n is selected from the set {1, 2, 3, . . ., 100} and the number 2ⁿ + 3ⁿ + 5ⁿ is formed. Total number of ways of selecting n so that the formed number is divisible by 4 is equal to 

If n is odd,

      $\begin{array}{l}{3}^{\mathrm{n}}=4{\mathrm{\lambda }}_1-1,5^{\mathrm{n}}=4{\mathrm{\lambda }}_2+1\\ \Rightarrow 2^{\mathrm{n}}+3^{\mathrm{n}}+5^{\mathrm{n}}\text{ is divisible by }4\text{ if }\mathrm{n}\ge 2\end{array}$

Thus, n = 3,5,7,9, ...,99, i.e., n can take 49 different values.

If n is even, $3^{\mathrm{n}}=4{\mathrm{\lambda }}_1+1,5^{\mathrm{n}}=4{\mathrm{\lambda }}_2+1$

$\Rightarrow 2^{\mathrm{n}}+3^{\mathrm{n}}+5^{\mathrm{n}}\text{ is not divisible by }4$

$\text{ as }{2}^{\mathrm{n}}+3^{\mathrm{n}}+5^{\mathrm{n}}\text{ will be in the form of }4\mathrm{\lambda }+2\text{. }$

Thus, the total number of ways of selecting 'n' is equal to 49 .