How many numbers have at least one zero in them between 100,000 and 112,000?

Answer 1

Well every number between 100,000 and 109,999 will have a zero in the 10 thousands place, so that is 10,000 right off the bat.

Similarly the numbers 110,000 to 110,999 all have a zero in the thousands place, so that is another 1,000.

That leaves the numbers between 111,000 and 112,000.

You could count them...
111,000 to 111,099 --> 100
111,100 to 111,109 --> 10
111,110 to 111,199 --> 9
111,200 to 111,209 --> 10
111,210 to 111,299 --> 9
...
111,900 to 111,909 --> 10
111,910 to 111,999 --> 9
112,000 --> 1
---------------
Total in this range: 100 + 9(19) + 1 = 272

But an easier way might be to use permutations. You have the last 3 digits. What is the chance that at least 1 of these digits is zero?

Well, the easier question is what is the chance that *none* of the digits are zero. That would be 9 choices for the first digit, 9 choices for the second digit and 9 choices for the last digit.
9 x 9 x 9 = 729. So 729 combinations have *no* zero. Since 111,000 to 112,000 is 1001 numbers, that leaves 272 numbers that must have at least one zero, same as what we counted manually.

Either way, the total would be 11,272, if you are inclusive of 100,000 and 112,000.

(Note: If you mean exclusively *between* 100,000 and 112,000, the answer is two less, or 11,270.)

Answer 2

Better: 1_ _ _ _ _

Count how many have no 0's in them.

First slot can only be 1.
Second slot can be 2 if following 3 are all 0, ignore this case, so second slot must be 1.
3rd slot can be 1-9
4rd slot can be 1-9
5rd slot can be 1-9
So number with no 0s is 1*1*9*9*9 = 729

Number with at least 1 0: 112000 - 100000 + 1 - 729 = 11272

Answer 3

1200

Answer 4

every tenth number will
every hundreth number will
every thousanth will
every ten thousandth will

so add um up. it'll only take all night

Answer 5

11,270

Answer 6

DONT KNOW