888, 88, 8, 8, 8.
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).
The 3-digit counting numbers are 100 through 999 = 900 numbers.Half them are multiples of 2 (even numbers).The other half are not . . . 450 of them.
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.
0.092, because in the tenths part of this number, the digit is 0 wheras in the other numbers it is more than 0.
yes
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.
Yes. By 1 digit, 2 digit and some even by other 3 digit numbers.
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.
Eight (8) of them do.They are1221243642486384 .
No, there are composite numbers that end in every other digit.
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).
9,000,000 if there are no other requirements.
Yes.