Question: The last digit in $a^{2}$ is....

Question

The last digit in $a^{2}$ is.

Answer 1

Solution

We have $nC_{r - 1}:^{n}C_{r} = 6:33$

Now, $\Rightarrow 2n - 13r + 2 = 0$

= $nC_{r}:^{n}C_{r + 1} = 33:110 \Rightarrow 3n - 13r - 10 = 0$

Thus the last digits of $(1 + x)^{12} =$ are $\left\lbrack 1 + 12x + \frac{12.11}{2.1}x^{2} + \frac{12.11.10}{3.2.1}x^{3} + .... \right\rbrack$ i.e., 1.

Question: The last digit in \(a^{2}\) is....

Solution