Intermediate How Many Ones?

[MAS1]

How many times does the digit “1” appear in all the whole numbers between 0 and 999,999? For example “1” appears 3 times in the number 1,101 and 2 times in 101,058.

[Mr. Hui]

This is an interesting problem. My first thought is looking at every place value and calculating how many times "1" can possibly appear. I'm sure there's a variety of strategies.

[Yiwan]

This kind of challenges make the people think........

I will try to find the answer.

[Sugengz]

Using Mathematica the answer is 600,000...

Here is the code:

For[x = 1; y = 0, x <= 999999, x++; y = y + StringCount[ToString[x], "1"]]; Print[y]

[MAS1]

Originally Posted by Sugengz

Using Mathematica the answer is 600,000...

Here is the code:

For[x = 1; y = 0, x <= 999999, x++; y = y + StringCount[ToString[x], "1"]]; Print[y]

Yes.
There are 1,000,000 numbers between 0 and 999,999.
We can think of each of these numbers as having six digits whether 000,001 or 999,973.
So if there are 1 million numbers with six digits each then there are six million total digits.
Since we are going from 000,000 to 999,999 and there are ten digits (0,1,2,…,8,9), then each digit is used the same number of times.
So there must be 6,000,000 / 10 = 600,000 of each digit used
Therefore, there are 600,000 “1s” between 0 and 999,999.

Also, there are:
1 "1" between 0 and 9
20 "1's" between 0 and 99
300 "1's" between 0 and 999
4,000 "1's" between 0 and 9,999
50,000 "1's" between 0 and 99,999
600,000 "1's" between 0 and 999,999
etc

[weepa]

That's cool, I couldn't figure it out.

[twlightfan152]

hello im new to xp math:

[jmw106462]

904 times idk?

[pineapple74]

Originally Posted by MAS1

Yes.
There are 1,000,000 numbers between 0 and 999,999.
We can think of each of these numbers as having six digits whether 000,001 or 999,973.
So if there are 1 million numbers with six digits each then there are six million total digits.
Since we are going from 000,000 to 999,999 and there are ten digits (0,1,2,…,8,9), then each digit is used the same number of times.
So there must be 6,000,000 / 10 = 600,000 of each digit used
Therefore, there are 600,000 “1s” between 0 and 999,999.

Also, there are:
1 "1" between 0 and 9
20 "1's" between 0 and 99
300 "1's" between 0 and 999
4,000 "1's" between 0 and 9,999
50,000 "1's" between 0 and 99,999
600,000 "1's" between 0 and 999,999
etc

Like that And will remember that.

[court4298]

Originally Posted by Sugengz

Using Mathematica the answer is 600,000...

Here is the code:

For[x = 1; y = 0, x <= 999999, x++; y = y + StringCount[ToString[x], "1"]]; Print[y]

thanks i needed help