Solution:

11.3135 + 9.3115+ 9.375

Solution:

There is no mention if you can use sign or not and thus the only possible way to get this is as follows:

(15 - 9) + (13 - 7) + (7 - 1) + (9 - 1) + (13 - 9)

If you solve inside the brackets, you will get the following equation
6 + 6 + 6 + 8 + 4
Adding all these numbers, you will get 30 which is the required sum.

Solution:

138,150,180

The series lists numbers that are flanked by two prime numbers.

4 (3 and 5 are prime)
6 (5 and 7 are prime)
12 (11 and 13 are prime)
18 (17 and 19 are prime)
30 (29 and 31 are prime)
42 (41 and 43 are prime)
60 (59 and 61 are prime)
72 (71 and 73 are prime)
102 (101 and 103 are prime)
108 (107 and 109 are prime)

thus

138 (137 and 139 are prime), 150 (149 and 151 are prime), 180 (179 and 181 are prime)

Solution:

151 (131+20)

Explanation:
1st 7 prime numbers post 100 are
101 , 103 , 107 , 109 , 113 , 127 , 131
...Adding 20 to numbers
121 , 123 , 127 , 129 , 133 , 147 , 151

Solution:

5

23-2=21
21+3=24
24-5=19
19+7=26
26-11=15
15+13=28
28-17=11
11+19=30
30-23=7
7+29=36
36-31=5


(2,3,5,7,11,13,17,19,23,29,31,37,41,43... are Prime Numbers)

Solution:

24

write the numbers in ascending order in words as
onE (Position of 'E' in alphabet is 5)
twO (Position of 'O' in alphabet is 15)
threE (Position of 'E' in alphabet is 5)
fouR (Position of 'R' in alphabet is 18)
and so one

Solution:

445

2=(1+1)1
6=(2+1)2
21=(6+1)3
88=(21+1)4

So 445=(88+1)5

Solution:

794

The nth term in the series is given by the sum of x^n for x in the range 1 to 3, i.e. 1^n + 2^n + 3^n.

Thus the first term is 1^1 + 2^1 + 3^1 = 1 + 2 + 3 = 6,
the second term is 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14,
and so on.

The sixth term is then 1^6 + 2^6 + 3^6 = 1 + 64 + 729 = 794.

Solution:

6526884.

To get from 4 to 12, you use the following formula. 12 = (4 squared)/2 + 4.
Continue on with (X^2)/2 + X.

Solution:

100

list of consecutive numbers that does not contain a 'T' in the spelling