I need to add up all the digits in the numbers $1$ to $10000$. How would I do that without using a calculator?
I don't get it one bit. We are doing something in our math class about this.
I need to add up all the digits in the numbers $1$ to $10000$. How would I do that without using a calculator?
I don't get it one bit. We are doing something in our math class about this.
As I understand it we are not told to sum up all the numbers from $1$ to $10\,000$ but the decimal digits of these numbers. Taking the numbers from $0000$ to $9999$ instead we have $10\,000$ numbers having four digits with an average value of $4.5$. The sum of all appearing digits therefore is $10\,000\cdot4\cdot 4.5=180\,000$. Add $1$ to this for the single number $10\,000$, and obtain $180\,001$ as final result.
First note that the sum of $0$ through $9$ is $45$. Also note that from $0000$ to $9999$, each digit appears exactly $1000$ times, for each of the $4$ positions. Therefore, the sum of all digits in the numbers $1$ through $10000$ is $$45\cdot 4\cdot 1000 + 1=180001$$ I hope this helps you intuitively understand.
How many times does a 1 appear in the last digit? How many times does a 2 appear in the last digit? Etc.
How many times does a 1 appear in the 10's position? How many times does a 2 appear in the 10's position? Etc.
How many times does a 1 appear in the 100's position? How many times does a 2 appear in the 100's position? Etc.
How many times does a 1 appear in the 1,000's position? How many times does a 2 appear in the 1,000's position? Etc.
How many times does a 1 appear in the 10,000's position? (Answer: Exactly once.) How many times does a 2 appear int 10,000's position? (Answer: never.)
Answer those questions.
Hint Add first all digits up to $9999$.
Hint 2 Ignore the zeroes.
Hint 3 Each digit $1-9$ appears how many times in each of the four possible positions?