Question

Prove the divisibility of the following numbers:
24^54·54^24·2^10 by 72^63

Answer 1

Factorize each composite base:

`24=2^3\cdot3`

`54=2\cdot3^3`

`72=2^3\cdot3^2`

Then

`24^(54)=2^(162)\cdot3^(54)`

`54^(24)=2^(24)\cdot3^(72)`

`72^(63)=2^(189)\cdot3^(126)`

Now cancel as many factors as possible:

`(24^(54)\cdot54^(24)\cdot2^(10))/(72^(63))=(2^(196)\cdot3^(126))/(2^(189)\cdot3^(126))=2^7`

The denominator vanishes, so
`72^(63)` does divide
`24^(54)\cdot54^(24)\cdot2^(10)`.