Question

The last digit of $(2137{)}^{754}$ is

• A. $1$
• B. $3$
• C. $7$
• D. $9$

Answer 1

The correct option is D $9$

The last digit of $(2137{)}^{754}$ is same as the last digit of $7^{754}$
Consider $7^{754}=(49{)}^{377}$
$=(50-1{)}^{377}$
$={}^{377}{C}_0(50{)}^{377}-{}^{377}{C}_1(50{)}^{376}+\cdots +{}^{377}{C}_{376}(50{)}^1-{}^{377}{C}_{377}(50{)}^0$
$=50k-1,k\in N$
For the last digit, we calculate $\left\{\frac{7^{754}}{10}\right\}$
$\left\{\frac{7^{754}}{10}\right\}=\left\{\frac{50k-1}{10}\right\}$
$=\left\{\frac{10(5k-1)+9}{10}\right\}=9$

Alternate:
End digit of $7^{4n+1}$ is $7$
End digit of $7^{4n+2}$ is $9$
End digit of $7^{4n+3}$ is $3$
End digit of $7^{4n}$ is $1$
$\therefore (2137{)}^{754}=(2137{)}^{4\left(188\right)+2}$
$\therefore$ End digit of $(2137{)}^{754}$ will be same as that of $7^{4n+2}$ i.e., $9.$