Question 1055383: Determine the largest integer less than or equal to 2016 that leaves a remainder of 3 when divided by 7 and leaves a remainder of 4 when divided by 11.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it looks like the number will be 1984.
i don't know if there's an easy way to do this.
this is how i did it.
divide 2016 by 7 and you get 288 with a remainder of 0.
subtract 4 from 2016 and you get 2012.
divide 2012 by 7 and you get 287 with a remainder of 3.
that's the largest number less than or equal to 2016 that is divisible by 7 with a remainder of 3.
divide 2016 by 11 and you get 183 with a remainder of 3.
add 1 to 2016 and you get 2017.
divide 2017 by 11 and you get 183 with a remainder of 4.
you got the remainder of 4 but 2017 is greater than 2016, so that's no good.
subtract 11 from 2017 and you get 2006.
divide 20006 by 11 and you get 182 with a remainder of 4.
that's the largest number less than or equal to 2016 that is divisible by 11 with a remainder of 4.
your two largest numbers are:
2012 and 2006.
you want to find the largest number that is divisible by both and leaves a remainder of 3 when divided by 7 and leaves a remainder of 4 when divided by 11.
work your way down each set.
2012 is reduced by 7 each time.
2006 is reduced by 11 each time.
you get:
2012 - 7 = 2005 - 7 = 1998 - 7 = 1991 - 7 = 1984
2006 - 11 = 1995 - 11 = 1984.
1984 is the magic number as best i can determine.
when it is divided by 7, you get 283 plus a remainder of 3.
when it is divided by 11, you get 180 plus a remainder of 4.
there may be an easier way to figure this out but i don't know it.
|
|
|
