How many times does the digit 1 occur between the numbers 15000 and 30000?

Answer 1

1 + 1 + 15 + 150 + 1500 = 1667

[edit]
It depends on how you count the numbers. I verified the answer above using a computer program, counting each '1' when only when it the digit changed from 0 to 1, i,e:
If it's the case that:
15000, number of ones = 1
15001, add 1 to the number of ones
15002, nothing added
15010, add 1 to the number of ones
15011, add 1 to the number of ones
15012, nothing added
Then the digit 1 'occurs' 4 times in the above example and my answer above is correct.

If you are supposed to count a one every time and in the above list there would be 11 '1's, then others who have said 10500 are correct.

Answer 2

In 000 to 999, the hundreds digit will be 1 exactly 100 times. Similarly the tens digit will be 1 exactly 100 times. And the units digit will be 1 exactly 100 times.

So for every thousand, the number 1 will appear 300 times.

15 x 300 = 4500.

Now in addition to that you have 1000 times when the thousands digit will be 1 (21000 to 21999).

4500 + 1000 = 5500

Finally, there are 5000 times when the ten-thousands digit will be 1 (15000 to 19999).

5500 + 5000 = 10500

So the digit 1 will appear a total of 10,500 times (between 15000-30000, inclusive).

If you mean between 15000-30000 but not including the end-points, then subtract 1 from this number.

Anyway, I figure the answer as: 10,500 times.

Answer 3

The correct answer is 10,500. Louise's counts are right for 15000 to 19999, not for 15001 to 19999. (3645+1215+135+5 = 5000 numbers, not 4999), So later she double counts the 15000 and gets 10,501.

Two answers before have the right number of 10,500.

Answer 4

Between 15001 and 19999 (inclusive),
1 appears exactly once in 3645 numbers. [5×9×9×9](1)
1 appears twice in 1215 numbers. [5×1×9×9](3)
1 appears three times in 135 numbers. [5×1×1×9](3)
1 appears four times in 5 numbers. [5×1×1×1](1)

Between 20000 and 29999,
1 doesn't appear at all in 6561 numbers. [9×9×9×9](1)
1 appears exactly once in 2916 numbers. [1×9×9×9](4)
1 appears twice in 486 numbers. [1×1×9×9](6)
1 appears three times in 36 numbers. [1×1×1×9](4)
1 appears four times in 1 number. [1×1×1×1](1)

The total number of 1's appearing strictly between 15000 and 30000 is
3645 + 2430 + 405 + 20 + 2916 + 972 + 108 + 4
= 10500 times.

If in asking how many times 1 occurs between 15000 and 30000 you mean to include those numbers as well, it's 10501 times.

Answer 5

If we count up from 15,000 to 30,000 the ones-place is going to change 15,000 times (this is how many ones we need to add to 15,000 to make it 30,000). As the ones-place changes it cycles through the ten digits 0-9. In other words the ones-place changes ten times per every cycle through the digits. this gives us a total of 1,500 cycles that the ones-place will go through, and every time it goes through a cycle another digit 1 is used. Therefore, there will be 1,500 instances of the digit 1 that appear in the ones-place as we count up to 30,000, and each instance will last for only one number.

The tens place will also cycle through the digits. the tens-place will change value once per every ones-place cycle. Since there will be 1,500 ones-place cycles, there will be 150 tens-place cycles. Again, each cycle will produce a digit 1 for a total of 150 instances of the digit 1 appearing in the tens-place. Each instance of the digit 1 in the tens-place will persist for ten numbers for a total of (150)(10) = 1,500 digits 1 appearing in the tens-place. (If we look at the series of numbers 23509; 23510; 23511; 23512; 23513; 23514; 23515; 23516; 23517; 23518; 23519; 23520, we see that a digit 1 appears in the tens place and then persists for ten numbers while the ones-place goes through a cycle. Therefore we need to count this instance of a digit 1 in the tens-place ten times.)

A similar analysis of the hundreds-place will show that it will go through 15 cycles for 15 instances of the digit 1 in the hundreds-place and each instance will persist for 100 numbers for a total of 1,500 digits 1 appearing in the hundreds-place.

the thousands-place will cycle only 1.5 times which will produce only one instance of the digit 1, and this instance will persist for 1,000 numbers, giving a total of 1,000 digits 1 appearing in the thousands-place.

the tenthousands-place has a digit 1 in it to begin with, and it will persist for 5,000 numbers before it rolls over to become a digit 2. This give us 5,000 digits 1 that appear in the tenthousands-place,

Summing all these up we have a grand total of:

1,500 + 1,500 + 1,500 + 1,000 + 5,000 = 10,500

Between 15,000 and 30,000 the digit 1 appears 10,500 times.

Answer 6

30000-15000=15000

15000:11 (1,11,21,31,41,51,61,71,81,91)=1364

Answer 7

you must really be tired guy. or bored out of your skull.

Answer 8

14000?