Question

The product of unit's digit in $(7^{95}-3^{58})$ and $(7^{95}+3^{58})$:

• A. is cube of $2$
• B. lies between $6$ and $10$
• C. is $6$
• D. lies between $3$ and $6$

Answer 1

Answer: (A), (B)

The correct options are
A is cube of $2$
B lies between $6$ and $10$

Unit's digit in $\left(7^{95}\right)$
$=$Unit's digit in $\left[\right(7^4{)}^{23}\times 7^3]$
$=$Unit's digit in $7^3$ (as unit digit in $7^4=1$)
$=$Unit's digit in 343
Unit's digit in $3^{58}=$ Unit's digit in $(3^4{)}^{14}\times 3^2$ [as unit's digit $3^4=1$]
$=$ Unit's digit is 9
So, Unit's digit in $(7^{95}-3^{58})$
$=$ Unit's digit in (343 -9)
$=$ Unit's digit in 334 $=$ 4
Unit's digit in $(7^{95}+3^{58})=$ Unit's digit in (343 + 9)
$=$ Unit's digit in $352=2$
So the product is $4\times 2=8$