Question

If $x$ and $y$ are the number of possibilities that $A$ can assume such that the unit digit of A and $A^3$ are same and the unit digit of $A^2$ and $A^3$ are same respectively ,then the value of $x-y$ is (where $A$ is a single digit number)

• A. $4$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (B)

$2$

Here, according to the question

if $A=1$, then $A^3=1$,

We see that unit digit number of $A$ and $A^3$ are same.

for, $A=2\Rightarrow A^3=8,$, unit digits are not same

$A=3\Rightarrow A^3=27,$ unit digits are not same

$A=4\Rightarrow A^3=64,$ unit digits are same

$A=5\Rightarrow A^3=125,$ unit digits are same

$A=6\Rightarrow A^3=216,$ unit digits are not same

$A=7\Rightarrow A^3=343,$ unit digits are not same

$A=8\Rightarrow A^3=512,$ unit digits are not same

$A=8\Rightarrow A^3=512,$ unit digits are same

So, there are five possible solutions where unit digits of $A$ and $A^3$ are same.

$\therefore x=5$

Again, if we take $A=1,A^2=1,A^3=1$, unit digits are same

$A=2\Rightarrow A^2=4,A^3=8,$ unit digits are not same

$A=3\Rightarrow A^2=9,A^3=27,$ unit digits are not same

$A=4\Rightarrow A^2=16,A^3=64,$ unit digits are not same

$A=5\Rightarrow A^2=25,A^3=125,$ unit digits are same

$A=6\Rightarrow A^2=36,A^3=216,$ unit digits are same

$A=7\Rightarrow A^2=49,A^3=343,$ unit digits are not same

$A=8\Rightarrow A^2=64,A^3=512,$ unit digits are not same

$A=9\Rightarrow A^2=81,A^3=729$ unit digits are not same

Here, we can see that there are three possible solutions where unit digits of $A^2,A^3$ are same.

So, $y=3$

Hence, $x-y=5-3=2.$