Solution:

Heroine

Solution:

20 + 20 + 20

55 + 5

Solution:

charles put the ring into the box, secure it with one of your locks, and send the box to the girl firend.
She should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back with her lock. Now she can unlock his own lock and retrieve the object.

Solution:

Bill. If you read the message upside down, you'll notice that the numbers resemble letters and that those letters form legible sentences. The message is 'Bill is boss. He sells oil.'

Solution:

Turn the grid upside down. Circle the numbers 1, 9, 1, 1 When you flip it upside down the 6 becomes 9

Solution:

You can never solve a problem on the level on which it was created.

Here is the answer key:
A-Z
B-Y
C-X
D-W
E-V
F-U
G-T
H-S
I-R
J-Q
K-P
L-O
M-N
N-M
O-L
P-K
Q-J
R-I
S-H
T-G
U-F
V-E
W-D
X-C
Y-B
Z-A

Solution:

Because the man is still living

Solution:

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.

Solution:

43

Explanation:
Here is the thing. You can buy any number of wings that can be divided evenly by three except for three by using the combinations of boxes of 6 and boxes of 9. (You must use at most the box of 9 and then multiple boxes of 6). Now, if there is a number that can't be divided by 3, use a box of 20. After purchasing the box of 20, if the number that remains is divisible by 3, then you are good to go. On the other hand if it is not, you will have to use a second box of 20. And the remaining number will definitely be divisible by 3.

Thus if we calculate the largest number that can't be bought is the one that requires two 20 pieces boxes before the remainder can be thinned down to a number divisible by 3. We know that 3 is the number that can't be purchased and can evenly be divisible, the largest possible number of wings that can't be bought are:
3 + 20 + 20 = 43.

Solution:

6

Explanation:
(2^x)(30^3) = (2^3) (3^3) (4^3) (5^3)

(2^x)(30^3) = (120^3)

(2^x) = (120^3)/ (30^3)

(2^x) = (4^3)

(2^x) = 64

ln(2^x) = ln(64)

x ln(2) = ln(64)

x = ln(64)/ln(2)

x = 6

Thus x = 6.