We are not told anything about repeats so I will assume you can repeat a digit. For example, you can have 111111 The first digits has 50 choices so does the second and the third..etc so we have 50^6 combinations.
450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.
then there would only be a 1. and there would be no 2,3,4,5,6,7,8,9 digit numbers onlya one digit number. and there would be only 1,2,3,4,5,6,7,8,9,numbers to use not 10, 20,30,40 ,50 ,60,70,80,90,100 because they need zeros
To find the product of a 2-digit number and a 4-digit number that is approximately 500,000, we need to consider the magnitude of the numbers involved. Since a 2-digit number ranges from 10 to 99 and a 4-digit number ranges from 1000 to 9999, their product will be in the range of 10,000 to 99,000,000. To get a product around 500,000, we can estimate that the 2-digit number is around 50 and the 4-digit number is around 10,000. Therefore, the product of a 50 and 10,000 is 500,000.
20, 40, 50
50 to the 5th power.156,250 combination.
We are not told anything about repeats so I will assume you can repeat a digit. For example, you can have 111111 The first digits has 50 choices so does the second and the third..etc so we have 50^6 combinations.
50 of them.
300
There are many such combinations. 1 x 50 for instance.
450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.
51
I count 14 of them.
there are 50 prime numbers between 1-100 and of them 2,3,5,7 are single digit so that would leave 46 double digit prime numbers between 1-100
The digit zeroes appear in the numbers 10, 20, 30, 40 and 50, making a total of 5 digit zeroes.
16
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)