Question

The digit at unit's place in the number ${17}^{1995}+{11}^{1995}-7^{1995}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

We have ${17}^{1995}+{11}^{1995}-7^{1995}$

$={(7+10)}^{1995}+{(1+10)}^{1995}-{\left(7\right)}^{1995}=[{\left(7\right)}^{1995}{+}^{1995}{C}_1{\left(7\right)}^{1994}\times 10{+}^{1995}{C}_2{\left(7\right)}^{1993}\times {10}^2+..{+}^{1995}{C}_{1995}\times {10}^{1995}]+[{}^{1995}{C}_0{+}^{1995}{C}_1\times 10{+}^{1995}{C}_2\times {10}^2+...{+}^{1995}{C}_{1995}\times {10}^{1995}]-{\left(7\right)}^{1995}=[{}^{1995}{C}_1{\left(7\right)}^{1994}\times 10+..+{10}^{1995}]+[{}^{1995}{C}_1\times 10+...{+}^{1995}{C}_{1995}\times {10}^{1995}]+1$

$=$ a multiple of $10+1$

Thus, the unit place digit is $1$.