find the greatest number which divides 18777,233 and279 leaving the same remainder in each case

Question

find greatest number which divides 187,233 and 279 leaving the same remainder in each case

Answer 1

18777=ax+r; 233=bx+r; 279=cx+r are three integer multiples, a, b and c, of the same integer x with the same remainder, r.

So 279-233=46=(c-b)x. Therefore x and c-b must be factors of 46: (1,46) or (2,23).

[Also 233-187=46=(b-a)x, if 18777 should read 187. The same factors apply.]

Take each of these possibilities in turn:

We can rule out x=1, so if x=46 then c-b=1. But 18777/46 has 9 as the remainder.

x=2 will work because all the numbers are odd, leaving a remainder of 1.

Try x=23: r=3 for 233 and 279, but 9 for 18777. [187 has r=3.]

So the number, x, appears to be 2, and r=1. However, if 18777 should be 187, as indicated in the context, but not the title, of the question, then 23 is the largest number because 23 divides into all of the numbers with r=3. The most probable answer is therefore 23.

Answer 2

?????????????? gratest ??????????? maebee yu want the BIGGEST smallest numb=233 & sqrt(233)=15.sumodd so the biggest divider=15 233/15=15.53333333333333333... it hav LEFTOVER =8 233/14 get LEFTOVER=9 233/13 get LEFTOVER=12...that gotta be biggest kauz leftover from 11 gotta be < 11