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$22^{22} * 33^{33} * 44^{44} * 55^{55}$

  1. 3
  2. 2
  3. 1
  4. 0
asked in Quantitative Aptitude by (176 points) 2 3 8 | 309 views
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$2^{1}=2 \ ; \ 2^{2} = 4 \ ; \ 2^{3}=8 \ ;\ 2^{4}=16$

$2^{5}=32 \ ; \ 2^{6} = 64 \ ; \ 2^{7}=128 \ ;\ 2^{8}=256$

From above it is clear that the Cyclicity of 2 is 4.

 

$3^{1}=3 \ ; \ 3^{2} =9 \ ; \ 3^{3}=27 \ ;\ 3^{4}=81$

$3^{5}=243 \ ; \ 3^{6} =729 \ ; \ 3^{7}=2187 \ ;\ 3^{8}=6561$

From above it is clear that the Cyclicity of 3 is 4.
 

 

$4^{1}=4 \ ; \ 4^{2} =16 \ ; \ 4^{3}=64 \ ;\ 4^{4}=256$

From above it is clear that the Cyclicity of 4 is 2.
 

 

$5^{1}=5 \ ; \ 5^{2} =25 \ ; \ 5^{3}=125 \ ;\ 5^{4}=625$

From above it is clear that the Cyclicity of 5 is 1.

 

Hence,

Last digit of $22^{2}=4$

Last digit of $33^{3}=3$

Last digit of $44^{4}=6$

Last digit of $55^{5}=5$

 

The last digit of $22^{2} \times 3^{33} \times 44^{4} \times 55^{5}=4 \times 3 \times 6 \times 5 = 360$

 

Hence,The last digit of $22^{2} \times 3^{33} \times 44^{4} \times 55^{5} = 0$

 

Hence, Option(d)0 is the correct choice.
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