\(2^{45}=\left(2^3\right)^{15}=8^{15}\)

8ⁿ with n \(\in N\) divide for 7 then we have remainder is 1

=> 2⁴⁵ divide for 7 then we have remainder is 1

We have:\(2^{45}=\left(2^5\right)^9=32^9\)

\(32^n\) with n ∈ N divide 7 then we have remainder is 1

So 245 divide for 7  then we have remainder is 1

2^45=(2^3)^15=8^15

8n with n ∈N∈N divide for 7 then we have remainder is 1

=> 245 divide for 7 then we have remainder is 1