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## Number System: Cyclicity Test-2

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Question 1 |

What will be the last digit of ${{2}^{{{3}^{4}}}}$

8 | |

5 | |

2 | |

4 |

Question 1 Explanation:

\[\begin{align}
& To\text{ }find\text{ }the\text{ }last\text{ }digit\text{ }or\text{ }unit\text{ }digit\text{ }of~{{2}^{{{3}^{4}}}} \\
& \begin{array}{*{35}{l}}
It\text{ }can\text{ }be\text{ }written\text{ }as\text{ }Last\text{ }digit\text{ }of\text{ }{{2}^{81}}. \\
As\text{ }we\text{ }know\text{ }that\text{ }cyclicity\text{ }of\text{ }2\text{ }is\text{ }4. \\
hence\text{ }81/4\text{ }gives\text{ }remainder\text{ }1. \\
Last\text{ }digit\text{ }of\text{ }{{2}^{81}}=\text{ }last\text{ }digit\text{ }of\text{ }{{2}^{1}}=\text{ }2 \\
\end{array} \\
\end{align}\]

Question 2 |

What will be the last digit of ${{2}^{{{34}^{{{54}^{{{56}^{{{78}^{9}}}}}}}}}}$

8 | |

5 | |

6 | |

4 |

Question 2 Explanation:

${{2}^{{{34}^{{{54}^{{{56}^{{{78}^{9}}}}}}}}}}$

2 has the even power and multiple of it .

Hence its nothing but in form of 2^4k.

Now we know that unit digit of 2 has cyclicity of 4 Hence the unit digit of expression.

${{2}^{{{34}^{{{54}^{{{56}^{{{78}^{9}}}}}}}}}}$ is 24K= 24 = 6.

2 has the even power and multiple of it .

Hence its nothing but in form of 2^4k.

Now we know that unit digit of 2 has cyclicity of 4 Hence the unit digit of expression.

${{2}^{{{34}^{{{54}^{{{56}^{{{78}^{9}}}}}}}}}}$ is 24K= 24 = 6.

Question 3 |

What will be the last digit of ${{77}^{{{45}^{67}}}}$

3 | |

0 | |

1 | |

7 |

Question 3 Explanation:

Unit digit of 77 is 7.

And cyclicity of 7 is 4 ,i.e 7, 9, 3, 1.

Hence, last digit of ${{77}^{{{45}^{67}}}}$ will be depends upon remainder of $\frac{{{45}^{67}}}{4}$

We know that all powers of a number ending with 5 will always end with 5.

So 45 raised to any power will end with 5 at the end.

This, when divided by 4, will always leave a remainder of 1.

Thus, the unit digit in this case will be the same as 77

And cyclicity of 7 is 4 ,i.e 7, 9, 3, 1.

Hence, last digit of ${{77}^{{{45}^{67}}}}$ will be depends upon remainder of $\frac{{{45}^{67}}}{4}$

We know that all powers of a number ending with 5 will always end with 5.

So 45 raised to any power will end with 5 at the end.

This, when divided by 4, will always leave a remainder of 1.

Thus, the unit digit in this case will be the same as 77

^{1}, that is 7.Question 4 |

What will be the last digit of ${{3}^{{{4}^{{{6}^{7}}}}}}\times {{6}^{{{3}^{7}}}}\times 1331$

3 | |

2 | |

6 | |

8 |

Question 4 Explanation:

Now we have expression which is multiple of 3 numbers:

Unit digit of ${{3}^{{{4}^{{{6}^{7}}}}}}$as the power of 3^4K, hence the unit digit = 1.

Unit digit of ${{6}^{{{3}^{7}}}}$= as we know cyclicity of 6 is 1 = hence the unit digit = 6.

Unit digit of 1331 =1

Hence the last digit / unit digit of the expression = 1x1x6 =6.

Unit digit of ${{3}^{{{4}^{{{6}^{7}}}}}}$as the power of 3^4K, hence the unit digit = 1.

Unit digit of ${{6}^{{{3}^{7}}}}$= as we know cyclicity of 6 is 1 = hence the unit digit = 6.

Unit digit of 1331 =1

Hence the last digit / unit digit of the expression = 1x1x6 =6.

Question 5 |

What will be the last digit of ${{47}^{{{53}^{{{67}^{64}}}}}}-{{9}^{{{34}^{45}}}}$

2 | |

6 | |

8 | |

none |

Question 5 Explanation:

Unit digit of ${{47}^{{{53}^{{{67}^{64}}}}}}$ = the expression is in for of 47

Hence unit digit of 47^53^67^64 will depend on the ${{47}^{\operatorname{Re}mainder\,\,of}}\frac{{{53}^{{{67}^{64}}}}}{4}$ .

$\frac{{{53}^{{{67}^{64}}}}}{4}$ = will give the remainder 1 as 53/4 = 1 remainder and any power of 1 will remain 1 , so the unit digit of 47

${{9}^{{{34}^{45}}}}$ = ${{9}^{even\,power\,is\,always\,ends\,with\,1}}$Hence the unit digit of expression is 7-1 = 6

So the right answer is (b)

^{odd power}, we know the cyclicity of unit digit of 47 i.e 7 is 4.Hence unit digit of 47^53^67^64 will depend on the ${{47}^{\operatorname{Re}mainder\,\,of}}\frac{{{53}^{{{67}^{64}}}}}{4}$ .

$\frac{{{53}^{{{67}^{64}}}}}{4}$ = will give the remainder 1 as 53/4 = 1 remainder and any power of 1 will remain 1 , so the unit digit of 47

^{1}= 7${{9}^{{{34}^{45}}}}$ = ${{9}^{even\,power\,is\,always\,ends\,with\,1}}$Hence the unit digit of expression is 7-1 = 6

So the right answer is (b)

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