Question

The remainder on dividing $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} $ by 2000 is.
a) 1
b) 1000
c) 100
d) 0

Answer 1

Hint: The given expression is a sum of exponential functions depending on ‘n’. Here we have to find the remainder after dividing it by 2000. We can check it to be zero after putting n=0,1 but after that it is difficult as well as time consuming to calculate the remainder for n=2,3…and so on. So the primary idea is to check if it is divisible by 2000. We will use the fact that $ {a^n} - {b^n} $ is divisible by $ a - b $ .

Complete step-by-step answer:
The given expression is $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} $
We will use the result that $ {a^n} - {b^n} $ is divisible by $ a - b $ .
 $ \Rightarrow {121^n} - {25^n} + {1900^n} - {( - 4)^n} $
 $ = {121^n} - {( - 4)^n} + {1900^n} - {25^n} $
 $ {121^n} - {( - 4)^n} + {1900^n} - {25^n} $ is divisible by $ \left[ {121 - ( - 4)} \right] + \left[ {1900 - 25} \right] $
 $ = \left[ {121 + 4} \right] + \left[ {1900 - 25} \right] $
 $ = 125 + 1875 = 2000 $
Clearly, 2000 is divisible by the number 2000.
Let , $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} = 2000 \times a $ , where ‘a’ is any integer.
Therefore, $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} $ is divisible by 2000.
So, the remainder after dividing $ 2000 \times a $ with 2000 will give remainder as ‘0’.
This means the remainder on dividing $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} $ by 2000 is 0.
Thus option ‘d’ is correct among all the other choices.
So, the correct answer is “Option D”.

Note: You can check the result to be true by taking n=0 or n=1. For n=0 we have the expression $ {121^n} - {25^n} + {1900^n} - {( - 4)^n} $ as zero which will generate the remainder 0 after division with 2000, while for calculating it for n=1 you will get the value as 2000. You can check it yourself. While calculating the remainder always try to factor the expression to a simpler form because it may lead to finding an answer as zero.