USA Today reporter Paul Wiseman described the old rules for the three-digit telephone area codes by writing about "possible area codes with 1 or 0 in the second digit (Excluded: codes ending in 00 or 11, for all toll-free calls, emergency services, and other special issues)." Codes beginning with 0 or 1 should also be excluded. How many different area codes were possible under these old rules?
2006-09-27 09:03:07 · 6 answers · asked by Jay A 2 in Science & Mathematics ➔ Mathematics
8 possible for first
2 for second
10 for last digit
160 area codes minus 16 (x00 (8) x11 (8))
144
2006-09-27 09:11:21 · answer #1 · answered by phillytocalifornia 3 · 1⤊ 1⤋
You have 3 digits to work with:
Digit one could be 2, 3, 4, 5, 6, 7, 8, or 9, so 8 possibilities
Digit two could be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 so 10 possibilities.
Digit three could be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 so 10 possibilities.
8*10*10 = 800, but this included numbers that end in 00 or 11
So first we have to subtract all the numbers with 00 at the end.
Digit 1 has 8 possibilities
Digit 2 has only 1 possibility.
Digit 3 has only 1 possibility.
So there are 8*1*1 numbers that end in 00 = 8
Same logic means that we have 8 numbers that end in 11
So 800 - 8 - 8 = 784 possible numbers.
Hope this helps. Good luck.
2006-09-27 09:13:25 · answer #2 · answered by SmileyGirl 4 · 1⤊ 2⤋
144. It is not 160 because you exclude codes ending in 00 or 11. That is, exclude the 16 codes 200, 300, 400, 500, 600, 700, 800, 900, 211, 311,411,511,611,711,811,911.
2006-09-27 09:06:29 · answer #3 · answered by Anonymous · 0⤊ 1⤋
128
2006-09-27 09:11:21 · answer #4 · answered by Bob 1 · 0⤊ 0⤋
I've got a boy figure. Straight down no curves. I wish i did have curves though, my clothes just hang off me :/
2016-03-18 02:01:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋
I think it is 160
2006-09-27 09:08:33 · answer #6 · answered by bretttwarwick 3 · 0⤊ 0⤋
who really cares?
2006-09-27 09:11:30 · answer #7 · answered by Suzzi 2 · 0⤊ 0⤋