kingjewel1 Posted January 16, 2005 Share Posted January 16, 2005 Mobile phone operator zim" in my country is saying it has more than four million clients. If a phone number is 9 digits long and always starts with a 6... How is this possible? Link to comment Share on other sites More sharing options...
Primarygun Posted January 16, 2005 Share Posted January 16, 2005 For a number with two digit, if the first is 6, there are ten evens out. 6,1, ..........6,0. 1x10. For a number with nine digit, the first is 6, so 10^8 possible evens. Link to comment Share on other sites More sharing options...
Mart Posted January 31, 2005 Share Posted January 31, 2005 There are 9 places to put digits in. The first place always has a 6. The second place can have any one of ten different digits using these: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 The third place can have any one of ten different digits using these: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 And so on . . . So there are 10 possible digits for the second place. 10 for the third and so on . . . up to the 9th place place . . . . . . = 2 . . 3 . . 4 . . 5 . . 6 . . 7 . . 8 . . 9 possible digits = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10^8 This is 10 000 000 = 10 million. They're telling it how it is This is not a factorial type problem If you need to know about this ask Link to comment Share on other sites More sharing options...
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