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Question

Mathematics Question on Binomial theorem

The remainder when 3100×25003^{100} \times 2^{500} is divided by 55 is

A

1

B

2

C

3

D

4

Answer

1

Explanation

Solution

31=3,32=9,33=27,34=81,35=2433^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81,3^{5}=243 Now, for power of 3 , the unit digit keep on repeating after power difference of 4 , So, unit digit for 34=38=312=.31003^{4}=3^{8}=3^{12}=\ldots .3^{100}. So unit digit is 1 21=2,22=4,23=8,24=16,25=322^{1}=2,2^{2}=4,2^{3}=8,2^{4}=16,2^{5}=32 Now, for power of 2 , the unit digit keep on repeating after power difference of 5 , So, unit digit for 24=28=212=.25002^{4}=2^{8}=2^{12}=\ldots .2^{500}.So, unit digit is 6 . So, the last digit of the product will be 6 . Dividing by 5 we will have 1 as a remainder.