Question

The digit at the unit place in the number ${19}^{2005}+{11}^{2005}-9^{2005}$ is:

• A. $2$
• B. $1$
• C. $0$
• D. $8$
• E. $9$

Answer 1

The correct option is B

$1$

Explanation of the correct answer:

The given expression is: ${19}^{2005}+{11}^{2005}-9^{2005}$

Binomial expansion is given as${\left(a+b\right)}^{\mathbf{n}}\mathbf{=}\mathbf{1}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{n}}{\mathbf{b}}^{\mathbf{0}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{1}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{b}}^{\mathbf{1}}\mathbf{+}\mathbf{\cdots }\mathbf{\cdots }\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}{\mathbf{a}}^{\mathbf{0}}{\mathbf{b}}^{\mathbf{n}}$

Simplifying the given expression

$$\begin{gathered} {19}^{2005}+{11}^{2005}-9^{2005} \\ ={11}^{2005}-9^{2005}+{19}^{2005} \\ ={(1+10)}^{2005}-{(10-1)}^{2005}+{\left(20-1\right)}^{2005} \\ =\left[{(1+10)}^{2005}+{(1-10)}^{2005}\right]-{\left(1-20\right)}^{2005} \end{gathered}$$

Applying binomial expansion, we get

$=2\left(1+{}^{2005}{C}_0{10}^2+{}^{2005}{C}_1{10}^4+...+{}^{2005}{C}_{2004}{10}^{2004}\right)-\left(1-{}^{2005}{C}_1{20}^1+{}^{2005}{C}_2{20}^2+\cdots \cdots \cdots \cdots +{}^{2005}{C}_{2005}{20}^{2005}\right)$

Taking ${10}^2$ common out from the first expansion and taking ${20}^1$ from the second term, now the entire expression can be simplified to

$$\begin{gathered} =2(1+100k)-(1+20m)\mathbf{(}\mathbf{\text{where}}\mathbf{}\mathit{m}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathit{k}\mathbf{}\mathbf{\text{are positive integers}}\mathbf{)} \\ =2+200k-1-20m \\ =(200k-20m)+(2-1) \end{gathered}$$

Therefore, the unit's digit is $\Rightarrow 2-1=1$

Hence, the correct answer is Option (B).