Question

If $n\in N$, then ${121}^n-{25}^n+{1900}^n-{(-4)}^n$ is divisible by which one of the following?

• A. $1904$
• B. $2000$
• C. $2002$
• D. $2006$

Answer 1

Answer: (A), (B)

The correct options are
A $1904$
B $2000$

$nϵN,{121}^n-{25}^n+{1900}^n-(-4{)}^n$

Let us use the fact that $(a-b)$ divides $a^n-b^n$.

Therefore in ${121}^n-{25}^n+{1900}^n-(-4{)}^n$

$(121-25)|(121{)}^n-(25{)}^n\Rightarrow 96|(121{)}^n-(25{)}^n\Rightarrow 16|({121}^n-{25}^n)$

$(1900-(-4\left)||\right(1900{)}^n-(-4{)}^n)\Rightarrow 1094|\left(\right(100{)}^n-(-4{)}^n)16|\left(\right(1900{)}^n-(-4{)}^n)$

$\therefore 16$ divides ${121}^n-{25}^n+{1900}^n-(-4{)}^n$

Now

$(121-(-4\left)\right)|\left(\right(121{)}^n-(-4{)}^n)\Rightarrow 125|\left(\right(121{)}^n-(-4{)}^n)$

$(1900-25)|\left(\right(1900{)}^m-\left(25\right)6n)\Rightarrow 1875|({1900}^n-{25}^n\left)125\right({1900}^n-{25}^n)$

Therefore $125$ also divides ${121}^n-{25}^n+{1900}^n-(-4{)}^n$

Hence, $16\times 125=2000$.