Assuming positive integers, and no leading zeros, the range of five digit numbers is 10000 to 99999. The ones that end in zero can be found by taking the four digit numbers: 1000 to 9999 and multiplying each by ten. {1000,1001, 1002, ...9999}, multiplied by ten is {10000,10010,10020,....99990}. There are 9000 of them.
9
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
8*10*10=800
1,000. The list looks just like the counting numbers from 000 to 999 .
There are 10000 such codes. Each of the numbers 0-9 can be in the first position. With each such first digit, each of the numbers 0-9 can be in the second position. With each such pair of the first two digits, each of the numbers 0-9 can be in the third position. etc.
There are 9 of them.
All of them. There are roughly 45,000 5-digit odd numbers. None of them end in 0.
There are no three didgit numbers but there are 63 three digit numbers.
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
100,000
Even.Work:Even numbers always end with a digit of 0, 2, 4, 6 or 8Odd numbers always end with a digit of 1, 3, 5, 7, or 9
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.
There are no 3 digit numbers between 0 and 9 because 0 and 9 are 1 digit numbers.
99.Because if you do 0 minus 99 whats the answer?99.So it has 99 numbers in a 2-digit number.
0 Look at the product of the first 3 prime numbers: 2 x 3 x 5 = 30. Any number multiplied by 30 will have a 0 in the units digit. So, no matter how many prime numbers you are multiplying, if once you have a number ending in 0, all of the rest will end in 0.
81
That would be all the numbers that end with 0 or 5. So:* For the first digit, you have 9 options (digits 0-9). * For the second digit, you have 10 options. * For the third digit, you have 10 options. * For the fourth digit, you have 2 options (the digits 0 or 5). You can multiply all those together.