find the number which is nearest to 3105 and is exactly divisible by 21
"on dividing 3105 by 21 , we get 18 as remainder .  number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . answer b 3108"
add(3105, subtract(21, reminder(3105, 21)))