Solveeit Logo

Question

Question: The remainder of \[{{2005}^{2002}}+{{2002}^{2005}}\] when divided by \(200\) is …………. A) 0 B) 1 ...

The remainder of 20052002+20022005{{2005}^{2002}}+{{2002}^{2005}} when divided by 200200 is ………….
A) 0
B) 1
C) 2
D) 3

Explanation

Solution

To solve the question like this we need to know the concept of binomial theorem. The formula used in binomial expansion is for (a+b)n{{\left( a+b \right)}^{n}} is nC0anb0+nC1an1b1+nC2an2b2.......................nCn1a2bn1+nCna1bn{}^{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 11 with a big natural number as it’s power then value tends to 00 .

Complete step by step solution:
The question asks us to find the remainder when the function 20052002+20022005{{2005}^{2002}}+{{2002}^{2005}} is divided by 200200. To solve this question we need to simplify the function given. Firstly we would simplify20052002{{2005}^{2002}}. On simplification we get:
20052002=(2000+5)2002=2002C02000052002+2002C12000152001........2002C20022000200250{{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 2002C02000052002{}^{2002}{{C}_{0}}{{2000}^{0}}{{5}^{2002}} is not divisible by 200200 , rest all the terms in the sum are divisible by 200200 due to the presence of 20002000 in each term.
Now Remainder =5200<1=\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
(5200)20021{{\left( \dfrac{5}{200} \right)}^{2002}}\ll 1
Similarly in this case two the remainder will become 00 , for (2005)2002{{\left( 2005 \right)}^{2002}}.
On expanding the other term (2002)2005{{\left( 2002 \right)}^{2005}} we get: 20022005=(2000+2)2005=2005C02000022005+2005C12000122004........2005C20052000200550{{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 2005C02000022005{}^{2005}{{C}_{0}}{{2000}^{0}}{{2}^{2005}} will be divisible by 200200 due to the presence of 20002000 in the other terms in the expansion.
So to find the remainder for 20022005{{2002}^{2005}} we will just consider only one term for finding the remainder.
Now Remainder =2200<1\dfrac{2}{200}<1
On putting the power 20052005 on the remainder we get (2200)20051{{\left( \dfrac{2}{200} \right)}^{2005}}\ll 1, which means the remainder will tend to 00 even for this case.
Now on adding remainder for both the function the remainder we get00.
\therefore So the remainder of 20052002+20022005{{2005}^{2002}}+{{2002}^{2005}} when divided by 200200 is A)0A)0.

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