Question - Finding Odd Numbers with Provided Digits
Solution:
To find out how many odd positive integers less than 100 can be represented using the provided digits (2, 3, 8, and 9), we need to look at the possible combinations of these digits that will give us two-digit odd numbers (since all single-digit odd numbers are less than 10 and would not use a combination of the provided digits).An integer is odd if its last digit (ones place) is odd. Among the given digits, the only odd digits are 3 and 9. These digits can only appear in the ones place for the number to be odd.For the tens place, we can use any of the remaining digits (2, 3, 8, and since we've used 9 in ones place, 9 can't be used for two consecutive digits numbers), but we need to ensure that we are not forming a number greater than or equal to 100.So, the possible tens-place digits when the ones place is 3 are 2, 8, and 9 (we can't use 3 at both places as it would form the number 33, which is not greater than any two-digit number). That gives us these numbers: 23, 83, and 93.When the ones place is 9, the possible tens-place digits are 2, 3, and 8, which give us these numbers: 29, 39, and 89.Adding these up, we have:For ones place = 3: 23, 83, 93 (3 numbers)For ones place = 9: 29, 39, 89 (3 numbers)In total, there are 3 + 3 = 6 odd positive integers less than 100 that can be formed using the digits 2, 3, 8, and 9.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com