Sunday, August 1, 2010

NMOS 2010 Q11

Q11. When each of the numbers 2010, 2377 and 3111 is divided by a number D, they give the same remainder R. Find the largest possible value of D+R.


2377 – 2010 = 367 -- D


2010 / 367 = 5 R 175


2377 / 367 = 6 R 175


3111 / 367 = 8 R 175



D + R = 367 + 175 = 542

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