Consecutive numbers without using a 7 ?

[starfish99]

Consecutive numbers without using a "7"?

I tried to calculate the number of consecutive integers between 10,000 and 99,999 that do not use the number 7. I got part of the way there.
There are 90,000 numbers(if you start with 10,000) and if you subtract 1/9 of 90,000 that will give you a total of 80,000 that don't begin with a seven. How do you handle the calculation of the the seven in the thousands, hundreds, tens and units place?

 

[Robert1986]

Let me see if I understand this correctly. Say that you wanted to do the same, but do it for numbers in the range 0 to 9. Then the numbers that you are talking about are just: {1,2,3,4,5,6,8,9}, right? Or, if you are doing it from 10 to 99 the numbers 10,11,12,13,14,15,16,18, would all be in the set (I'm not saying these are all the numbers, just some) but the numbers 7,17,27,74 would not (again, there are more that aren't in the set, I just want to make sue I understand), correct?

You have a five digit number. Let's write it this way: a_4*10^4 + a_3*10^3 + a_2*10^2 +a_1*10^1 + a_0*10^0. Now, there are certain restrictions we have on the a_i's, right? What are they? For instance, none of them can be "7", right? Can you think of any other restrictions for any of the other numbers (consider a_4, in particular.) Now, given the restrictions on each a_i, how many possibilities are there for each a_i? That is, how many choices do you have for a_4, a_3,...,a_0?

Now do you see what to do? In other words, make this a completely combinatorial problem.

 

[starfish99]

So a(4) can be any of eight numbers(1 thru 9) excepting 7 and a(3) a(2) a(1) and a(0) can be any of nine numbers (0 thru 9) excepting 7. The ans. is 90 000 minus 8x9x9x9.
Thank you

 

[starfish99]

Sorry Robert1986
8X9X9X9 is the number of 5 digit numbers without a seven.

90 000 minus 8X9X9X9 is the number of 5 digit numbers with a seven.
Thanx again