Best Answer

a lot

* * * * *

Possibly, though that answer is relative.

The correct answer is 900.

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

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

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.)

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

6 possible 3 digit combonations

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

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

Related questions

450

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.)

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.

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

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

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.

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

6 possible 3 digit combonations