Q30) There are 10 consecutive positive integers written on a blackboard. One number is erased. The sum of remaining nine integers is 2011. Which number was erased?

for any weight we have 3 possibilities:-
Place it on the weight panPlace it on other pan, along with something whose weight we need to findWe don't use itSo, it is similar to base 3, hence by taking weight of form 3^n, we can ensure that we are using least number of weights and when we just have two possibilities. so 3^n >=300 so n=6

Can start by understanding that the numbers will be close to the cube root of 190740 which should lie between 50 and 60 (between 125000 and 216000). Can start with the smallest prime numbers and check. It will be satisfied for 53 * 59 * 61. Total of 53 + 59 + 61 = 173.

999 = 37 * 27
9999 = 99 * 101
999999 = 999* 1001 = 999 * 11 * 91 = 999 * 11 * 13 * 7
2004 is a multiple of 3, 4 and 6.
The given number can be divided into groups of three 9's, four 9's or six 9's at a time.
The given number is divisible by 7, 13, 37 and 101.