Question

If $n\in N$, then $121^n-25^n+1900^n-(-4{)}^n$ is divisible by

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

Answer 1

Answer: (B)

$121^n-25^n=(96+25{)}^n-25^n$ is divisible by $96$.
$1900^n-(-4{)}^n=(1904-4{)}^n-(-4{)}^n$ is divisible by $1904$. Hence, both the above are divisible by $16$.
Further, $121^n-(-4{)}^n$ is divisible by $125$ and $1900^n-25^n$ is divisible by $1875$. Hence, both the above are divisible by $125$.
$\therefore$ The given number is divisible by $16\times 125=2000$.