Question

Write the units digit of 9⁹⁹.

with full explanation!​

Answer 1

We've to find unit digit of $9^{99}.$

We can write $9^{99}$ as,

$\longrightarrow9^{99}=(10-1)^{99}$

According to binomial theorem,

• $\displaystyle (a-b)^n=\sum_{r=0}^n\ (-1)^r\cdot\,^n\!C_r\,a^{n-r}\,b^r$

Taking $a=10,\ b=1,\ n=99,$

$\displaystyle\longrightarrow 9^{99}=\sum_{r=0}^{99}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{99-r}\cdot 1^r$

$\displaystyle\longrightarrow 9^{99}=\sum_{r=0}^{99}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{99-r}$

Let me take final term in the sum (99th term) out of it.

$\displaystyle\longrightarrow 9^{99}=\left(\sum_{r=0}^{98}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{99-r}\right)+(-1)^{99}\cdot\,^{99}\!C_{99}\cdot10^{99-99}$

$\displaystyle\longrightarrow 9^{99}=\left(\sum_{r=0}^{98}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{98-r+1}\right)-1$

$\displaystyle\longrightarrow 9^{99}=\left(\sum_{r=0}^{98}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{98-r}\cdot10\right)-1$

We can take that 10 common out of the sum.

$\displaystyle\longrightarrow 9^{99}=10\left(\sum_{r=0}^{98}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{98-r}\right)-1$

Taking $\displaystyle\sum_{r=0}^{98}\ (-1)^r\cdot\,^{99}\!C_r\cdot10^{98-r}=k,$

$\displaystyle\longrightarrow 9^{99}=10k-1$

$\displaystyle\longrightarrow 9^{99}=10(k-1+1)-1$

$\displaystyle\longrightarrow 9^{99}=10k-10+10-1$

$\displaystyle\longrightarrow 9^{99}=10(k-1)+10-1$

$\displaystyle\longrightarrow 9^{99}=10(k-1)+9$

Now $9^{99}$ is in the form of $10x+y$ whose unit digit is $y.$

Here $x=k-1$ and $y=9.$

Thus we can conclude that the unit digit of $9^{99}$ is $\bf{9}.$

Hence 9 is the answer.

Answer 2

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}} </p><p>$

$\displaystyle{9}^{99}$

$= \displaystyle{(10 - 1)}^{99}$

On Expansion we get

$= {10}^{99} - \binom{99}{1} {10}^{98} + \binom{99}{2} {10}^{97} + .......... + \binom{99}{98} {10}^{1} - \binom{99}{99} \times 1$

$= 10 \{ {10}^{98} - \binom{99}{1} {10}^{97} + \binom{99}{2} {10}^{96} + .......... + \binom{99}{98} \}- 1$

Since the expression

$10 \{ {10}^{98} - \binom{99}{1} {10}^{97} + \binom{99}{2} {10}^{96} + .......... + \binom{99}{98} \}$

is a multiple of 10. So the unit digit of the expression

$10 \{ {10}^{98} - \binom{99}{1} {10}^{97} + \binom{99}{2} {10}^{96} + .......... + \binom{99}{98} \}$

must be 0

Hence the unit digit of the expression

$10 \{ {10}^{98} - \binom{99}{1} {10}^{97} + \binom{99}{2} {10}^{96} + .......... + \binom{99}{98} \}- 1$

must be 9

So the unit digit of ${9}^{99}$ is 1