Aptitude Discussion

**Common Information**

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

Q. |
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.**

**ABHIJEET**

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**Balu**

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i dont understand this solution.

**Rohit Sen**

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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

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

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..

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

so divide the following by 9 to get your answer

∑9 + 18x∑9 + 270∑9 + 1000