Question

The minimum positive integral value of $m$ such that $(1073{)}^{71}-m$ may be divisible by $10$ , is

• A. 1
• B. 3
• C. 7
• D. 9

Answer 1

Answer: (C)

$(1073{)}^{71}-m=(73+1000{)}^{71}-m$
$={}^{71}{C}_0(73{)}^{71}+{}^{71}{C}_1(73{)}^{70}(1000)+{}^{71}{C}_2(73{)}^{69}(1000{)}^2+\dots \dots ..+{}^{41}{C}_{71}(1000{)}^{71}-m$
Above will be divisible by 10 if ${}^{71}{C}_0(73{)}^{71}$ is divisible by $10$
Now ${}^{71}{C}_0(73{)}^{71}=(73{)}^{70}\cdot 73={\left(73^2\right)}^{35}.73$
The last digit of $73^2$ is $9$ , so the last digit of ${\left(73^2\right)}^{35}$ is $9$
$\therefore$ Last digit of ${\left(73^2\right)}^{35}.73$ is $7$
Hence, the minimum positive integral value of $m$ is $7$ , so that it is divisible by $10$ .