think about it.
2007-06-21 11:36:33 · 33 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
20
10 9s in the ones position
10 9s in the tens position
Alternatively, considering the two digit numbers from 00 to 99. There are 100 numbers, hence a total of 200 digits. And each of the digits from 0 to 9 appear equally frequently, so each appears 200/10 = 20 times.
Using similar reasoning, the number of 9's from 0 to 10^n is
n * 10^(n-1)
2007-06-21 11:38:33 · answer #1 · answered by Dr D 7 · 6⤊ 0⤋
1) It could be 20, if we are counting the number of times the digit 9 appears in the integers between 1 and 100. It could be 40 if you count a 6 as an appearance of 9 (upside down).
2) It could be infinite, since you did not actually specify integers, and we might have to include the following:
1.9
1.09
1.009
etc... (which gives us an infinite number of nines before we even get from 1 to 2)
3) It could also be larger, if you want the integers to be counted, but don't require us to use standard representations, which would allow us to write:
1 = 9/9
2 = 18/9
3 = 27/9
etc...
or any other whacky representation (all the numbers over the denominator 9999999 for example).
4) It could be none, if we count the appearance of the digit 9 while working in a numerical base less than 9. Especially in binary, which means your question is how many 9s appear between 1 and 4.
5) It could vary wildly, if we were working in a numerical base bigger than 9. (See previous case.)
6) It could be one, if you meant how many times does the number 9 itself appear in the interval [1,100].
So the answers I've come up with are:
none, 1, 20, 40, infinity
and a number of other answers that I won't determine, from cases 3 and 5.
2007-06-21 12:19:48 · answer #2 · answered by сhееsеr1 7 · 0⤊ 0⤋
10
2007-06-21 11:40:22 · answer #3 · answered by sesshy 2 · 0⤊ 2⤋
10
2007-06-21 11:39:41 · answer #4 · answered by Dave 2 · 0⤊ 2⤋
20
2007-06-21 11:40:37 · answer #5 · answered by Anonymous · 1⤊ 1⤋
Anybody can see that there's 10 numbers that end with 9, 10 numbers that begin with 9, 10 more numbers that end with 6 which is 9 upsidedown, and 10 more that begin with 6. So, there's 40 altogether, you just have to know where to look for them.
2007-06-21 12:15:47 · answer #6 · answered by Scythian1950 7 · 0⤊ 0⤋
1 there is only one number 9 period. If ur asking how many times the the digit 9 appear the answer is 20.
2007-06-21 11:41:31 · answer #7 · answered by Da Man 4 · 2⤊ 0⤋
If you have 1, you would have to add 99 in order to get from 1 to 100.
There are eleven 9's in 99, so the answer is 11.
2007-06-21 11:40:20 · answer #8 · answered by Anonymous · 0⤊ 2⤋
20 nines
There are 10 9s from 1-100 in the tens place, and you add the ten nines in the 90s. Add then all up and you get 20.
2007-06-21 11:53:09 · answer #9 · answered by Kaitlyn 2 · 0⤊ 0⤋
You have 9,19,29,...,89 = 9 nines
then 90, 91,..., 98 = 9 nines
then 99 = 2 nines
20 all together
2007-06-21 11:41:00 · answer #10 · answered by Kathleen K 7 · 2⤊ 0⤋