phone flub. math help?

Question

debbie forgot lindas pnone number. she did remember the first three digits were 297- and that all the rest of the digits in the phone number were 2s and 9s and 7s. debbie is determined to get in touch with linda. "ill just dial phone numbers that fit this description untill i reach her. there cant be THAT many, can there?" she said. how many wrong numbers might she dial?

Answer 1

Worst case, she would dial every wrong number and finally get it on the last one.

Since you have 4 digits, and 3 possible choices for each digit, there are 3x3x3x3 = 3^4 = 81 possible phone numbers.

So worst case she'd dial 80 wrong numbers before getting the right one.

Answer 2

There are four remaining digits. For each of the four digits there are three possible numbers (2, 9, and 7) so the total number of combinations would be 3^4 = 81

Answer 3

There are 3^4 or 81 4-digit numbers composed of 2s, 7s, and 9s. Therefore, one would expect to dial 39.5 (half of 81 minus 1) wrong numbers before hitting the right one.

Answer 4

If there are four remaining digits (297-xxxx), and there are three possible choices for each digit (2, 7, 9) then the possible combinations of digits are:

3^4 = 81

There are 81 combinations of phone number.

Answer 5

Only the last 4 digits are in question and they can each be any of three numbers. So the number of possible combinations is:

3^4 = 81

Of those, 80 are wrong numbers and one is right.

Answer 6

So sorry: There are 7 remaining digits, at least for
phone numbers in the USA
So there are 3^7 = 2187 possibilities. She might
dial 2186 wrong numbers before getting
the right one.
In the USA the first 3 digits give the area code.

Answer 7

what you can do is call 411 can give them her last name

Answer 8

a lot