how can i make prime-number division remainder?

[deerlittle7]

where in, a number will be divided by a given prime divisor

(example:97) and extract the final last two digit decimal places.
example:
input number:24964
24964/97
=257.36/*the .36 value will be minus 1*/
=35 is the final answer

[Bazzy]

the % operator returns the remainder of an integral division
24964%97 returns 35