25 ^ 190 - 19 ^ 190 - 8 ^ 190 + 2 ^ 190 is divisible by

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$0$</p><p>Now, \left({25}^{190} - {8}^{190}\right) \text{ and }$\left(- {19}^{190} + {2}^{190}\right)$are divisible by$17$,${a}^{n} - {b}^{n}$is divisible by$a - b$if$n$is even, <br />So given number is divisible by$34$as it is divisible by$2 \text{ and } 17$both,</p><p>Now,${25}^{190} = {\left(7 \times 3 + 4\right)}^{190}$, dividing by$7$we get,</p><p>${\left(7 \times 3 + 4\right)}^{190} = {4}^{190} = {16}^{95} = {2}^{95} = 4 \cdot {8}^{31}$</p><p>Now when$4 \cdot {\left(7 + 1\right)}^{31}$divided by$7$will give remainder as$4 \cdot 1 = 4$</p><p>And${19}^{190} = {\left(7 \times 3 - 2\right)}^{190}$, when divided by$7$we get,</p><p>${\left(7 \times 3 - 2\right)}^{190} = {\left(- 2\right)}^{190} = {2}^{190} = 2 \times {8}^{63} = 2 \times {\left(7 + 1\right)}^{63}$</p><p>So, when$2 \times {\left(7 + 1\right)}^{63}$divided by$7$will give remainder as$2$</p><p>Similarly${8}^{190} = {\left(7 + 1\right)}^{190}$will give$1$remainder </p><p>And${2}^{190} = 2{\left(7 + 1\right)}^{63}$will give$1$remainder,</p><p>So combining all we get,${25}^{190} - {19}^{190} - {8}^{190} + {2}^{190} = 4 - 2 - 1 + 1 = 5$<br />$⇒$Not divisible by$7$, so it will not be divisible by$14