(n=1,2,3, \ldots 20\), we recognize that the last two digits of \(3^{19}\) are 67 and those of \(3^{20}\) are 01.

Since \(3^{999}=\left(3^{20}\right)^{49} \cdot 3^{19}\), the last two digits of \(3^{999}\) are 67.

By calculating the last two digits of \(2^{m}\) for \(m=1,2,3, \ldots, 22\), we recognize that the last two digits of \(2^{22}\) are 04. It follows that:

\[
2^{999}=\left(2^{22}\right)^{45} \cdot 2^{9}
\]

The last two digits of \(\left(2^{22}\right)^{45}\) are the same as those of \(4^{45}=2^{90}=\left(2^{22}\right)^{4} \cdot 2^{2}\). The last two digits of \(\left(2^{22}\right)^{4}\) are the same as those of \(4^{4}=2^{8}\), which are 56, and the last two digits of \(2^{9}\) are 12.

Therefore, the last two digits of \(2^{999}\) are the same as the last two digits of the product \(56 \cdot 4 \cdot 12\). The last two digits of this product are 88, and thus the last two digits of \(3^{999}-2^{999}\) are 79.

Adapted from \([2]\)