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Q: How many 3-digit positive integers are odd and do not contain the digit 5?

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18 positive integers and 36 integers (negative and positive)

There are 10 one digit positive integers (0 - 9) and 9 one digit negative integers (-9 to -1) making 19 in all.

90

44

4

Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.

the answer would be 900

-21

Positive integers are greater than negative integers. For positive integers: * The integer with more digits is larger. * If two integers have the same length, compare the first digit. If the first digit is the same, compare the second digit, then the third, etc., until you find a difference. In each case, the integer with the larger digit (at the first position where you find a difference) is the larger one.

54

There are 2,000 of them ... every positive integer (counting number) from 2,000 to 2,999 .

there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.

First, separate the negative and positive integers (put them into two separate groups). If there is a zero, you can put it in its own group - or put it into the same group with the positive integers. Negative integers come first, then zero, then positive integers.For positive integers:An integer with less digits comes before an integer with more digits.For integers with the same number of digits, look at the first digit. The integer with the smaller digit in this position comes first.If the first digit is the same, look at the second digit. If those are equal, look at the third digit, etc.For negative integers, it is the other way round - for example, an integer with MORE digits comes first.

There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.

There are one thousand. Starting at 2000 and going to 2999

900 This explains it. A positive integer is a palindrome if it reads the same forward and backwards such as 1287821 and 4554. Determine the number of 5-digit positive integers which are NOT palindromes. We start by counting the total number of 5 digit positive integers. The first digit is between 1 and 9, so we have 9 choices. Each of the other 4 digits can be anything at all (10 choices for each). This gives us 9(10)4 = 90000 five-digit positive integers. Now we need to count the number of 5 digit palindromes. Again, we have 9 choices for the first digit and 10 choices for each of the next two. The tens and units digits however are fixed by our choices so far. Therefore, there are only 900 five-digit palindromes. Therefore, the total number of five-digit positive integers which are not palindromes is 90000-900 = 89100.

There are twelve instances where the integers from 1 to 200 contain the digit 1 at least twice:-11,101,110,111,121,131,141,151,161,171,181,191.

22 of them.

They are 3-digit positive integers.

10 of them.

1 - 3 - 7 - 9 - so ..... 9 * 10 * 4 + 9 * 4 * 6 + 4 * 6 *6 =360+216+104 = 680

100

900 numbers from 100 to 999

72 is one

It is 19.