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Q: How many five-digit odd numbers are possible if the leftmost digit cannot be zero?

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No, they cannot with real numbers. With complex numbers it is possible.

-1200 and -800, I thought at first, but I cannot think of 2 numbers possible. if 1200 is positive, 960000 has to be negative

A sudoku box cannot contain 5 prime numbers, so it is obviously an error. And once you accept that the numbers in the box are simply nonsense numbers, anything is possible.

No, the product of two positive mixed numbers can never be less than one.

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.)

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No, they cannot with real numbers. With complex numbers it is possible.

450

You cannot turn whole numbers into reduced fractions. Whole numbers are as reduced as possible. The fraction is 26/1

-1200 and -800, I thought at first, but I cannot think of 2 numbers possible. if 1200 is positive, 960000 has to be negative

A sudoku box cannot contain 5 prime numbers, so it is obviously an error. And once you accept that the numbers in the box are simply nonsense numbers, anything is possible.

No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.

No, the product of two positive mixed numbers can never be less than one.

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.

There are 4,500 combinations.

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

If they are not on the same dimension then you cannot do this. They will be different numbers and will not go together.