How many digits of 3 are there between 1 and 10^12, and what is the percentage of these 3's out of the total of 10^12 numbers, and why? Please explain. Thank you for any help.
+312 For my solution, 1=000000000001, 2=000000000002 and so on (we will write every number's digits up to the 12nd digit even if that digit is 0)
the numbers between 1 and 1012 are all the combination of 12 digits (10 posibilities for every digit because there are 10 digits) except for the combination 000000000000 and including the 13 digits combination 1000000000000. But both of these combinations dont contain any 3, so for us they are the same. so the question is "how many 3's are there between 0 and 1012-1.
We have exactly 1012 combinations. Lets look at the digit in the nth place (the units digit is in the first place, the tens digit is in the second place and so on). there are exactly 1/10*(the number of combinations=1012) where that digit is 3 (because there are 1011 combinations for the other 11 digits, and if we want it to have a 3 as its digit in the n place we need to put 3 as the digit in that place) meaning there are 1011 numbers between 0 and 1012-1 that contain 3 as their digit in the n place. there are 12 places, so we repeat this 12 times (we count the times 3 appears as the first digit, then count the times 3 appears as the second digit.... and so on until the 12nd digit). So, there are 12*1011 3's between 0 and 1012-1, meaning there are 12*1011 between 1 and 1012.
Your result doesn't take into account the fact that half the matrix is set to zero:
000000000000
000000000001
000000000034
000000000400
000000006868
000000048049
000000536107
000007379780
000056289484
000869347589
005626115408
042828679713
679952567873
So the correct result is 10^12 rows times 12 columns, divided by ten (limit to a single digit in the set {0-9}) and divided by two to account for the empty half of the matrix = 6 * 10^11 threes.
Since we are using all the 10 digits from 0 - 9 in the number 10^12, and we are excluding one digit(3 in this case), we therefore have: 9^12 digits left. And the difference between: 10^12 - 9^12 =717,570,463,519 must be 3s. The percentage would be:717,570,463,519 / 10^12 =71.75%.
As the number n becomes larger and larger, any one digit tends towards 1 or 100%. So, in 10^100, the digit 3 is present in 99.99% of the numbers. And this true of any of the 10 digits from 0-9.
Ehrlich: I think the question is: How many threes are there between 1 and 10^12?. Take a simpler example: How many threes are there between 1 and 1,000. Answer=271 threes!. OR:10^3 - 9^3 =
1,000 - 729 =271 threes. The same principle applies to ALL numbers that are powers of 10. So, 10^12 is a power of 10. And the calculation is EXACTLY the same as in 10^3 or 1,000!. There should be no dispute about this, because it is very clear.
+312 But by that he meant he wants to know the number of times the digit "3" appears. Lets test MY theory, but except for the number of 3's between 1 and 1012, the number of 3's between 1 and 100. when it was between 1 and 1012 the answer was 1011*12, and between 1 and 102 the answer will be 101*2=20. You can count the 3's, the anwer will be 20 (The number of numbers containing 3 as a digit will be 102-92=19).
Heres another way to think about it: Every number is supposed to appear an equal number of times (if we count the zeroes after that number, the way i suggested writing the numbers in my first comment) . imagine we're counting the number of times ANY digit appears. every number has 12 digits on total, meaning our answer for the number of digits will be 12*1012. Every digit appears an equal amount of times, meaning we have to divide by 10 (the number of digits) to get 12*1011, the number of times ANY of the digits appear.
No!. Count how many threes are there between 1 and 100? The answer is 19 threes. That is what the questioner wants. Simply count the number of threes between 1 and 10^12. Very simple!.
+312 ok. Lets count together:
1
2
3 NUMBER OF 3's=1
4
5
.
.
.
13 NUMBER OF 3's=1
.
.
23 NUMBER OF 3's=1
.
.
.
30 NUMBER OF 3's=1
31 NUMBER OF 3's=1
32 NUMBER OF 3's=1
33 NUMBER OF 3's=2
34 NUMBER OF 3's=1
35 NUMBER OF 3's=1
36 NUMBER OF 3's=1
37 NUMBER OF 3's=1
38 NUMBER OF 3's=1
39 NUMBER OF 3's=1
.
.
.
43 NUMBER OF 3's=1
.
.
.
53 NUMBER OF 3's=1
.
.
.
63 NUMBER OF 3's=1
.
.
.
73 NUMBER OF 3's=1
.
.
.
83 NUMBER OF 3's=1
.
.
.
93 NUMBER OF 3's=1
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2=20
The way to approach this problem is as follows:
Write each number as 12 places, as Ehrlich did, as follows:
000,000,000,001, 000,000,000,002, 000,000,000,003...and so on.
Since we have 10^12, or a trillion numbers, and each one of them consists of 12 digits, we, therefore, have a TOTAL OF: 12 x 10^12 =12 trillion INDIVIDUAL DIGITS.
And since each of the 10 digits(0 - 9) has an equal representation in the above 12 trillion individual digits, we therefore will have:
12 trillion(or 1.2 x 10^13) digits / 10 =1.2 trillion =1.2 x 10^12 threes in 10^12 numbers.
One problem. The array of numbers, 10^12 rows high and 12 digits wide, is half zeros:
000000000000
000000000001
000000000034
000000000400
000000006868
000000048049
000000536107
000007379780
000056289484
000869347589
005626115408
042828679713
679952567873
Therefore the correct result is 1 * 10^12 (height) * 12 (width) / 10 (limit result to threes) and finally / 2 (to account for the empty half of the matrix) = 6 * 10^11
+312 Half zeroes? could you prove that? And How did you infer from that that there are 6*1011 threes? it doesnt make sense. I know alot of the digits are zeroes, but it doesnt change anything.
Please post your proof.
In your original method, you suggested that the number matrix be filled out with leading zeros so that all the numbers have the same width. In a set of integers between 0 and 1 * 10^12, this would create a square matrix with zeros clustered at the uper left, like this:
000000000000
000000000003
000000000014
000000000481
000000004067
000000011202
000000765025
000006570349
000052570231
000569504785
007221909937
010323716328
838219253314
This change makes it easier to understand the problem, but it doesn't change the result, which is true for a set of sequential 1 * 10^12 integers that aren't padded in this way:
0
6
46
426
1435
55123
230105
2873680
63725050
640899375
0060833473
53442421828
372859331893
The point? The differing length of the integers between 0 and 1 * 10^12 needs to be taken into account. So the number of threes in the set of integers between 0 and 1 * 10^12 is 1 * 10^12 (height) times 12 (width) divided by 10 (isolate any single digit) divided by two (to account for the missing digits in the upper left of the martix as shown above).
You still did not address the question of "Guest #13". That is, if this method works for the total number of 3's from 1 to 1,000, which comes to 300 threes, and from 1 to 10,000, which comes to 4,000 threes(of which there is no doubt, since both have been counted by computer programs), then why is it different for the total number of 3's from 1 to 10^12?
+312 1. The matrix is not a square matrix, its length is 1012 and its width is 12.
2. In my solution i HAD to count those zeroes (in the second one. In the first one i just used them to show we can look at it as a set of combinations)
3. You keep claimimg that the zeroes i added are exactly one half of the digits. Your calculation is wrong, because you are assuming A TON of things you think are right. I calculated it, and the number of added zeroes is less than 1/100 of all of the digits (after the zeroes were added)
You're right -- I became enamored of my simple model and overlooked something important about the set of integers {0 , 1 * 10^12-1}, which is that (unlike my model) nearly all of of them are 12 digits long.
Granted its limitations, a Python model shows perfect correspondence with the idea that the total number of '3' digits in the set {0,1 * 10^12-1} is equal to 1 * 10^12 (height) * 12 (width) / 10 (isolate a single digit) = 12 * 10^11
A general rule can be written for any chosen upper bound m:
t = (log(m)/log(10)) * m / 10
t = number of occurrences of a single digit in the set {1,9}
This rule doesn't work for zero, for a reason I haven't figured out.
That sounds right and thereby settles the answer to this question as: 6 x 10^11 threes !!.
If the method used for the number of 3's between 1 and 1,000 and between 1 and 10,000 is exactly the same as between 1 and 10^12, which everybody agrees that it is 300, and 4,000 threes respectively, without dividing by 2, then what is so special about the number of 3's between 1 and 10^12????. Some rigorous explanation is needed!!!.
+312 What do you NOT understand from my solution? tell me and ill explain it further
+33614 I don't have any great insights to offer! I agree the total number of 3s is 12*10^11.
There is a distinction to be made between number of individual digits that are 3s and number of integers that contain 3s. For example there are 20 individual digits that are 3s between 1 and 100, but only 19 different integers that contain 3s (because 33 counts two towards individual digits, but just one towards individual integers). The figure of 12*10^11 refers to individual digits, not individual integers.
Your solution and mine posted as "Guest #6", which begins with:"The way to approach this problem is as follows:" are exactly the same!! So, I do agree with your solution! My beef is also with the people who are dividing by 2. I asked Alan and heureka to take a look at it and see if they have anything NEW to add. Got it?.
+26367 How many 3s..........
How many digits of 3 are there between 1 and 10^12, and what is the percentage of these 3's out of the total of 10^12 numbers, and why? Please explain.
Thank you for any help.
I agree with Alan the total count of 3s is in general
between 1 and 10n = \(n\cdot 10^{n-1}\)
between 1 and 1012 = \(12\cdot 10^{11}\)
My empirical solution:
\(\small{ \begin{array}{|lcr|c|rrrrrrrrrrrrr|r|} \hline &&& n & &12th&11th&10th&9th&8th&7th&6th&5th&4th&third&second&first & \text{Sum of 3's} \\ &&& & \text{3's in Postion} \\ \hline 1 &\ldots& 10 & 1 & & & & & & & & & & & & &1\times 3 & 1\\ 1 &\ldots& 100 & 2 & & & & & & & & & & & &10\times 3 &10\times 3 & 20 \\ 1 &\ldots& 1000 & 3 & & & & & & & & & & &100\times 3&100\times 3&100\times 3 & 300 \\ 1 &\ldots& 10000 & 4 & & & & & & & & & &1000\times 3&1000\times 3&1000\times 3&1000\times 3 & 4000 \\ \cdots \\ 1 &\ldots& 1000000000000 & 12 & &100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3 & 1200000000000 \\ \hline \end{array} } \)
