Aptitude Discussion

Q. |
The numbers 1 to 29 are written side by side as follows $1234567891011......2829$. If the number is divided by 9, then what is the remainder? |

✖ A. |
0 |

✖ B. |
1 |

✖ C. |
2 |

✔ D. |
3 |

**Solution:**

Option(**D**) is correct

Sum of digits from $1$ to $10 = 46$

sum of digits from $11$ to $20 =56$

sum of digits from $21$ to $29 =63$

sum of the digits from numbers $= 46+56+63 =165$

Sum of digits in the number $165 =12$ which gives a **remainder of 3** when divided by $9$.

**Edit:** For an alternative solution, check comment by **Sravan Reddy.**

**Sagar Bhatt**

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**Sravan Reddy**

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Firstly, as mentioned by others, its not sum of first 29 numbers. so simply can't use $\dfrac{n(n+1)}{2} = 450!!$ This is wrong.

What was asked was sum of digits of all 1st 29 natural numbers which will be the one as mentioned in solution.

Alternate way of summation:

Sum of 1 to 9 is 45 - repeated 3 times (1,9),(11,19),(21,29) - so total $45*3 = 135$

Now sum of 10 1's and 10 2's which appear in ten's digit is $10+20=30$

So, total sum is 165 which upon dividing by 9 gives 3 as reminder.

Sorry, it's 435 not 450 in my first statement.

**Hemanta**

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sum will be 435 not 165..

poor people they dont know how to do addition

Sum will be 165 only, as it is not the sum of FIRST 29 numbers.

So using $\dfrac{n(n+1)}{2}$ is wrong.

**Gannu**

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the easy method is to employ sum to 'n' terms, b'cause all we have to do is to add the numbers to check the divisibility of the given number

Just add first n natural number, sums to 435, divide it by 9 remainder = 3 . Done !!