how many numbers between 100 and 500 have a 4 as at least one of the digits?
2007-12-03 12:41:27 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let's ignore 500 for a moment. It obviously doesn't.
So we have 400 numbers to consider (100 to 499)
Let's count the numbers in this range that *don't* have any fours in them.
That would mean the first digit is 1, 2 or 3.
The second digit is 0, 1, 2, 3, 5, 6, 7, 8 or 9.
The last digit is 0, 1, 2, 3, 5, 6, 7, 8 or 9.
3 x 9 x 9 = 243 combinations that don't have any fours.
Subtracting this from 400 possible numbers we get:
400 - 243 = 157
2007-12-03 12:50:50 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
Everything between 100-499
The first digit can't be a 1, 2 or 3.
The middle digit can't be 1, 2, 3, 5, 6, 7, 8, and 9.
The last digit can't be a 1, 2, 3, 5, 6, 7, 8, and 9.
So 3 x 9 x 9 which equals 243.
499 - 243 equals 256.
So the answer would be 256 numbers that have a least one of the digits as a 9.
2007-12-03 20:52:30 · answer #2 · answered by Anonymous · 0⤊ 1⤋
157
2007-12-03 20:49:16 · answer #3 · answered by cubsfan3812 2 · 0⤊ 0⤋
obviously everything from 400-499
from 100-200 you have:
104,114,124,134,140,141,142,143,144,
145,146,147,148,149,154,164,174,184,
194 which is 19 numbers.
so 19+19+19+100 = 157
2007-12-03 20:47:11 · answer #4 · answered by grompfet 5 · 0⤊ 0⤋
150?
2007-12-03 20:45:57 · answer #5 · answered by emeraldheart 2 · 0⤊ 1⤋