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Number System: Cyclicity Test-2

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Question 1
What will be the last digit of ${{2}^{{{3}^{4}}}}$
A
8
B
5
C
2
D
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 ${{2}^{{{34}^{{{54}^{{{56}^{{{78}^{9}}}}}}}}}}$
A
8
B
5
C
6
D
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.
Question 3
What will be the last digit of ${{77}^{{{45}^{67}}}}$
A
3
B
0
C
1
D
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 771, that is 7.
Question 4
What will be the last digit of ${{3}^{{{4}^{{{6}^{7}}}}}}\times {{6}^{{{3}^{7}}}}\times 1331$
A
3
B
2
C
6
D
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.
Question 5
What will be the last digit of ${{47}^{{{53}^{{{67}^{64}}}}}}-{{9}^{{{34}^{45}}}}$
A
2
B
6
C
8
D
none
Question 5 Explanation: 
Unit digit of ${{47}^{{{53}^{{{67}^{64}}}}}}$ = the expression is in for of 47odd 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 471 = 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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