Decryption Time (Problem $1$ )

In numerical form, k represents $11$ .

Now, according to my encryption formula (to be posted), $(HF)^2$ represents Highest Factor squared, which means $\frac{x}{(HF)^2}$ $= 11$

Now, $x = 44, HF = 2$ as $\frac{44}{4}$ $= 11$ .

So, now we are going to show all the addition pairs of $44$ (excluding pairs $0, 44$ and $44, 0$ )

$1, 43$

$2, 42$

$3, 41$

$4, 40$

$5, 39$

$6, 38$

$7, 37$

$8, 36$

$9, 35$

$10, 34$

$11, 33$

$12, 32$

$13 ,31$

$14, 30$

$15, 29$

$16, 28$

$17, 27$

$18, 26$

$19, 25$

$20, 24$

$21, 23$

$22, 22$

$23, 21$

$24, 20$

$25, 19$

$26, 18$

$27, 17$

$28, 16$

$29, 15$

$30, 14$

$31, 13$

$32, 12$

$33, 11$

$34, 10$

$35, 9$

$36, 8$

$37, 7$

$38, 6$

$39, 5$

$40, 4$

$41, 3$

$42, 2$

$43, 1$

Now (not shown in the main problem due to the list being incomplete), here is the list of characters used:

$a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0$

Using the hints:

$8, 36$

The pair corresponds to the letters $hJ$

Now (not shown in the main problem due to uncertainity):

The number pair may (or may not) show the shifts. If it doesn't, divide by $2$ . If that doesn't work, subtract by any number that has a relationship to the corresponding number in the number pair.

$h - 8 = a$

$36$ doesn't.

Therefore:

$\frac{36}{2}$ $= 18$

$J - 18 = s$

Concatenate the letters to obtain the answer:

Answer: $\fbox{as}$