My approach is this.. I am sure there would be some better and faster way of doing this
Even multiples of 15 between 295 and 615 will give 300 as first no and 600 as the last no. The diff between 2 consecutive no in this series will be 30. This gives, total nos in the series = 11.
Sum of n numbers in arithmetic progression is given by
(n/2)[2a+(n-1)d]
So, in this case sum will be (11/2)[600+300]= 11*450=11*5*3*3*5*2
Greatest prime factor = 11 ANS
MGMAT 11
This topic has expert replies
Source: Beat The GMAT — Problem Solving |












