Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

${25}^{190}-{19}^{190}-8^{190}+2^{190}$

Now  $\left({25}^{190}-{19}^{190}\right), 8^{190}, 2^{190}$ are divisible by $2$, as ${25}^{190}$ will give remainder as $1$ and ${19}^{190}$ will give remainder as $1$ when divided by $2$, so combined ${25}^{190}-{19}^{190}$ will give remainder $0$ 

Now, $\left({25}^{190}-8^{190}\right) \& \left(-{19}^{190}+2^{190}\right)$ are divisible by $17$, $a^n-b^n$ is divisible by $a-b$ if $n$ is even, 
So given number is divisible by $34$ as it is divisible by $2 \& 17$ both,

Now,${25}^{190}={\left(7\times 3+4\right)}^{190}$, dividing by $7$ we get,

${\left(7\times 3+4\right)}^{190}=4^{190}={16}^{95}=2^{95}=4·8^{31}$

Now when $4·{\left(7+1\right)}^{31}$ divided by $7$ will give remainder as $4·1=4$

And ${19}^{190}={\left(7\times 3-2\right)}^{190}$, when divided by $7$ we get,

${\left(7\times 3-2\right)}^{190}={\left(-2\right)}^{190}=2^{190}=2\times 8^{63}=2\times {\left(7+1\right)}^{63}$

So, when $2\times {\left(7+1\right)}^{63}$ divided by $7$ will give remainder as $2$

Similarly $8^{190}={\left(7+1\right)}^{190}$ will give $1$ remainder 

And $2^{190}=2{\left(7+1\right)}^{63}$ will give $1$ remainder,

So combining all we get,${25}^{190}-{19}^{190}-8^{190}+2^{190}=4-2-1+1=5$
$\Rightarrow$Not divisible by $7$, so it will not be divisible by $14$