Question
Question: The remainder of \[{{2005}^{2002}}+{{2002}^{2005}}\] when divided by \(200\) is …………. A) 0 B) 1 ...
The remainder of 20052002+20022005 when divided by 200 is ………….
A) 0
B) 1
C) 2
D) 3
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 is nC0anb0+nC1an−1b1+nC2an−2b2.......................nCn−1a2bn−1+nCna1bn . 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 20052002+20022005 is divided by 200. To solve this question we need to simplify the function given. Firstly we would simplify20052002. On simplification we get:
20052002=(2000+5)2002=2002C02000052002+2002C12000152001........2002C20022000200250
On analysing we can see that only the first value in the sum which is 2002C02000052002 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 =2005<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
(2005)2002≪1
Similarly in this case two the remainder will become 0 , for (2005)2002.
On expanding the other term (2002)2005 we get: 20022005=(2000+2)2005=2005C02000022005+2005C12000122004........2005C20052000200550.
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 will be divisible by 200 due to the presence of 2000 in the other terms in the expansion.
So to find the remainder for 20022005 we will just consider only one term for finding the remainder.
Now Remainder =2002<1
On putting the power 2005 on the remainder we get (2002)2005≪1, which means the remainder will tend to 0 even for this case.
Now on adding remainder for both the function the remainder we get0.
∴ So the remainder of 20052002+20022005 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 (a+b)n is (n+1). 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.