4,5 and
They are 24, 25 and 26.
That isn't possible. The three consecutive number are assumed to be integers; the sum of three consecutive integers is always a multiple of 3 (try it out).
29
119
125
889 898 988 997 979 799
There are hundreds of possible answers. One triplet is {101, 102, -178}.
100+25+25 = 150
They are 24, 25 and 26.
25, 27, 29
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
There are infinitely many possible sets.For example, using 2 numbers,25 + 25, 24 + 26, 23 + 27, ... 1 + 49, 0 + 50, -1 + 51, -2 + 52, ....Then you can look at numbers with 1 digit after the decimal point:24.9 + 25.1, 24.8 + 25.2, and so onNext you can look at numbers with 3 digits, 4 digits, ... , infinitely many digits after the decimal point.Then move to triplets of numbers, sets of 4 numbers, 5, 6, ... infinitely many numbers.
25 digit numbers would be septillions.
Yeah, of course you can: 25, 34, 43, 52, 61, 70
8 won't to start with. The only thing that will is 5.
the mean of four numbers is 25 and three of the numbers are 17 23 and 25 what is the fourth number?
The first digit must be 7 or greater, because the digits of 699 only sum to 24. Of the numbers beginning with 7, only 799 sums to 25, because if either of the other two digits are less than 9, the sum will be less than 25. That's one so far. For number beginning with 8, the other two digits must sum to 17. That means they must be 9 and 8. So there are two suitable numbers beginning with 8, namely 889 and 898. That's a total of three so far. Finally, we have numbers beginning with 9. The other two digits must sum to 16, so the candidates are: 9 and 7, or 8 and 8. That gives us three numbers that fit the bill: 997, 979 and 988. That's six altogether. So the answer is six.