Question

${19^{2005}} + {11^{2005}} - {9^{2005}}$ is a number. What is the unit digit of this number?
A. $2$
B. $1$
C. $0$
D. $8$
E. $9$

Answer 1

Hint: Break down all the numbers, 19,11, and 9 in terms of 1 and another number. This will result in each number being in the form of either $(1 + x)$ or $(1 - x)$ respectively. Apply binomial expansions of ${(1 + x)^n}$ and $(1 - x){}^n$ to proceed further

Formula used:
By binomial expansions we have:
${(1 + x)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^r}} = 1 + {}^n{C_1}x + \ldots + {}^n{C_n}x{}^n$
${(1 - x)^n} = \sum\limits_{r = 0}^n {{{( - 1)}^r}{}^n{C_r}{x^r}} = 1 - {}^n{C_1}x + \ldots + {( - 1)^n}{}^n{C_n}x{}^n$

Complete step by step Solution:
The given number is:
${19^{2005}} + {11^{2005}} - {9^{2005}}$
Breaking down the numbers,
${19^{2005}} + {11^{2005}} - {9^{2005}} = {(20 - 1)^{2005}} + {(10 + 1)^{2005}} - {(10 - 1)^{2005}}$
Taking the minus sign outside,
${19^{2005}} + {11^{2005}} - {9^{2005}} = - {(1 - 20)^{2005}} + {(1 + 10)^{2005}} + {(1 - 10)^{2005}}$ … (1)
By binomial expansions we have:
${(1 + x)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^r}} = 1 + {}^n{C_1}x + \ldots + {}^n{C_n}x{}^n$ … (2)
${(1 - x)^n} = \sum\limits_{r = 0}^n {{{( - 1)}^r}{}^n{C_r}{x^r}} = 1 - {}^n{C_1}x + \ldots + {( - 1)^n}{}^n{C_n}x{}^n$ … (3)
Using these binomial expansions in equation (1),
${19^{2005}} = - (1 - 20{}^{2005}{C_1} + {20^2}{}^{2005}{C_2} - \ldots - {20^{2005}}{}^{2005}{C_{2005}})$ … (4)
${11^{2005}} = (1 + 10{}^{2005}{C_1} + {10^2}{}^{2005}{C_2} + \ldots + {10^{2005}}{}^{2005}{C_{2005}})$ … (5)
$ - {9^{2005}} = (1 - 10{}^{2005}{C_1} + {10^2}{}^{2005}{C_2} - \ldots - {10^{2005}}{}^{2005}{C_{2005}})$ … (6)
On adding equation (5) and equation (6), all the even terms cancel out. This gives
${11^{2005}} - {9^{2005}} = 2(1 + {10^2}{}^{2005}{C_2} + {10^4}{}^{2005}{C_4} + \ldots + {10^{2004}}{}^{2005}{C_{2004}})$ … (7)
Adding equation (4) and equation (7),
${19^{2005}} + {11^{2005}} - {9^{2005}} = - (1 - 20{}^{2005}{C_1} + {20^2}{}^{2005}{C_2} - \ldots - {20^{2005}}{}^{2005}{C_{2005}}) + 2(1 + {10^2}{}^{2005}{C_2} + \ldots + {10^{2004}}{}^{2005}{C_{2004}})$
Simplifying further,
${19^{2005}} + {11^{2005}} - {9^{2005}} = 2 - 1 - ( - 20{}^{2005}{C_1} + {20^2}{}^{2005}{C_2} - \ldots - {20^{2005}}{}^{2005}{C_{2005}}) + 2({10^2}{}^{2005}{C_2} + \ldots + {10^{2004}}{}^{2005}{C_{2004}})$
Now all the terms in the brackets result in terms having the unit digits as 0, which makes (2-1) the unit digit.
Thus, the unit digit is $2 - 1 = 1$ .

Therefore, the correct option is B.

Note: In the above question it is important to note that the terms apart from (2-1), all have 20 or 10 as their multiples, which makes 10 their common multiple. One of the important properties of the number 10 is that when multiplied by any number, it adds zeroes on the beginning digits accordingly. This property makes sure that those terms will have a zero at the unit digit, making (2-1) the number deciding the unit digit respectively.