Finding last digits - Practice Questions & MCQ

Working rule to find the last digit of an expression

If the given expression is $a^n$ then write the expression in the form of $(10 k \pm 1)^m$, where k and m are positive integers

Now take 10 commons for getting the last digit, 100 for getting the last two digits, 1000 for getting the last 3 digits and so on ...

After expanding $(10 \mathrm{k} \pm 1)^{\mathrm{m}}$ it will look like

$
\begin{aligned}
(10 \mathrm{k} \pm 1)^{\mathrm{m}}= & (10 \mathrm{k})^{\mathrm{m}}+{ }^{\mathrm{m}} \mathrm{C}_1(10 \mathrm{k})^{\mathrm{m}-1}( \pm 1)+{ }^{\mathrm{m}} \mathrm{C}_2(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^2 \\
& +{ }^{\mathrm{m}} \mathrm{C}_3(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^3 \ldots \ldots \ldots+{ }^{\mathrm{m}} \mathrm{C}_{\mathrm{m}-1}(10 \mathrm{k})( \pm 1)^{\mathrm{m}-1}+( \pm 1)^{\mathrm{m}}
\end{aligned}
$
Hence, the number is $10 \alpha+( \pm 1)^m$ (this last part decides the last digit) The number can also be written as $100 \beta+{ }^m C_{m-1}(10 k)( \pm 1)^{m-1}+( \pm 1)^m$ (last 2 terms decide the last 2 digits) $\alpha, \beta$ are Integers.

Finding Remainder Using Binomial Theorem

If the number is given in the form of ' $a^n$ ' and which is divided by ' $b$ '. To find the remainder, adjust the power of ' $a$ ' to $a^m$ such that it is very close to ' $b$ ' with a difference of 1 (i.e. $b+1$ or $b-1$ ).
Also, when the number of the type $7 \mathrm{k}-1$ is divided by 7, the remainder cannot be -1, as the remainder is always positive.
So, in such cases, we have $\frac{7 k-1}{7}=\frac{7 k-7+6}{7}=k-1+\frac{6}{7}$
Hence, the reminder is $6(=7-1)$
For example
If $32^{30}$ is divided by 7, then the remainder is
We know that $32=2^5$, so, $32^{30}$ can be written as

$
\begin{aligned}
\left(2^5\right)^{30} & =2^{150}=\left(2^3\right)^{50}=8^{50}=(7+1)^{50} \\
& =\left[(7)^{50}+{ }^{50} \mathrm{C}_1(7)^{49}+{ }^{50} \mathrm{C}_2(7)^{48}+\ldots+1\right] \\
& =7\left[(7)^{49}+{ }^{50} \mathrm{C}_1(7)^{48}+{ }^{50} \mathrm{C}_2(7) 47+\ldots\right]+1 \\
& =7 \mathrm{k}+1
\end{aligned}
$

$\Rightarrow$ remainder is 1.