Question

The remainder of \[{{2005}^{2002}}+{{2002}^{2005}}\] when divided by $200$ is ………….
A) 0
B) 1
C) 2
D) 3

Answer 1

Hint: To solve the question like this we need to know the concept of binomial theorem. The formula used in binomial expansion is for ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}.......................{}^{n}{{C}_{n-1}}{{a}^{2}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{1}}{{b}^{n}}$ . It is used to describe the algebraic expansion of powers of a binomial. A number less than $1$ with a big natural number as it’s power then value tends to $0$ .

Complete step by step solution:
The question asks us to find the remainder when the function \[{{2005}^{2002}}+{{2002}^{2005}}\] is divided by \[200\]. To solve this question we need to simplify the function given. Firstly we would simplify\[{{2005}^{2002}}\]. On simplification we get:
${{2005}^{2002}}={{\left( 2000+5 \right)}^{2002}}={}^{2002}{{C}_{0}}{{2000}^{0}}{{5}^{2002}}+{}^{2002}{{C}_{1}}{{2000}^{1}}{{5}^{2001}}........{}^{2002}{{C}_{2002}}{{2000}^{2002}}{{5}^{0}}$
On analysing we can see that only the first value in the sum which is ${}^{2002}{{C}_{0}}{{2000}^{0}}{{5}^{2002}}$ is not divisible by $200$ , rest all the terms in the sum are divisible by $200$ due to the presence of $2000$ in each term.
Now Remainder \[=\dfrac{5}{200}<1\]
If a positive number like 2002 is powered on a number less than 1 then the value of that number tends to zero, like
\[{{\left( \dfrac{5}{200} \right)}^{2002}}\ll 1\]
 Similarly in this case two the remainder will become $0$ , for \[{{\left( 2005 \right)}^{2002}}\].
On expanding the other term \[{{\left( 2002 \right)}^{2005}}\] we get: \[{{2002}^{2005}}={{\left( 2000+2 \right)}^{2005}}={}^{2005}{{C}_{0}}{{2000}^{0}}{{2}^{2005}}+{}^{2005}{{C}_{1}}{{2000}^{1}}{{2}^{2004}}........{}^{2005}{{C}_{2005}}{{2000}^{2005}}{{5}^{0}}\].
Expansion is done with the help of binomial expansion, in which we need to know about the combination.
Here again all the terms except \[{}^{2005}{{C}_{0}}{{2000}^{0}}{{2}^{2005}}\] will be divisible by \[200\] due to the presence of $2000$ in the other terms in the expansion.
So to find the remainder for \[{{2002}^{2005}}\] we will just consider only one term for finding the remainder.
Now Remainder =\[\dfrac{2}{200}<1\]
On putting the power $2005$ on the remainder we get \[{{\left( \dfrac{2}{200} \right)}^{2005}}\ll 1\], which means the remainder will tend to $0$ even for this case.
Now on adding remainder for both the function the remainder we get$0$.
$\therefore $ So the remainder of \[{{2005}^{2002}}+{{2002}^{2005}}\] when divided by $200$ is $A)0$.

Note: If the power is negative the term results in a big number. Total number of terms in the expansion of ${{\left( a+b \right)}^{n}}$ is $\left( n+1 \right)$. The sum of the indices of $a$ and $b$ in each term is $n$ . It is a correct expansion when the terms are complex numbers. Terms that are equidistant from both ends will have coefficients that are equal.