For the beginners... Remainders...

no one?

well from eulers theorem since 5 and 11 r coprime so 5^{\varphi (11)}\equiv 1mod11
\varphi (11)=10
so 5^10≡1mod11
similarly 3 and 11 r coprime so 3^10=1mod11
so 5^10-3^10=(1-1)mod11≡0mod11

5³â‰¡4(mod 11)
(5³)³â‰¡4³(mod 11)
5¹⁰≡4³X5(mod 11)

4³X5≡1(mod 11)

Therefore,5¹⁰≡1(mod 11)

3⁵≡1(mod 11)
(3⁵)²â‰¡1²(mod 11)
3¹⁰≡1(mod 11)

5¹⁰-3¹⁰≡1-1(mod 11)
5¹⁰-3¹⁰≡0(mod 11)

OR

Since

5¹⁰≡1(mod 11)

Let 5¹⁰ be 11 k+1

3¹⁰≡1(mod 11)

Let 3¹⁰ be 11 m+1

Therefore

5¹⁰ - 3¹⁰ =11k+1 -11m-1
=11(k+m)

Hence it is divisible by 11

There was just a small error which I have corrected:

11k+1-11m-1
=11(k-m)