Hard Puzzles 

Workings

Hint
This is a simple substitution cipher.

Answer
50.

This puzzle is a simple test to see how easy a ROT13 part encrypted puzzle can be. Every other letter has been converted to ROT13 and the other letters are left as they were. But, you may ask, what is the puzzle? Well, how many words are in this puzzle!

ROT13 replaces each character with the one 13 places away.
A = N, B = O, C = P, D = Q, E = R, F = S, G = T, H = U, I = V, J = W, K = X, L = Y, M = Z, N = A, O = B, P = C, Q = D, R = E, S = F, T = G, U = H, V = I, W = J, X = K, Y = L, Z = M.

Workings

Hint
The list is related to numbers.

Answer
Llama.

The length of each word is the same length as the numbers one, two, three, four, five, six, etc.

The next number is SEVEN, which has 5 letters, and llama is the only word with 5 letters.

Workings

Hint

Answer
Purchase 1: £11.50, £12.00, £12.00, £12.50 = £48.00 (which gives a 24% discount on the next purchase). Purchase 2: £13.00, £13.00, £14.50 = £40.50 minus 24% discount = £30.78. For a total cost of £78.78.

Workings

Hint
The answer begins with 3.

Answer
64 x 58 = 3712.

Reasoning
Since no digit is duplicated, neither number can end in 1, otherwise, the last digit of the answer would already have been used.

Neither number can end in 5 because the answer would then end in 5 or 0.

So the first number can only be 62, 63, 64, 67, or 68.

We can now look at what the second number can end with, and we find that...

   if the first number was 62, the second number can only end in 4 or 7.

Why...
   not 1 as previously explained
   not 2 because we've already used that in the 62
   not 3 because 62 x *3 would end in 6, which is already in the 62
   not 5 as previously explained
   not 6 because we've already used that in the 62
   not 8 because 62 x *8 would end in 6, which is already in the 62

We can repeat this for the other possible first numbers and find that...

   if the first number was 62, the second number can only end in 4 or 7.
   if the first number was 63, the second number can only end in 4, 7, or 8.
   if the first number was 64, the second number can only end in 2, 3, 7, or 8.
   if the first number was 67, the second number can only end in 2, 3, or 4.
   if the first number was 68, the second number can only end in 3 or 4.

Let's check these in turn...

If the first number was 62, the only possible values are:

   62 x 14 = 868
   62 x 34 = 2108
   62 x 54 = 3348
   62 x 74 = 4588

or

   62 x 17 = 1054
   62 x 37 = 2294
   62 x 57 = 3534
   62 x 87 = 5394

Only 4 calculations are required for each option, as we didn't need to check 62 x 84 as we already know that the answer would end in 8, or 62 x 47 as we already know that the answer would end in 4.

All of these answers fail as they don't contain all of the digits.

Similarly, we can look at 63, then 64, etc. For each of the numbers, we have to check the possible endings, but in each case, only 4 calculations are required (as seen in the examples above).

We (quickly?) find 64 x 58 = 3712.