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Q.
The minimum positive integral value of $m$ such that $(1073)^{71}- m$ may be divisible by $10$ , is
Binomial Theorem
Solution:
$(1073)^{71}- m =(73+1000)^{71}- m$
$={ }^{71} C _{0}(73)^{71}+{ }^{71} C _{1}(73)^{70}(1000)+{ }^{71} C _{2}(73)^{69}(1000)^{2}+\ldots \ldots . .+{ }^{41} C _{71}(1000)^{71}- m$
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 $10$ .