2006-11-20 07:51:43 · 8 answers · asked by david w 1 in Science & Mathematics ➔ Mathematics
1000 - 1100 and every 100s set to 1800 would be
1 every 10s to 80 (89)
10 every 90 set (90-99)
1*8 + 10*1 = 18 every 100s set.
18 * 8 set (1000-1100; ... 1700-1800) = 144
For the 1900 its 100 in the hundred position because a 9 is there for every answer [1999-1899]. The 10s amd 1s positions would repeat like the other sets.
144 [first eight 100s set] + 100 [from 1900s hundreds position] + 18 [10s and 1s positions]
= 262 9s from 1000-2000
2006-11-20 08:17:30 · answer #1 · answered by bourqueno77 4 · 0⤊ 1⤋
I put the numbers 1000 to 2000 into a spreadsheet and came up with an answer of 300.
So whoever answered first is correct
The guy who answered before me got the first part wrong. There are 20 9s per set of 100 between 1000 and 1899 (he forgot to include the 00s and there are 11 between 90 and 99 and not 10.)
2006-11-20 18:40:58 · answer #2 · answered by rakesh18uk 2 · 0⤊ 1⤋
There are 100 occurences of 9 in the units column (1 in each 10, and there are 100 tens between 1000 and 2000)
There are 100 occurences in the tens columns (10 in each hundred, and there are 10 hundreds betwen 1000 and 2000)
There are 100 occurences in the hundreds columns.
There are 300 occurences in total.
2006-11-21 08:05:24 · answer #3 · answered by coolman9999uk 2 · 0⤊ 0⤋
about 538
2006-11-20 16:03:57 · answer #4 · answered by reem h 2 · 0⤊ 2⤋
I think 201
2006-11-20 16:03:41 · answer #5 · answered by dre 2 · 0⤊ 2⤋
300 even
2006-11-20 15:53:56 · answer #6 · answered by shinglezach 2 · 1⤊ 1⤋
There is actually in infinite number of 9's.
Example....
1000.99999999999999999999999999999999999999999999999..etc.
2006-11-20 15:55:35 · answer #7 · answered by 10 Point 2 · 0⤊ 2⤋
Just enough, any less and they wouldn't reach.
2006-11-20 15:59:43 · answer #8 · answered by lulu 6 · 0⤊ 2⤋