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

To obtain the product of the last two digits of $$(1919)^{1919}$$ we must first find those two digits, i.e. evaluate $$(1919)^{1919} \pmod{100}$$.

Step 1 : Reduce the base modulo 100
$$1919 \equiv 19 \pmod{100}$$
Hence $$ (1919)^{1919} \equiv 19^{1919} \pmod{100} $$.

Step 2 : Reduce the exponent using Euler’s theorem
Since $$\gcd(19,100)=1$$, Euler’s theorem gives $$19^{\phi(100)} \equiv 1 \pmod{100}$$.
Factorising $$100 = 2^{2}\cdot5^{2}$$, $$\phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40$$.

Therefore the powers of 19 repeat every 40:
$$19^{40} \equiv 1 \pmod{100}$$.

Now reduce the exponent 1919 modulo 40:
$$1919 = 40\times47 + 39 \quad\Longrightarrow\quad 1919 \equiv 39 \pmod{40}$$.

Thus
$$19^{1919} \equiv 19^{39} \pmod{100} \quad -(1)$$

Step 3 : Shorten the exponent further (optional but quicker)
We shall show that $$19^{10} \equiv 1 \pmod{100}$$, which will cut the power down even more.

Compute successive powers of 19 (always keeping only the last two digits):
$$19^{1}=19$$
$$19^{2}=361\equiv61$$
$$19^{3}=61\cdot19=1159\equiv59$$
$$19^{4}=59\cdot19=1121\equiv21$$
$$19^{5}=21\cdot19=399\equiv99$$
$$19^{6}=99\cdot19=1881\equiv81$$
$$19^{7}=81\cdot19=1539\equiv39$$
$$19^{8}=39\cdot19=741\equiv41$$
$$19^{9}=41\cdot19=779\equiv79$$
$$19^{10}=79\cdot19=1501\equiv01\equiv1$$

Since $$19^{10}\equiv1\pmod{100}$$, rewrite the required power:

$$19^{39}=19^{10\cdot3}\cdot19^{9}\equiv(19^{10})^{3}\cdot19^{9}\equiv1^{3}\cdot19^{9}\equiv19^{9}\pmod{100}$$.

Step 4 : Evaluate $$19^{9}\pmod{100}$$ (already calculated above)
From the list, $$19^{9}\equiv79\pmod{100}$$.
Hence the last two digits of $$(1919)^{1919}$$ are 7 and 9.

Step 5 : Required product
Product of the digits $$7 \times 9 = 63$$.

Final Answer : 63