The product of the last two digits of (1919)^1919 is Answer: 63

Question

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

Answer: 63

Accepted Answer

$$1919^{1919} = (1920-1)^{1919} = \sum_{k=0}^{1919}\binom{1919}{k}(1920)^{1919-k}(-1)^k$$

Modulo 100, all terms with $(1920)^{1919-k}$ for exponent $\ge2$ vanish (since $1920^2$ is divisible by 100). So

$$1919^{1919}\equiv \binom{1919}{1918}(1920)^1(-1)^{1918}+\binom{1919}{1919}(-1)^{1919}\pmod{100}$$

$$\equiv 1919\times1920 - 1 \equiv 3684480-1 = 3684479\pmod{100}$$

The last two digits are $79$. Product of last two digits $=7\times9=63$.

Solutions · 1