Ta có:
`\qquad 2^9 ≡ 12 (mod 100)`
`=>2.2^9≡2.12 (mod 100)`
`=>2^{10}≡24 (mod 100)`
`=>(2^{10})^3≡24^3 (mod 100)`
`=>2^{30}≡24 (mod 100)`
`=>(2^{30})^3≡24^3 (mod 100)`
`=>2^{90}≡24 (mod 100)`
`=>2^9 . 2^{90}≡12 . 24 (mod 100)`
`=>2^{99}≡88 (mod 100)`
Do đó: `2^9+2^{99}≡12+88 (mod 100)`
`=>2^9+2^{99}≡0 (mod 100)`
Vậy `(2^9+2^{99})` chia hết cho $100$