Number System
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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

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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.


(3) Comment(s)


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..