34, 45 etc.
There are ninety 2-digit numbers all together. 45 of them are odd numbers. The other 45 of them are even numbers.
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
Twelve of them.
The numbers 6, 5 and 7 have many things in common with each other. The main thing is that they are all numbers and only contain one digit. Another similarity is that they are very close together.
That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
To make this simpler, I'll assume that numbers with a leading zero like 0179 are "four digit numbers" for the purposes of this question. The number of such numbers that can be formed by those numbers without repeating one of the digits is 4x3x2x1 = 24. We could list them all and add them up, but let's instead just realize that we can choose any digit and put it in any position, and the other digits can then be arranged in 3x2x1 = 6 ways. So the sum for any position is (0+1+7+9)*6 = 102, and the sum of the set of "four digit" numbers is 102x1000+102x100+102x10+102 = 113322. You can figure out for yourself what the sum would be if you exclude the numbers that are really three digit numbers (hint: use the technique above to find the sum of the three digit numbers that can be formed using the digits 1, 7, and 9; then subtract that from the other total).
No, there are composite numbers that end in every other digit.
9,000,000 if there are no other requirements.
Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities. If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.
In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.
You multiply the one digit number on the bottom to every number on the top starting at the right and so on with every other number on the bottom.
For every digit, you multiply the total by the number of possibilities in that one space. So assuming you are only using numbers, that would be 10^9 possibilities. If your pin is using letters, that'd be 26^9. If it's numbers AND letters, 36^9. If it were numbers AND letters and was case sensitive, THAT would skyrocket possibilities to 62^9 (in other words: 13,537,086,500,000,000 combinations). In other words, if you are depositing $, you are quite safe, and if you're trying to steal identities, sucks for you!
Half of them. Every other number from 0 to 99998 is even. Since there are 90,000 five digit numbers, that means that 45,000 of them are even.
The first digit can by selectd in one of two ways, since a number starting with zero would not be a 5-digit number. After that, each of the other four digits can be selected in 3 ways. So the number of 5-digit numbers is 2*34 = 162
you say one number first then the other one for example 789 you say seven eight and then nine
Same as for other numbers. You sum them together and divide the result by the number of fractions.Same as for other numbers. You sum them together and divide the result by the number of fractions.Same as for other numbers. You sum them together and divide the result by the number of fractions.Same as for other numbers. You sum them together and divide the result by the number of fractions.
Sure. Any 1-digit number contains only one-digit numbers as factors. Any larger number, on the other hand, contains itself as a factor, so these can be excluded.
They are written as numbers usually are. The place value of the digit immediately to the left of the decimal point is ones and the place value of all other digits is ten times the value of the digit to their right.
No digit that is 5 places long is equal to 6 times 2 other than 12.000
Only one positive prime number has a 5 in the ones digit. That prime number is 5. All other numbers with a 5 in the ones digit are composite because they will be divisible by 5.
For a 2-digit prime number (which are all odd) to be the sum of two prime numbers, one of the prime numbers will have to be 2. That means the difference between the sum and the other addend will have to be 2. Prime numbers that differ by 2 are called twin primes. There are six pairs of 2-digit twin primes. Your numbers are 13, 19, 31, 43, 61 and 73.
As a digit in other numbers it appears 20 times