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If $n \in \mathrm{N}$, then $121^{\mathrm{n}}-25^{\mathrm{n}}+1900^{\mathrm{n}}-(-4)^{\mathrm{n}}$ is divisible by which one of the following?
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Verified Answer
The correct answer is:
2000
So, $z=1+\sqrt{5}$
$121^{n}-25^{n}+1900^{n}-\left(-4^{n}\right)$
Let us substitute $\mathrm{n}=1$ We get,$(121)^{1}-(25)^{1}+(1900)^{1}-\left(-4^{1}\right)$
$=121-25+1900+4$
$=2025-25$
$=2000$
So, given expression is divisible by 2000
$121^{n}-25^{n}+1900^{n}-\left(-4^{n}\right)$
Let us substitute $\mathrm{n}=1$ We get,$(121)^{1}-(25)^{1}+(1900)^{1}-\left(-4^{1}\right)$
$=121-25+1900+4$
$=2025-25$
$=2000$
So, given expression is divisible by 2000
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