Question

The product of the last two digits of $ (1919)^{1919} $ is:

Hint:

When computing large powers modulo 100, use Euler's theorem to reduce the exponent and then compute the smaller power directly. This can simplify the calculations significantly.

Answer 1

Correct Answer: 63

Since \(1919 \equiv 19 \pmod{100}\), we consider \(19^{19} \pmod{100}\). We can use Euler's totient theorem.
\(\phi(100) = \phi(2^2 \cdot 5^2) = 100 \cdot \left(1 - \frac{1}{2}\right) \cdot \left(1 - \frac{1}{5}\right) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40\).

Since \(\gcd(19, 100) = 1\), by Euler's totient theorem, \(19^{40} \equiv 1 \pmod{100}\).
\(1919 = 40 \cdot 47 + 39\)
So, \(19^{1919} \equiv 19^{40 \cdot 47 + 39} \equiv (19^{40})^{47} \cdot 19^{39} \equiv 1^{47} \cdot 19^{39} \equiv 19^{39} \pmod{100}\).
Now we need to find \(19^{39} \pmod{100}\).
Note that \(19^2 = 361 \equiv 61 \pmod{100}\).
\(19^4 \equiv 61^2 \equiv 3721 \equiv 21 \pmod{100}\)
\(19^8 \equiv 21^2 \equiv 441 \equiv 41 \pmod{100}\)
\(19^{16} \equiv 41^2 \equiv 1681 \equiv 81 \pmod{100}\)
\(19^{32} \equiv 81^2 \equiv 6561 \equiv 61 \pmod{100}\)

Then, \(19^{39} = 19^{32+4+2+1} = 19^{32} \cdot 19^4 \cdot 19^2 \cdot 19 \equiv 61 \cdot 21 \cdot 61 \cdot 19 \pmod{100}\)
\(\equiv (61 \cdot 21) \cdot (61 \cdot 19) \pmod{100}\)
\(61 \cdot 21 = 1281 \equiv 81 \pmod{100}\)
\(61 \cdot 19 = 1159 \equiv 59 \pmod{100}\)
\(19^{39} \equiv 81 \cdot 59 \pmod{100}\)
\(\equiv 4779 \equiv 79 \pmod{100}\)

The last two digits are 79.
The product of the last two digits is \(7 \cdot 9 = 63\).