Question

Find the remainder when ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$ is divided by 2000.
A. 1 B. 1000 C. 100 D. 0

Answer 1

Hint: We use the factored form of ${{a}^{n}}-{{b}^{n}}$ to get the divisors of the total form. We break 2000 into two different co-primes factors to get the factors. Then we use the 4 different given numbers to make the co-primes. The co-primes divide the total function. So, their multiplication will also divide the function. As it’s totally divisible we can easily find the remainder.

Complete step-by-step answer:
We can see every individual term in the given function is of the form ${{x}^{n}}$. So, we try to take two terms at a time to find the factored form of ${{a}^{n}}-{{b}^{n}}$. The factored form will be ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+...+a{{b}^{n-2}}+{{b}^{n-1}} \right)$.
So, we can tell that $\left( a-b \right)$ divides ${{a}^{n}}-{{b}^{n}}$.
Now we break 2000 into multiple of two co-primes. So, $2000=16\times 125$. 16 and 125 are co-primes.
We break the given function into parts ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}=\left\{ {{121}^{n}}-{{(-4)}^{n}} \right\}+\left\{ {{1900}^{n}}-{{25}^{n}} \right\}$ to get the form of ${{a}^{n}}-{{b}^{n}}$.
Here, $\left\{ {{121}^{n}}-{{(-4)}^{n}} \right\}$ will be divisible by $121-\left( -4 \right)=121+4=125$.
Also, $\left\{ {{1900}^{n}}-{{25}^{n}} \right\}$is divisible by $1900-25=1875$.
As 1875 is a multiple of 125, $\left\{ {{1900}^{n}}-{{25}^{n}} \right\}$ is divisible by 125.
From these two statements we can say the two parts in addition will be divisible by 125 as they are both divisible by 125.
So, $\left\{ {{121}^{n}}-{{(-4)}^{n}} \right\}+\left\{ {{1900}^{n}}-{{25}^{n}} \right\}$ is divisible by 125 which means ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$ is divisible by 125.
Now we break the given function into parts in a different combination.${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}=\left\{ {{121}^{n}}-{{25}^{n}} \right\}+\left\{ {{1900}^{n}}-{{(-4)}^{n}} \right\}$ to get the form of ${{a}^{n}}-{{b}^{n}}$.
Here, $\left\{ {{121}^{n}}-{{25}^{n}} \right\}$ will be divisible by $121-25=96$.
Also, $\left\{ {{1900}^{n}}-{{(-4)}^{n}} \right\}$is divisible by $1900+4=1904$.
As both 96 and 1904 are multiple of 16, both $\left\{ {{121}^{n}}-{{25}^{n}} \right\}$ and $\left\{ {{1900}^{n}}-{{(-4)}^{n}} \right\}$ is divisible by 16.
From these two statements we can say the two parts in addition will be divisible by 16 as they are both divisible by 16.
So, $\left\{ {{121}^{n}}-{{25}^{n}} \right\}+\left\{ {{1900}^{n}}-{{(-4)}^{n}} \right\}$ is divisible by 16 which means ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$ is divisible by 16.
Both co-primes divides ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$. So, their multiplication also divides ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$.
So, $16\times 125=2000$ divides ${{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{(-4)}^{n}}$.
If a number is totally divisible then the remainder is 0.
In this case also the remainder will be 0.

Note: We need to remember that the multiplication of those two co-primes could divide the whole function only because of them being coprime. If their H.C.F wasn’t 1 then their multiplication could or couldn’t divide the function. We only used the $\left( a-b \right)$ part of the factorisation of ${{a}^{n}}-{{b}^{n}}$ as the rest of that is of no use in solving the problem.