Common Information Question: 2/2
Find the remainder when $P$ is divided by 9?
Solution:Option(C) is correct
We need to find sum of S of the digits of $P$.
Till 999 each of the digits from 1 to 9 occurs equal number of times.
Total digit of all the numbers from 1 to 999 will be multiple of $\sum 9 = 45.$
$ S= 45k+(1+0+0+0)$
⇒ Required remainder = 1.
Error(s) Found !!!
Balu (Apr 28'14 at 23:35)
i dont understand this solution.
Rohit Sen (Sep 30'13 at 15:17)
simler technique is to use the formula:
n(n+1)/2 for first n natural nos. here n = 1000.. so 1000*1001/2= 500500. Divide it by 9 to get the remainder 1.
but sum of first n natural number is adding
here 10,11...till 1000 are the addition of 2 and three digits numbers
but in the question its has been asked for the sum of all the digits like $1+2+3+4+5+6+7+8+9+1+0+1+1.....1+0+0+0$
therefore these two are different cases..
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