# Difficult Number System Solved QuestionAptitude Discussion

Common Information

$P =1234567891011.............9989991000$

 Q. Common Information Question: 2/2 Find the remainder when $P$ is divided by 9?
 ✖ A. 8 ✖ B. 7 ✔ C. 1 ✖ D. 6

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 $\sum 9 = 45.$

Hence,

$S= 45k+(1+0+0+0)$
⇒ Required remainder = 1.

## (5) Comment(s)

Ravi
()

There are equal number of 1s,2s etc till 9s....and less 0s.

So when we add (1+2+3+4...+9)... n times , the addition will be a 45 which is a multiple of 9 and then adding 1 of 1000 ...And now that's your remainder.

ABHIJEET
()

1 to 9 has 9 digits

10 to 99 has 90 x 2 = 180 digits

100 to 999 has 900 x 3 = 2700 digits

all the digits from 1 to 9 are equally distributed in each of the sections

∑9 + 18x∑9 + 270∑9 + 1000

Balu
()

i dont understand this solution.

Rohit Sen
()

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.

Nitin
()

but sum of first n natural number is adding

$1+2+3+4+5+6+7+8+9+10+11....+1000$

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..