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Best answer
11 votes
11 votes

We know that

For unit digit $1$,

  • ${1}^{n}=1$ where $n$ is whole number

For unit digit $2,$

  • $2^{1}=2$
  • $2^{2}=4$
  • $2^{3}=8$
  • $2^{4}=16$
  • $2^{5}=3{\color{Red}{2}}$
  • Cyclicity of $2$ is $4.$

For unit digit $3,$

  • $3^{1}=3$
  • $3^{2}=9$
  • $3^{3}=27$
  • $3^{4}=81$
  • $3^{5}=24{\color{Magenta}{3}}$
  • Cyclicity of $3$ is $4.$

For unit digit $4,$

  • $4^{1}=4$
  • $4^{2}=16$
  • $4^{3}=64$
  • $4^{4}=256$
  • $4^{5}=102{\color{Blue}{4}}$
  • Cyclicity of $4$ is $2.$

For unit digit $6,$

  • $6^{1}=6$
  • $6^{2}=36$
  • We can write last digit of $6$ is $6^{n}=6$ where $n\in N$


$(2171)^{7}+(2172)^{9}+(2173)^{11}+(2174)^{13}$

We want to find last digit, so we can focus only last digit.

$\implies(1)^{7}+(2)^{9}+(3)^{11}+(4)^{13}$

$\implies 1+(2^{4})^{2}\cdot 2^{1}+(3^{4})^{2}\cdot 3^{3}+(4)^{odd}$

$\implies 1+(6)^{2}\cdot 2^{1}+(1)^{2}\cdot 3^{3}+4$

$\implies 1+6\cdot 2+ 7+4$

$\implies 1+2+ 7+4$

$\implies 14$

So, last digit is $4$

The answer $(B)$ is the correct choice

selected by
2 votes
2 votes
2171 (last digit 1) -> 1^n=1

2172 (last digit 2) -> 2 repeats its last digits as : 2,4,8,6,2..... => 9=4+4+1 => last digit = 2

2173 (last digit 3) -> 3 repeats its last digits as : 3,9,7,1,3..... => 11=4+4+3 => last digit = 7

2174 (last digit 4) -> 4 repeats its last digits as : 4,6,4..... => 13=2*6 + 1 => last digit = 4

1 + 2 + 7 + 4 =14

Answer : b
2 votes
2 votes
Divide all the powers by 4
2171^7=If you divide 7÷4 you get remainder 3. Now take units digit of 2171 i.e 1. Now take the remainder 3 as power of 1 i.e 1^3=3.
Now take 2172^9 divide 9 by 4 you get remainder as 1 then 2^1
So if you divide all the powers by 4 you get like this 1^3+2^1+3^3+4^1=34 we need units place so answer is 4
Option (b) is correct
Answer:

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