Ta có:
`\qquad 5^{100}-125^{33}-25^{49}`
`=5^{100}-(5^3)^{33}-(5^2)^{49}`
`=5^{100}-5^{99}-5^{98}`
`=5^{98} .5^2-5^{98} . 5-5^{98}.1`
`=5^{98}. (5^2-5-1)`
`=5^{98}.19`
Vì `19\ \vdots\ 19`
`=>5^{98}. 19\ \vdots\ 19`
Vậy `5^{100}-125^{33}-25^{49}` chia hết cho `19`