A 7-digit phone number is to be chosen so that the first and last digits are not 0 and the last digit is even. How many different numbers can be chosen? (repetition of a digit is allowed)
2007-05-10 16:54:00 · 5 answers · asked by Alexis's Love Potion #9 4 in Science & Mathematics ➔ Mathematics
EDIT: Dangit! They both beat me to it!
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9 x 10 x 10 x 10 x 10 x 10 x 4 = 3,600,000 or 3.6 Million
2007-05-10 17:07:19 · answer #1 · answered by art_is_my_religion 3 · 2⤊ 0⤋
Lets see.....7 digits, 0, even number, carry the two, divide by four, square root of 34897596846, carry the 1, at 22%, that's............................ impossible!
2007-05-11 22:40:58 · answer #2 · answered by Hot Coco Puff 7 · 0⤊ 0⤋
There are ten digits between 0 and 9 inclusive.
[__] [__] [__] – [__] [__] [__] [__]
[9] [10] [10] – [10] [10] [10] [4]
9*(10^5)*4 = 3,600,000
2007-05-11 00:01:50 · answer #3 · answered by gtmooney14 3 · 5⤊ 0⤋
You have to multiply the options for each digit together
ABC-DEFG
A 9 options (1,2,3,4,5,6,7,8,9)
B-F 10 options (0-9)
G 4 options (2,4,6,8)
so therefore your answer is 9x10x10x10x10x10x4=3600000
2007-05-11 00:06:13 · answer #4 · answered by angela 1 · 2⤊ 0⤋
There are ten digits between 0 and 9 inclusive.
[__] [__] [__] – [__] [__] [__] [__]
[9] [10] [10] – [10] [10] [10] [4]
9*(10^5)*4 = 3,600,000
2007-05-11 00:13:26 · answer #5 · answered by Elisa 2 · 0⤊ 2⤋