Mathematics - Binomial Theorem Question with Solution | TestHub

We want to find the last two digits of $23^{14}$.

We can rewrite $23^{14}$ as $(23^2{)}^7=(529{)}^7$.

Now, we can express $529$ as $530-1$. So, we have $(530-1{)}^7$.

Using the binomial expansion $(a-b{)}^n={\sum }_{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}(-b{)}^k$:

$$\begin{gathered} \text{displaylines}(530-1{)}^7=(\genfrac{}{}{0ex}{}{7}{0})(530{)}^7-\left(\genfrac{}{}{0ex}{}{7}{1}\right)(530{)}^6+(\genfrac{}{}{0ex}{}{7}{2})(530{)}^5-\left(\genfrac{}{}{0ex}{}{7}{3}\right)(530{)}^4+ \\ (\genfrac{}{}{0ex}{}{7}{4})(530{)}^3-\left(\genfrac{}{}{0ex}{}{7}{5}\right)(530{)}^2+(\genfrac{}{}{0ex}{}{7}{6})(530{)}^1-\left(\genfrac{}{}{0ex}{}{7}{7}\right)(1{)}^7 \end{gathered}$$

We are interested in the last two digits, which means we need to find the value modulo 100.

Notice that any term with $(530{)}^2$ or higher powers of 530 will be a multiple of $530^2=280900$, which is a multiple of 100.

Therefore, we only need to consider the last two terms:

$$-\left(\genfrac{}{}{0ex}{}{7}{5}\right)(530{)}^2+(\genfrac{}{}{0ex}{}{7}{6})(530{)}^1-\left(\genfrac{}{}{0ex}{}{7}{7}\right)(1{)}^7(\mathrm{mod}100)$$

However, since $-\left(\genfrac{}{}{0ex}{}{7}{5}\right)(530{)}^2$ is a multiple of 100, we only need to consider:

$$\left(\genfrac{}{}{0ex}{}{7}{6}\right)(530)-\left(\genfrac{}{}{0ex}{}{7}{7}\right)(1)(\mathrm{mod}100)$$

We know that $\left(\genfrac{}{}{0ex}{}{7}{6}\right)=7$ and $\left(\genfrac{}{}{0ex}{}{7}{7}\right)=1$.

So, the expression becomes:

$$7\times 530-1(\mathrm{mod}100)$$

$$3710-1(\mathrm{mod}100)$$

$$3709(\mathrm{mod}100)$$

The last two digits are $09$.