To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
300 times.
As a digit in other numbers it appears 20 times
81 lots of 10000 equal 810000. And since a decimal number is simply a way of representing a number in such a way that the place value of each digit is ten times that of the digit to its right, the required decimal representation is 810000.
If you count 11 as 2 instances, the digit 1 appears 18 times if you don't count 10, 19 times if you do. 10,11,12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91
4000 times.
5 times, simple maths
30 times because when you count up in tens this is the answer
ten thousand times greater
To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
11 times
19 times.
Including the one in ' 1 ' and the one in '100', there are 21 1s.Every other digit 2 - 9 appears 20 times between 1 and 100 .
Assuming you mean in the numbers 1, 2, 3, ..., 998, 999, 1000 then the digit 0. (The digit 1 appears 301 times, the digits 2-9 all appear 300 times each, but the digit 0 only appears 192 times.)
If we throw 1000000 out of our calculations (this is okay because it doesn't have any nines in it) then we're left with all the numbers with 6 or less digits. There are 10 choices for each digit. If the 100000's digit is 9, then there are 105=10000 ways to choose the remaining digits, so there are 10000 numbers that have 9 in the 100000s place. Similarly, there are 10000 numbers that have 9 in the 10000s place, 10000 numbers with 9 in the 1000s place, and so on. There are 6 different places 9 could be in, so the digit 9 is used 10000*6 or 60000 times. Improved... well your missing a zero in your answer.... the total amount of 9's used should be 600 000 , not 60 000 because already in the hundred thousands digit 9 is being used 100 000 times becasue from 900 000- 999 999, the number nine is being used atleast 100 000 times so already your answer is wrong. And if you did you method corectly, it should be the same for every other digit, which means the answer will be that 9 will appear 600 000 times between 1 and 1 000 000. Correct me if im wrong, but im pretty sure that that is the answer. (sources):http://www.cemc.uwaterloo.ca/contests/past_contests/2008/2008GaussSolution.pdf (question 25)
11 times
300 times.