Question

$4^{109}+6^{109}$ is divided by 25, the remainder is

• A. 20
• B. 10
• C. 15
• D. 5

Answer 1

The correct option is C 15

Rem $\left[\frac{4^{109}+6^{109}}{25}\right]=Rem\left[\frac{4^{109}}{25}\right]+Rem\left[\frac{6^{109}}{25}\right]Rem\left[\frac{4^{109}}{25}\right]\Rightarrow Rem\left[\frac{4^4.(4^5{)}^{21}}{25}\right]\Rightarrow Rem\left[\frac{256\times (1025-1{)}^{21}}{25}\right]Rem\left[\frac{-256}{25}\right]\Rightarrow -6.......\left(i\right)$
Now, Taking the other part,
$Rem\left(\frac{6^{109}}{25}\right)\Rightarrow Rem\left(\frac{6.(6^4{)}^{27}}{25}\right)\Rightarrow Rem\left[{\frac{6.(1300-4)}{25}}^{27}\right]Rem\left[\frac{6.(-4{)}^{27}}{25}\right]\Rightarrow Rem\left[\frac{-6\times 16\times (1025-1{)}^5}{25}\right]\Rightarrow Rem\left[\frac{96}{25}\right]\Rightarrow \mathrm{21..........}\left(ii\right)$
Adding i) & ii), we get 21 - 6 = 15
Hence, 15 is the answer.