JEE Main 2023 — Mathematics Algebra

Now to divide ${5}^{99}$599 by $11$11 we need to check the cyclicity of power of $5$5 when divided by $11$11,

So remainder when $\frac{{5}^{1}}{11}\rightarrow 5\text{as remainder}$1151​→5as remainder

Similarly,

$$\frac{{5}^{2}}{11}=\frac{25}{11}\rightarrow 3\text{as remainder}$$1152​=1125​→3as remainder

$$\frac{{5}^{3}}{11}=\frac{125}{11}\rightarrow 4\text{as remainder}$$1153​=11125​→4as remainder

$$\frac{{5}^{4}}{11}=\frac{625}{11}\rightarrow 9\text{as remainder}$$1154​=11625​→9as remainder

And $\frac{{5}^{5}}{11}=\frac{3125}{11}\rightarrow 1\text{as remainder}$1155​=113125​→1as remainder

So, rewriting the expression we get, ${({5}^{5})}^{19}\times {5}^{4}$(55)19×54

Now when $\frac{{({5}^{5})}^{19}\times {5}^{4}}{11}\rightarrow 1\times \frac{{5}^{4}}{11}$11(55)19×54​→1×1154​

Now again simplifying we get, $\frac{{5}^{4}}{11}\rightarrow 9\text{as remainder}$1154​→9as remainder.