0
There are 1,368 numbers of three digits that have the number "zero" in them.
Counting positive whole numbers, 171.
100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190 - is 19
200 - 19
300 - 19
400 - 19
500 - 19
600 - 19
700 - 19
800 - 19
900 - 19
Counting negative whole numbers, then there's another 171.
Counting decimal numbers of the .001 variety there is another 171, and another 171 of the -.001 variety.
But there is also another 171 of the 1.01 variety and 171 of the -1.01 variety. And finally another 171 of the 10.1 variety and the -10.1 variety.
This works out to 8 types of three digit numbers with 171 examples in each. All having at least one zero in them, none actually ending in a zero after the decimal point.
You could have several thousand more examples, if one used the square root symbol, or did all the imaginary sets (i), or did squares, cubes, etc., or did factorials (!). And if you started to really mix and match, such as (-8.01i!), well, then it would just get silly!
What else can I help you with?
How many three-digit numbers can you make from 0 6 0 9 2 0 1 2?
There are no three didgit numbers but there are 63 three digit numbers.
How many three-digit area codes are possible if the first digit cannot be a 0?
There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)
How many palindromes are from 0 to 1000?
There are 199 palindromic numbers between 0 and 1000. These include single-digit numbers (0 to 9), two-digit numbers (e.g., 11, 22, ... 99), and three-digit numbers (e.g., 101, 111, ... 999). Each of these categories contributes to the total, with the three-digit palindromes being in the form of ABA, where A and B are digits.
How many palindromic numbers between 0 and 1000?
There are 199 palindromic numbers between 0 and 1000. These include single-digit numbers (0-9), which total 10, and two-digit numbers (11, 22, ..., 99), which add up to 9. Additionally, there are 90 three-digit palindromic numbers, ranging from 101 to 999, that follow the format aba (where a and b are digits). Thus, the total is 10 (single-digit) + 9 (two-digit) + 90 (three-digit) = 109 palindromic numbers.
How many even 3-digit numbers can be formed from digits 0 1 2 4 5 7 9 if each digit can be used only once?
To solve this question, two cases must be considered:Case 1: The three-digit even number ends with 0.If the three-digit number ends with 0, then the number must be even. After the digit 0 is used, six digits remain. After any one of these six digits is chosen to be the first digit, five digits remain. To complete the number, any of the five remaining digits could be chosen to be the second digit of the number.___6___ * ___5___ * ___1___ = 30 three-digit even numbers ending in 0.1st digit-------2nd digit----3rddigit (0)Case 2: The three-digit even number does not end with 0.If the three-digit number does not end with 0, it can end with 2 or 4. Therefore, there are two possibilities for the third digit. The first digit could be any of the remaining digits, EXCEPT 0: if 0 were the first digit, it would not be a three-digit number. Therefore, there would be five possible digits for the first digit. 0 could be used as the second digit, along with the four remaining digits that were not previously used.___5___ * ___5___ * ___2___ = 50 three-digit even numbers not ending in 0.1st digit-------2nd digit-----3rddigit (0)The total number of three-digit even numbers is 80, since 30 three-digit even numbers ending in 0 plus 50 three-digit even numbers not ending in 0 equals 80.
How many three-digit numbers can you make from 0 6 0 9 2 0 1 2?
There are no three didgit numbers but there are 63 three digit numbers.
How many three digit numbers can you make with 4 9 and 0?
144
How many three-digit area codes are possible if the first digit cannot be a 0?
There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)
How many palindromic numbers between 0 and 1000?
There are 199 palindromic numbers between 0 and 1000. These include single-digit numbers (0-9), which total 10, and two-digit numbers (11, 22, ..., 99), which add up to 9. Additionally, there are 90 three-digit palindromic numbers, ranging from 101 to 999, that follow the format aba (where a and b are digits). Thus, the total is 10 (single-digit) + 9 (two-digit) + 90 (three-digit) = 109 palindromic numbers.
How many even 3-digit numbers can be formed from digits 0 1 2 4 5 7 9 if each digit can be used only once?
To solve this question, two cases must be considered:Case 1: The three-digit even number ends with 0.If the three-digit number ends with 0, then the number must be even. After the digit 0 is used, six digits remain. After any one of these six digits is chosen to be the first digit, five digits remain. To complete the number, any of the five remaining digits could be chosen to be the second digit of the number.___6___ * ___5___ * ___1___ = 30 three-digit even numbers ending in 0.1st digit-------2nd digit----3rddigit (0)Case 2: The three-digit even number does not end with 0.If the three-digit number does not end with 0, it can end with 2 or 4. Therefore, there are two possibilities for the third digit. The first digit could be any of the remaining digits, EXCEPT 0: if 0 were the first digit, it would not be a three-digit number. Therefore, there would be five possible digits for the first digit. 0 could be used as the second digit, along with the four remaining digits that were not previously used.___5___ * ___5___ * ___2___ = 50 three-digit even numbers not ending in 0.1st digit-------2nd digit-----3rddigit (0)The total number of three-digit even numbers is 80, since 30 three-digit even numbers ending in 0 plus 50 three-digit even numbers not ending in 0 equals 80.
How many three-digit numbers can be made with 0 1 2 and 3 and the same digit may be used more than once?
999
How many 3 digit numbers can be made from 0123456789?
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
How many three digit even numbers with different digits can be written using the digits 0 1 and 2?
To form a three-digit even number with different digits using the digits 0, 1, and 2, the last digit must be even, which can only be 0 or 2. If the last digit is 0, the first digit can be either 1 or 2 (2 options), and the middle digit will take the remaining digit (1 option). This gives us 2 valid numbers: 120 and 210. If the last digit is 2, the first digit can only be 1 (since it cannot be 0), and the middle digit must be 0. This gives us 1 valid number: 102. In total, there are 3 different three-digit even numbers: 120, 210, and 102.
How many 2-digit numbers contain the digit?
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
How many three digit numbers have the property that the middle digit is the arithmetic mean of the other two digits?
To find the three-digit numbers where the middle digit is the arithmetic mean of the other two digits, we denote a three-digit number as ( abc ), where ( a, b, c ) are its digits. The condition means ( b = \frac{a + c}{2} ), which implies ( a + c ) must be even. For ( a ) (1-9) and ( c ) (0-9), ( b ) must be an integer and a digit (0-9). The pairs ( (a, c) ) that yield valid integers for ( b ) can be counted, leading to 45 valid combinations for three-digit numbers. Thus, there are 45 such three-digit numbers where the middle digit is the arithmetic mean of the other two.
How many 5 digit numbers can be formed from the numbers 0-9?
100,000
How many 2-digit numbers end in 0?
There are 9 of them.
