* Question 6 Go/1 pt O Details Here is the whole story about... "God, Remainders, Ancient Alchemy, and Your Modern Job ET". So, the agency you're working for developed a technology (described in the story) to reshape natural distribution of electron shells (2, 8, 18, 32, 32, 18, 8) into man-made (or mad-made ?): (2, 4, 8, 8, 18, 40, 18, 8, 8, 4). Anyway, most of the discoveries appear "crazy" at first, and besides,.. they pay really well to their employees and researchers. Your assignment is to predict by theoretical calculations how such a transformation will affect... Nickel, Ni (column 10 in the periodic table from the story). In particular, how many active (valence) electrons the transformed element will have: (Nickel, Ni) Mod(2, 4, 8, 8, 18, 40, 18, 8, 8, 4) = 0 x o - How many electrons the transformed element will have on its external orbit: 5 x of . What column of the periodic table (refer to the one presented in the story) the transformed element will belong to: 10 x Describe in (a few) words how significantly (or not) chemical/ mechanical properties of Nickel, Ni might be changed. X Question 7 Go/1 pt O Details Here is the whole story about... "God, Remainders, Ancient Alchemy, and Your Modern Job D". So, the agency you're working for developed a technology (described in the story) to reshape natural distribution of electron shells (2, S, 18, 32, 32, 15, 8) into (58", 40m, 26*, #' ) , which means for any (heavy enough) element with atomic weight NV : the first m orbits (chelle) are filled with 58 electrons each (if 58 goes into N n time:), next m shells have 40 electrons each (if 40 goes into the leftover (N - 58n) m times), next k shells have 26 electrons each (as 26 goes into (NV - 58n - 40m) k time:) the last I chelle have 9 electrons each, and eventually, it might be one more external orbit with a few extra free valence electrons (and that "remainder", i.e. left-over electrons on the external shell will define 100% chemical and biological properties of the element) Your assignment is to develop such a formula/mechanism that would calculate N Mod (58", 40m, 26*, 9') for given N (i.e, number of valence electrons for an element with atomic weight NV ). For example, Kripton, Kro normally inert gas with 36 electrons in total, and zero valence electrons, under experimental conditions: 36 Mod (54", 38", 15*, 8') = 0-54 + 0- 38 + 2 . 15 + 0 .8 + 16 = 6, i.e. it will get 6 valence electrons, and therefore, may exhibit properties similar to Oxygen (see periodic table in the story, column 16 )- Super heavy (to be created) elements with atomic weights 265 or 598 would behave as the guys with 7 or 0 valence electrons (under specified experimental conditions): 265 Mod (64", 32", 18*, 8') = 3- 64 + 1 -32 + 1 . 18 + 2 .8 + 17 =7, 598 Mod (100", 30m, 15*, 4') = 5- 100 + 3 -30 + 0 - 15 + 2 -4 + 80 = 0 Now, to support the experimental research of the agency (you're working for), calculate the number of free/valence electrons for an element with atomic weight 49: 49 Mod (58", 407, 26*, 9') =Solve 6x + 8 = 10 mod 13 for I Show your work through using modular Arithmetic, and/or modular inverse. Correct answer without reasoning has no value (except hinting of possible cheating). Make sure to (attentively) type in the answer below as well: T. mod 13 ! The answer should be positive and less than 13 X Question 2 Go/1 pt 0) Details Solve the following system of two congruencies for I: I = 1 mod 5 I = 2 mod 3 Show your work through using modular Arithmetic, and/or modular inverse, and/or substitution method. Correct answer without reasoning has no value (except hinting at possible cheating). Make sure to (attentively) type in the answers below as well: I mod x Question 3 Go/1 pt O) Details Solve the following system of three congruences for I: I = 6 mod 7 I = 1 mod 5 I = 2 mod 3 Show your work through using modular Arithmetic, and/or modular inverse, and/or substitution method. Correct answer without reasoning has no value (except hinting at possible cheating). Make sure to (attentively) type in the answers below as well: mod Question 4 Go/1 pt 0 Details Find 81910 mod 7 using Fermat's Little Theorem. Show your understanding of the theorem. Correct answer without reasoning has no value (except hinting at possible cheating). Make sure to (attentively) type in the answer below as well: 818146 " mod 7 = X Question 5 Go/1 pt 0 Details solve for z the following congruence: 5x + (13 "mod 5) = 11 mod 17 Show your work in detail. Correct answer without reasoning has no value (except hinting at possible cheating). Make sure to type in the answer below as well: B x mod 17 ! The answer should be positive and less than 17 X Question 6 Go/1 pt O Details