(a) Find the last two digits of $9*7*8$.

(b) Find the last two digits of $2^{3^4}$. (By convention, exponent towers are evaluated from the top down, so $2^{3^4} = 2^{(3^4)}$.)

1 answer

### Part (a)
To find the last two digits of \( 9 \times 7 \times 8 \), we compute the product modulo \( 100 \):

\[
9 \times 7 = 63
\]

Next, we multiply \( 63 \times 8 \):

\[
63 \times 8 = 504
\]

To find the last two digits of \( 504 \), take \( 504 \mod 100 \):

\[
504 \mod 100 = 4
\]

So, the last two digits of \( 9 \times 7 \times 8 \) are:

\[
\boxed{04}
\]

### Part (b)
To find the last two digits of \( 2^{3^4} \), we need to compute \( 2^{3^4} \mod 100 \). First, we determine \( 3^4 \):

\[
3^4 = 81
\]

Now, we must find the last two digits of \( 2^{81} \). We use Euler's theorem and the properties of modular arithmetic.
Euler's theorem states that for any integer \( a \) co-prime to \( n \):

\[
a^{\phi(n)} \equiv 1 \mod n
\]

where \( \phi(n) \) is the Euler's totient function. For \( n = 100 \):

\[
\phi(100) = 100 \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{5} \right) = 100 \times \frac{1}{2} \times \frac{4}{5} = 40
\]

Since \( 2 \) and \( 100 \) are co-prime:

\[
2^{40} \equiv 1 \mod 100
\]

We can express \( 81 \) in terms of a multiple of \( 40 \):

\[
81 = 2 \times 40 + 1
\]

So,

\[
2^{81} = 2^{2 \times 40 + 1} = (2^{40})^2 \times 2^1 \equiv 1^2 \times 2 \equiv 2 \mod 100
\]

Thus, the last two digits of \( 2^{81} \) are:

\[
\boxed{02}
\]