For example, how many FIVE-DIGIT ZIP CODES are there? And how to calculate the same to arrive at the answer?
2006-09-18 10:16:27 · 14 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
OK, I thought so, the problem was, my daughter is doing homework and she asked me a question about 00000-99999, I gave her the answer and she then tells me I'm wrong. I study on it from the only angle I can find and then she comes back with "No Daddy, the other question...." So, SORRY to bother you. We're back on the same page over here, now. Thank you all....
2006-09-18 10:27:23 · update #1
There are 100,000 five-digit numbers from 00000 to 99999.
10 x 10 x 10 x 10 x 10 = 100,000
2006-09-18 10:18:20 · answer #1 · answered by Deep Thought 5 · 4⤊ 0⤋
The answer is 10*10*10*10*10
or 10^5 which is equal to a 1 with 5 zeros on the end.
A hundred thousand
Think about it, one digit is simply 10. Two digits is simply 100. O wait a second, is there a pattern emerging here? Wow so there is because 3 digits is 1000, with 3 zeros on the end!
See, as simple as that.
Now how many would there be with a ten digit number.... hmmmmm..
2006-09-18 10:20:06 · answer #2 · answered by Chris 3 · 0⤊ 0⤋
Well.... if you are talking all valid 5 digit numbers such that you cant have leading zeroes (i.e. 00005 isnt allowed) you would have
9*10*10*10*10 = 9x10^4
if however you can have leading zeros its equal to
10^5
most of the above are wrong because 1, 123, 5933, etc are not valid 5 digit numbers. You have to think of it like this:
In the first spot I can have 9 possible values (one through nine), and in each subsequent spot i can have 10 possible values (zero through nine) and then you multiply all of them together (nine times ten to the four)
Allen
2006-09-18 10:19:44 · answer #3 · answered by Allen G 3 · 0⤊ 0⤋
Hmmm...
There are 10 choices for each digit.. ,
The number for each digit can be chosen independently (i.e. digit #2's value isn't dependent on digit #1's value, etc, etc.).
So for independent events, multiply the choices or
10 * 10 * 10 * 10 * 10 = 10^5 = 100 000
I think that's it.
Good luck...
2006-09-18 10:28:21 · answer #4 · answered by Mark B 2 · 0⤊ 0⤋
Assuming you're using base ten, there are 10000 unique five-digit numbers. This makes sense, because there are 10 unique one-digit numbers, if you include the zero.
2006-09-18 10:19:58 · answer #5 · answered by Doctor Why 7 · 0⤊ 0⤋
10^5. However, if you consider ABCDE and DEBAC to be the same, there are 10*9*8*7*6 numbers.
2006-09-18 10:21:29 · answer #6 · answered by Fudge 2 · 0⤊ 0⤋
The range is from 00000 to 99999.
The total is 10,000.
2006-09-18 10:18:54 · answer #7 · answered by Jay 6 · 0⤊ 1⤋
have ten possible numbers for the first slot
0 -9 ten possible for the second slot again 0 -9
mutiply ten * ten *ten *ten *ten *ten = 100,000....or you could jsut add 1 too 99,999 (the highest possible # in this situation ) you add 1 too accountt for the number 00000
2006-09-18 10:19:29 · answer #8 · answered by brotherwolf 2 · 0⤊ 0⤋
9 choices in the 10000th position
9 in 1000th
8 in 100th
7in 10th
and 6 in units
so 9*9*8*7*6=27216 unique numbers
2006-09-18 10:23:21 · answer #9 · answered by raj 7 · 0⤊ 2⤋
100000. 00000-99999, I'm pretty sure it totals 100000 different unique numbers.
2006-09-18 10:20:38 · answer #10 · answered by Jethro 5 · 0⤊ 0⤋