Question

Let two distinct numbers $a$ and $b$ are selected from the set $\{1,2,3,\dots ,9,10\}.$ Then the probability that the last digit of the number $a^b$ will be $6,$ is

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $\frac{1}{45}$
• D. $\frac{8}{45}$

Answer 1

The correct option is D $\frac{8}{45}$

If $a=6$, then the last digit of $a^b$ will always be $6$.
So, for $a=6$, the probability is $\frac{1}{10}\times \frac{9}{9}=\frac{1}{10}$

Last digit of the numbers $1^b,3^b,5^b,7^b,9^b,{10}^b$ will never be $6.$

Now, last digit of the number $2^b$ will be $6$, if $b$ is a multiple of $4.$
So, for $a=2$, the probability is $\frac{1}{10}\times \frac{2}{9}=\frac{1}{45}$

Last digit of the number $4^b$ will be $6$, if $b$ is a multiple of $2.$
So, for $a=4$, the probability is $\frac{1}{10}\times \frac{4}{9}=\frac{2}{45}$

Last digit of the number $8^b$ will be $6$, if $b$ is a multiple of $4.$
So, for $a=8$, the probability is $\frac{1}{10}\times \frac{1}{9}=\frac{1}{90}$

$\therefore$ Required probability
$=\frac{1}{10}+\frac{1}{45}+\frac{2}{45}+\frac{1}{90}$
$=\frac{9+2+4+1}{90}$
$=\frac{16}{90}=\frac{8}{45}$