To find the number of odd numbers between 1000 and 9999 with distinct digits, we first note that the last digit must be one of the odd digits: 1, 3, 5, 7, or 9. This gives us 5 options for the last digit. For the first digit (thousands place), we can choose from 1 to 9 (excluding the last digit), giving us 8 options. The second digit (hundreds place) can be any digit from 0 to 9, excluding the first and last digits, which gives us 8 options. Finally, the third digit (tens place) can be any digit from 0 to 9, excluding the first, last, and second digits, yielding 7 options.
Thus, the total count is:
5 (last digit) × 8 (first digit) × 8 (second digit) × 7 (third digit) = 2240.
Therefore, there are 2240 odd numbers between 1000 and 9999 with distinct digits.
16
91.
solution: we know that there are 25 prime numbers are between 1-100 and 168 prime numbers less than 1000. 100 x 100=10000(5 digits) 999 x 999=998001(6 digits) 1000 x 1000=1000000(7 digits) so our answer should be same as the number of prime numbers between 100 to 999. hence, 168-25=143. 143 prime numbers will be there less than 1000 whose square has 5 or 6 digits.
952 of them.
it s 243..
There are 120 of them.
123
Not couting 1 and 1000, there would be 998 numbers.
9*9*8*7 = 4536
16
91.
252 of them.
solution: we know that there are 25 prime numbers are between 1-100 and 168 prime numbers less than 1000. 100 x 100=10000(5 digits) 999 x 999=998001(6 digits) 1000 x 1000=1000000(7 digits) so our answer should be same as the number of prime numbers between 100 to 999. hence, 168-25=143. 143 prime numbers will be there less than 1000 whose square has 5 or 6 digits.
952 of them.
it s 243..
There are 999 numbers between 1 and 1000, which includes all the integers from 1 to 999. If you're asking about the count of individual digits used in writing these numbers, they collectively comprise a total of 2887 digits. This is calculated by considering the number of digits in one-digit (1-9), two-digit (10-99), and three-digit numbers (100-999).
An infinite amount