The question is from CAT Number Theory - Remainders. A large number is given and we need to find out the remainder when the first 100 digits is divided by 9. Get practice by solving CAT questions from Remainders.

Question 11: Consider a large number N = 1234567891011121314………979899100. What is the remainder when first 100 digits of N is divided by 9?

- 0
- 8
- 1
- 5

5

N is nothing but first 100 natural numbers written in ascending order!

The first 99 natural numbers will give us 9 + (90 * 2) =189 digits.

So let us consider first 49 numbers. Number of digits = 9 + (40 * 2) = 89

11 more is needed to make 100 digits. So we can include 50, 51, 52, 53, 54 and 5

Hence the first 100 digits of N goes like this 1234567891011121314.......5253545

We need to find the sum of the above number to check for divisibility by 9.

From 1 to 50, each of the digits 1 to 4 occurs 14 times, digit 5 occurs 6 times and each of 6 to 9 occurs 5 times

Sum of digits from 1 to 4 = 10

Sum of digits from 6 to 9 = 30

Therefore, sum of digits from 1 to 50 = 10 * 15 + 5 * 6 + 30 * 5 = 330

Sum of digits from 51 to 54 plus a 5 = 5 * 4 + ( 1 + 2 + 3 + 4) + 5 = 20 + 15 = 35

Total sum = 330 + 35 = 365

When 365 is divided by 9, we get 5 as the remainder!

The question is **"What is the remainder when first 100 digits of N is divided by 9?"**

Choice D is the correct answer.

Try it free!

Register in 2 easy steps and start learning in 30 seconds!

Copyrights © 2019 All Rights Reserved by 2IIM.com - A Fermat Education Initiative.

Privacy Policy | Terms & Conditions

CAT^{®} (Common Admission Test) is a registered trademark of the Indian Institutes of Management. This website is not endorsed or approved by IIMs.

2IIM Online CAT Coaching

A Fermat Education Initiative,

10-C, Kalinga Colony, Bobbili Raja Salai

K.K.Nagar, Chennai. India. Pin - 600 078

**Phone:** (91) 44 4505 8484

**Mobile:** (91) 99626 48484

**WhatsApp:** WhatsApp Now

**Email: **prep@2iim.com