SOLUTION: Find the smallest that must be subtracted from 5912 to make it divisible by 15

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: Find the smallest that must be subtracted from 5912 to make it divisible by 15       Log On


   

Question 1042575: Find the smallest that must be subtracted from 5912 to make it divisible by 15


Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the smallest that must be subtracted from 5912 to make it divisible by 15
-----
5912/15 =~ 394.133
15*394 = 5910
subtract 2

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
divide 5912 by 15 to get 394.13333333
take away the fractional part and multiply 394 by 15 to get 5910
subtract 5910 from 5912 and the difference is 2.
the smallest that must be subtracted from 5912 is 2.
the result is 5910 which is divisible by 15.