SAMHITAC wrote:
A hundred and twenty digit number is formed by writing the first x natural numbers in front of each other as
12345678910111213...... Find the remainder when this number is divided by 8.
a)6
b)7
c)2
d)0
Please provide the solution with explanation.
12345678910111213...... Find the remainder when this number is divided by 8.
a)6
b)7
c)2
d)0
Please provide the solution with explanation.
First recall the divisibility rule of 8. If the last 3 digits are divisible by 8, so is the number. If you get a remainder when you divide last 3 digits by 8, it will be the remainder you get when you divide the entire number by 8. Imagine a large number:
145124 = 145000 + 124
If last 3 digits are 000, this number is divisible by 8. So the remainder obtained when you divide 124 by 8 will be the remainder you get when you divide 145124 by 8.
So what you need here is the last 3 digits of this number.
You will first write the first 9 one digit numbers. Then there are 90 (from 10 to 99) two digit numbers which will give 180 digits. But we need only 120 digits
120 - 9 = 111. We need to write first 55 two digit numbers and then the first digit of the 56th number. What will be the 55th two digit number? It will be 55 + 9 = 64 (because first two digit number is 10). So the 56th two digit number will be 65.
The given number is 123456...63646
When you divide 646 by 8, you get remainder 6.