Hexadecimal Equations (Problem $5$ , Version $3$ )

$E = 14$

$F = 15$

$2 \times 4 = 8$

$8! = 40320$

$2 \times 5 = 10$

$10! = 3628800$

$40320E + 3628800F$

$4 \times 16^5 + 0 \times 16^4 + 3 \times 16^3 + 2 \times 16^2 + 0 \times 16^1 = 4 \times 1048576 + 3 \times 4096 + 2 \times 256 = 4194304 + 12288 + 512 = 4207104 + 14 = 4207118$

$3 \times 16^7 + 6 \times 16^6 + 2 \times 16^5 + 8 \times 16^4 + 8 \times 16^3 + 0 \times 16^2 + 0 \times 16^1 = 3 \times 268435456 + 6 \times 16777216 + 2 \times 1048576 + 8 \times 65536 + 8 \times 4096 = 805306368 + 100663296 + 2097152 + 524288 + 32768 = 908623872 + 15 = 908623887$

$908623887 + 4207118$

$912831005$

Answer $\rightarrow \fbox{912831005}$

For the Bonus:

$912831005$

Do division by $16$ to obtain the L.H.S of the modulo calculations.

$912831005 (\text{mod}(16)) = 13$

$57051937 (\text{mod}(16)) = 1$

$3565746 (\text{mod}(16)) = 2$

$222859 (\text{mod}(16)) = 11$

$13928 (\text{mod}(16)) = 8$

$870 (\text{mod}(16)) = 6$

$54 (\text{mod}(16)) = 6$

$3 (\text{mod}(16)) = 3$

Read the remainders from bottom to top:

$3668112113$

Since $11 = B, 13 = D$

$3668B21D$

Bonus Answer $\rightarrow 3668B21D$