How many positive odd integers between 100 and 10000 can be formed using the digits 2,3,4,5 and 6
in the ones place two. 3, 5
note: --- in higher places can be zero, like 010 = 10 so 10 of them
in the tens place ten so 20 to date
in the hundreds place ten so 200
in the thousands place ten so 2000
10003 is too big so none allowed in ten thousands place
so I get 2000
To find the number of positive odd integers that can be formed using the digits 2, 3, 4, 5, and 6 between 100 and 10,000, we need to consider two cases:
Case 1: The number has three digits.
In this case, the first digit must be a 2, 3, 4, 5, or 6. Since the number is odd, the last digit must be 1, 3, 5, or 7. There are 5 choices for the first digit and 4 choices for the last digit. For the middle digit, any of the five given digits (2, 3, 4, 5, 6) can be used. So, for each choice of the first and last digits, there are 5 options for the middle digit. Therefore, the total number of odd integers with three digits is 5 × 5 × 4 = 100.
Case 2: The number has four digits.
In this case, the first digit can be any of the five given digits (2, 3, 4, 5, 6). The last digit must be 1, 3, 5, or 7. For the second and third digits, any of the five given digits can be used. So, there are 5 choices for the first digit, 5 choices for the second digit, 5 choices for the third digit, and 4 choices for the last digit. Therefore, the total number of odd integers with four digits is 5 × 5 × 5 × 4 = 500.
Combining both cases, the total number of positive odd integers between 100 and 10,000 that can be formed using the digits 2, 3, 4, 5, and 6 is 100 + 500 = 600.