252
46000
i think alot like 100 at least
17
There are 9000 4-digit numbers. 8*9*9*9 = 5832 of them do not contain a 5 The remaining 3168 contain a 5.
27: 077,177,277,377,477,577,677,877,977,717,727,737,747,757,767,777,787,797
1
There are 6804 such numbers.
The pages of a book are numbered from 1 to 128. How many page numbers contain the digit 6?
648
There are 90 two digit numbers, 10 through 99. There are 18 two digit numbers containing a 9. They are 19, 29, 39, 49, 59, 69, 79, 89, and 90 through 99. Therefore, there are 72 two digit numbers that do not contain a 9.
18
We are to calculate the number of 1-digit to 6-digit numbers (i.e. numbers less than one million) that have at least one 1. Let's divide this into six cases by the number of digits in the number.For one digit numbers, the only number is 1 - 1 such number.For two digit numbers, we can find the number which do not contain a 1 more easily. There are 8 possible digits for the tens place (0 and 1 are not allowed), and 9 digits for the ones place (1 is not allowed). Thus we have 8 * 9 = 72 which do not work. There are 90 two digit numbers, so 90 - 72 = 18 such numbers.For three digit numbers, we will proceed in the same manner. There are 900 total 3 digit numbers, of which 8 * 9 * 9 = 648 numbers which do not contain a 1. The remaining 900 - 648 = 252 numbers contain at least 1 one - 252.For four digit numbers, continuing in the same manner we have 8 * 9 * 9 * 9 = 5832 which are not good out of the 9000 total. So, 9000 - 5832 = 3168.For five digit numbers, we have 8 * 9 * 9 * 9 * 9 = 52488. The remaining 90000 - 52488 = 37512 are good.For six digit numbers, we have 8 * 9 * 9 * 9 * 9 * 9 = 472392. The reamining 900000 - 472392 = 427608 are good.Finally, adding all of the totals, we have 1 + 18 + 252 + 3168 + 37512 + 427608 = 468,559 numbers between 1 and 999999 which contain at least 1 one.
There are 22680 numbers, excluding those with leading 0s.
There are 61328 of them.
19 of them
13