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Q: How many three digit even numbers are possible if the left digit cannot be zero?

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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.)

a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.

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.

There are 900 three digit numbers. (99 - 1000) (# of possible numbers in the first position = 9) (# of possible numbers in the second position = 10) (# of possible numbers in the third position = 10) 9 *10 *10 = 900

450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.

102 = 100 which is the first possible three digit number that is a perfect square. 312 = 961 which is the last possible three digit number that is a perfect square. So there are 22 three digit positive numbers that are perfect squares.

A three digit number cannot be divisible by a 5 digit number - in any base.

Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.

Total number of possible 3-digit numbers = 9!x10!10!

None. The only way for it to be possible would be 3 zeros which is not considered a 3 digit numbers.

6 possible 3 digit combonations

It is possible to create a 3-digit number, without repeated digits so the probability is 1.

There are 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.

"Sum" requires at least two numbers. So there cannot be athree digit number that sums to anything.

105

There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.

There are only 999 three digit whole numbers.

111 and 201

There are 900 three-digit numbers, ranging from 100 to 999.

Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.

There are no three didgit numbers but there are 63 three digit numbers.

21

A zero

There are 450 such numbers.

There is only one three digit number for a gross, and that is 144.