2007-06-06 04:20:27 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
12*10^11
From 0-10, there is 1 three.
From 1-100, there are 10 unit threes, and 10 threes in the tens position. ie 20.
From 1 - 1000,
there are 10*20 = 200 threes in the units and tens, and 100 in the hundreds position. ie 300 total.
And so on. Generally from 1 - 10^n,
there are n*10^(n-1) threes.
So for n = 12, as in this case, there are
12*10^11 threes.
2007-06-06 04:35:11 · answer #1 · answered by Dr D 7 · 3⤊ 0⤋
There is only one number 3 up to 1000000000000. It comes between 2 and 4.
Different questions you might have meant to ask:
How many numbers between 1 and 1000000000000 are exactly divisible by three?
How many numbers between 1 and 1000000000000 contain at least one digit which is 3?
How many total digits of 3 are there in the numbers between 1 and 1000000000000?
Y.
2007-06-07 11:46:50 · answer #2 · answered by professor_yaffle1 1 · 0⤊ 0⤋
Do you want to know how many times the number 3 goes into
1000000000000? Well first you have to add some commas after every three numbers so it will make sense to read. Thus you get: 100,000,000,000.0. If you divide that number by 3, the answer you will get is 333333333333.33333333333333333333 or 32 three's. Not sure if that is what you want. Not even sure if you know what you want. :-(((
2007-06-06 11:41:34 · answer #3 · answered by Sicilian Godmother 7 · 0⤊ 0⤋
There are 10^12 different 12-digit numbers, if you count leading 0's. (i.e., 9 = 000000000009). If you add up the number of digits, you get 12 x 10^12. Each digit will occur 1/10 of the time, so the 3 will show up 12 X 10^11 times.
2007-06-06 12:25:55 · answer #4 · answered by Anonymous · 1⤊ 0⤋
in the number 1000000000000 there is no 3.... now if you want to know how many times 3 will go into 1000000000000 then divide... have fun
2007-06-06 11:30:58 · answer #5 · answered by gadsdencpl2005 2 · 0⤊ 0⤋
Look at sequences of 12 digits (everything between 000000000000 and 999999999999).
There are 12Cn * 9^(12-n) sequences with exactly n threes.
Therefore, the total number of threes is
... SUM n (12Cn) * 9^(12-n)
and, because n (12Cn) = 12 (11Cn),
... = 12 * 9 * SUM (11Cn) * 9^(11-n)
... = 108 (1 + 9)^11 = 108 * 10^11
... = 10 800 000 000 000.
2007-06-06 11:49:07 · answer #6 · answered by dutch_prof 4 · 0⤊ 0⤋
I don't understand your question,what is "maney"and what does "are the up to"mean?
Anyway are you really bothered?
2007-06-06 11:29:14 · answer #7 · answered by Anonymous · 0⤊ 0⤋
On my computer there is a calculator facility , use the one on yours to solve your question you lazy git asking us.
2007-06-06 11:33:51 · answer #8 · answered by john r 4 · 0⤊ 0⤋
3s your lucky number?
2007-06-06 11:52:15 · answer #9 · answered by Anonymous · 0⤊ 0⤋
33333333333
2007-06-06 18:09:39 · answer #10 · answered by UNIQUE 3 · 0⤊ 0⤋