MAXkerade Extra Credit Challenge O'Doom!!

[Johnny Canuck]

You don't need to have been previously involved in MAXkerade to complete this challenge. In fact, you don't even need to know what it is (although if you're curious, an overview is given here). However, a gauntlet of final exam revisions currently blocks me from putting out live challenges in the chat, and therefore, I've posted this MAXkerade Extra Credit Challenge O'Doom (MECCO) to tide you all over. For anyone who hasn't heard of my tourney, this is just a sweet puzzle you can work on in your spare time.

The MECCO is similar to a Countdown Wiki scavenger hunt. Proceed through the problems, working out the values of the variables A through X, and then try to solve the final puzzle. If you complete this last puzzle, please send your solution for it to me via PM. The first person to solve this will receive a reward that they are able to put in their forum signature.

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~~~ THE RULES ~~~
Please note: In all the puzzles below, "number" refers to a nominal number or index (for instance, the "seed number" of Dylan Taylor in Series 69 would be 1, since he was the #1 seed), while "quantity" refers to a cardinal number or count (the "quantity of points" achieved by Taylor would be 974). Some of the intermediary answers may not be integers. All the answers are available on the wiki, with the exception of "U", for which I may give a hint if one is needed. Write down the values of all your variables as you go!

~~~ THE VARIABLES ~~~
Let A equal the last digit of the first target CECIL was ever seen to generate on the show.
Let B equal the quantity of preliminary games won by the winner of the episode in which the above target generation occurred, excluding draws.
Let C equal the total quantity of tie-break conundrums that appeared in Series [B cubed].
Let D equal the number, translated from Roman numerals, of the only Championship of Champions in which the quantity of contestants was NOT an integer power of C.
Let E equal the seed number of the runner-up of the last completed series whose number was a multiple of D.
Let F equal the number of the only other past series, besides Series E, that was not represented by any contestant in Series 33 (which consists of CoC VIII + Supreme Championship).
Let G equal the length, in letters, of the first valid word spotted in the grand final of the series whose number is the product of those of the two unrepresented series described above (i.e., Series [E * F]).
Let H equal the quantity of points amassed by the winner of this grand final, excluding points obtained from the aforementioned G-letter word.
Let I equal the total quantity of letters in the first and last names of the first contestant who scored exactly H points in a "new" 15-round format game.
Let J equal the quantity of special episodes broadcast during the summer break that occurred during the Ith series not to have been presented by Richard Whiteley.
Let K equal the number of the most recent series in which a player made the grand final after having won exactly J games (excluding draws).
Taking "A" = 1, "B" = 2, ..., "Z" = 26 (wow, confusing as hell; at least they're not bolded), let L be the sum of the values of all the letters in the conundrum that the champion of Series K solved during his/her CoC grand final.
Let M equal the episode number of the most recent quarter-final, semi-final or grand final, excluding 30th Birthday Championship episodes, in which a contestant achieved a score of exactly L points.
Let N equal the conundrum solve time, in seconds as recorded by the wiki, for Episode [M modulo 1000].
Let O equal the total quantity of victories achieved, throughout his/her entire Countdown career, by the champion of the series whose number is closest to [N cubed].
Let P equal the score that the first-ever O-year-old series champion achieved on his/her debut episode.
Let Q equal the total quantity of invalid words declared by both contestants during the Pth Masters episode.
On one occasion during the show's history, it was possible to reach the target in a numbers game using exactly Q mathematical operators (an operator being +, -, * or /). Let R equal this target.
Let S equal the quantity of wins achieved by the player whom the Rth 15-round octochamp in Countdown history beat on his/her debut.
Series [S squared] contained an even quantity of preliminary games (heats). Let T equal the sum of the winners' (winner's?) scores from the two chronologically centremost heats of the series.
Let U equal the highest integer such that the product of U and T is a valid target in the Preposterous variant, as described on the forum in the CobliviLon 2012 thread.
Let V equal the seed number achieved by the winner of the most recent episode whose episode number ended in U.
Let W equal the sum of all the losing scores achieved during Championship of Champions V (not the Roman numeral V, the variable V!).
Let X equal the sum of all "large numbers" that the winner of Jeff Stelling and Rachel Riley's Wth non-special episode used in his/her solutions during this episode.

~~~ THE FINAL PUZZLE ~~~
Construct a Hyper numbers game that has M as the target and A, X, K, E, R, A, D, E as the working numbers. Solve it as closely as possible.

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The deadline for entering the MECCO is Friday, 15 August, at which time full solutions will be posted. However, only the first person to solve it correctly will get the signature prize. So get cracking!

[Johnny Canuck]

Eternal congratulations -- as usual, belated -- go out to Giles and Innis, who together formed the only team that even attempted this puzzle. A full solution will now be posted. If, for whatever reason, you still want to attempt this challenge, please don't scroll down, because below the spoiler space, the answers are written in full view.

SPOILER SPACE 0101 0111 SPOILER SPACE 0110 1111 SPOILER SPACE 0111 0111 SPOILER SPACE 0010 0111 SPOILER SPACE
SPOILER SPACE 0010 0000 SPOILER SPACE 0111 0110 SPOILER SPACE 0110 0101 SPOILER SPACE 0111 0010 SPOILER SPACE
SPOILER SPACE 0111 1001 SPOILER SPACE 0010 0000 SPOILER SPACE 0110 0111 SPOILER SPACE 0110 1111 SPOILER SPACE
SPOILER SPACE 0110 1111 SPOILER SPACE 0110 0100 SPOILER SPACE 0010 1110 SPOILER SPACE 0010 0000 SPOILER SPACE
SPOILER SPACE 0101 0011 SPOILER SPACE 0110 1000 SPOILER SPACE 0110 0001 SPOILER SPACE 0110 1101 SPOILER SPACE
SPOILER SPACE 0110 0101 SPOILER SPACE 0010 0000 SPOILER SPACE 0111 0100 SPOILER SPACE 0110 1000 SPOILER SPACE
SPOILER SPACE 0110 1001 SPOILER SPACE 0111 0011 SPOILER SPACE 0010 0000 SPOILER SPACE 0110 0100 SPOILER SPACE
SPOILER SPACE 0110 1111 SPOILER SPACE 0110 0101 SPOILER SPACE 0111 0011 SPOILER SPACE 0110 1110 SPOILER SPACE
SPOILER SPACE 0010 1100 SPOILER SPACE 0111 0100 SPOILER SPACE 0010 0000 SPOILER SPACE 0111 0011 SPOILER SPACE
SPOILER SPACE 0110 0001 SPOILER SPACE 0111 1001 SPOILER SPACE 0010 0000 SPOILER SPACE 0110 0001 SPOILER SPACE
SPOILER SPACE 0110 1110 SPOILER SPACE 0111 1001 SPOILER SPACE 0111 0100 SPOILER SPACE 0110 1000 SPOILER SPACE
SPOILER SPACE 0110 1001 SPOILER SPACE 0110 1110 SPOILER SPACE 0110 0111 SPOILER SPACE 0010 0000 SPOILER SPACE
SPOILER SPACE 0110 1001 SPOILER SPACE 0110 1101 SPOILER SPACE 0111 0000 SPOILER SPACE 0110 1111 SPOILER SPACE
SPOILER SPACE 0111 0010 SPOILER SPACE 0111 0100 SPOILER SPACE 0110 0001 SPOILER SPACE 0110 1110 SPOILER SPACE
SPOILER SPACE 0111 0100 SPOILER SPACE 0010 1110 SPOILER SPACE 0010 1110 SPOILER SPACE 0010 1110 SPOILER SPACE

MAXkerade Extra Credit Challenge O'Doom Answers!!
A
Problem: Let A equal the last digit of the first target CECIL was ever seen to generate on the show.
Answer: A = 8 (or A = 3; see explanation)
Source: http://wiki.apterous.org/Episode_1
Explanation: The article's introductory text clearly states that "Before the first numbers game, CECIL initially generated a target of 988." Since this was the first problem, however, I was lenient and I think I also accepted 493 as CECIL's first target and thus 3 as an answer.

B
Problem: Let B equal the quantity of preliminary games won by the winner of the episode in which the above target generation occurred, excluding draws.
Answer: B = 4
Source: http://wiki.apterous.org/Episode_1 and http://wiki.apterous.org/Michael_Goldman
Explanation: The first target was, of course, generated in Episode 1, which was won by Michael Goldman. Goldman won three further games (four in all) and drew another.

C
Problem: Let C equal the total quantity of tie-break conundrums that appeared in Series [B cubed].
Answer: C = 2
Source: http://wiki.apterous.org/Series_64
Explanation: The cube of B = 4 is 64; Series 64 contained two tie-break conundrums.

D
Problem: Let D equal the number, translated from Roman numerals, of the only Championship of Champions in which the quantity of contestants was NOT an integer power of C.
Answer: D = 10
Source: http://wiki.apterous.org/Championship_of_Champions and http://wiki.apterous.org/Championship_of_Champions_X
Explanation: All Championships of Champions have featured either eight or sixteen contestants (both of which are powers of two), with the exception of CoC X, which was shortened to one week and featured only six players.

E
Problem: Let E equal the seed number of the runner-up of the last completed series whose number was a multiple of D.
Answer: E = 3
Source: http://wiki.apterous.org/Series_70
Explanation: Series 70 was the most recent series whose number was a multiple of 10. Its runner-up was Andy Naylor, the #3 seed.

F
Problem: Let F equal the number of the only other past series, besides Series E, that was not represented by any contestant in Series 33 (which consists of CoC VIII + Supreme Championship).
Answer: F = 14
Source: http://wiki.apterous.org/Series_33
Explanation: The introductory text of the article states that Series 33 "included entrants from all previous series except for Series 3 and Series 14."

G
Problem: Let G equal the length, in letters, of the first valid word spotted in the grand final of the series whose number is the product of those of the two unrepresented series described above (i.e., Series [E * F]).
Answer: G = 9
Source: http://wiki.apterous.org/Series_42 and http://wiki.apterous.org/Episode_2797
Explanation: The product of E and F is 42, and the first valid word offered in the Series 42 final was the niner MYOGLOBIN.

H
Problem: Let H equal the quantity of points amassed by the winner of this grand final, excluding points obtained from the aforementioned G-letter word.
Answer: H = 83
Source: http://wiki.apterous.org/Episode_2797
Explanation: Michael Calder won the Series 42 final, and achieved 101 points therein: 18 from MYOGLOBIN and 83 from his remaining offerings.

I
Problem: Let I equal the total quantity of letters in the first and last names of the first contestant who scored exactly H points in a "new" 15-round format game.
Answer: I = 12
Source: http://wiki.apterous.org/Series_68 (and I guess http://wiki.apterous.org/15_round_format_(new) if you didn't know when the change occurred)
Explanation: The first player to get exactly 83 points in the "new" 15-round format was Joe McGonigle, against Ian Burn in Episode 5695.

J
Problem: Let J equal the quantity of special episodes broadcast during the summer break that occurred during the Ith series not to have been presented by Richard Whiteley.
Answer: J = 1
Source: http://wiki.apterous.org/Series_65 (and I guess http://wiki.apterous.org/Richard_Whiteley)
Explanation: Richard Whiteley's last series was Series 53, making Series 65 the twelfth series without him. The introductory text of this series' article states that "1 special episode [was] shown during the break."

K
Problem: Let K equal the number of the most recent series in which a player made the grand final after having won exactly J games (excluding draws).
Answer: K = 9
Source: http://wiki.apterous.org/Mick_Keeble, http://wiki.apterous.org/Series_9 and probably many other series seed tables.
Explanation: As mentioned in Mick Keeble's article, "Keeble and Joyce Cansfield were the only players to ... make Grand Finals, having won only one game." Joyce Cansfield appeared earlier than Keeble, however -- in Series 1.

L
Problem: Taking "A" = 1, "B" = 2, ..., "Z" = 26 ... Let L be the sum of the values of all the letters in the conundrum that the champion of Series K solved during his/her CoC grand final.
Answer: L = 88
Source: http://wiki.apterous.org/David_Trace and http://wiki.apterous.org/Episode_601
Explanation: David Trace was the champion of Series 9, and lost the Championship of Champions III grand final to Harvey Freeman. Therein, he solved the conundrum AERIALGUN (NEURALGIA), which has a letter value sum of 1 + 5 + 18 + 9 + 1 + 12 + 7 + 21 + 14 = 88.

M
Problem: Let M equal the episode number of the most recent quarter-final, semi-final or grand final, excluding 30th Birthday Championship episodes, in which a contestant achieved a score of exactly L points.
Answer: M = 5516
Source: Many series pages, going backwards in time...
Explanation: Peter Lee (R.I.P. -- this question was In Memoriam) was, at the time of writing, the most recent player to get 88 points in a finals-stage game -- namely, Episode 5516, which was his Series 66 semi-final against Jack Worsley.

N
Problem: Let N equal the conundrum solve time, in seconds as recorded by the wiki, for Episode [M modulo 1000].
Answer: N = 3.5
Source: http://wiki.apterous.org/Episode_516 and maybe http://en.wiktionary.org/wiki/modulo
Explanation: "Y modulo Z" is an operator which gets the remainder when Y is divided by Z. Thus, M modulo 1000 can be found by taking the last three digits of M -- namely, 516. In Episode 516, Ted Kimmons solved the conundrum on 3.5 seconds.

O
Problem: Let O equal the total quantity of victories achieved, throughout his/her entire Countdown career, by the champion of the series whose number is closest to [N cubed].
Answer: O = 15
Source: http://wiki.apterous.org/Series_43 and http://wiki.apterous.org/Graham_Nash
Explanation: The cube of 3.5 is 42.875, and the closest integer to this is 43. Graham Nash won both Series 43 and Championship of Champions XI, racking up a total of fifteen victories.

P
Problem: Let P equal the score that the first-ever O-year-old series champion achieved on his/her debut episode.
Answer: P = 73
Source: http://wiki.apterous.org/Wayne_Summers
Explanation: Wayne Summers was the first 15-year-old to win a series (Series 24), and he achieved 73 points in his first game (Episode 1263).

Q
Problem: Let Q equal the total quantity of invalid words declared by both contestants during the Pth Masters episode.
Answer: Q = 0
Source: http://wiki.apterous.org/Episode_M73
Explanation: No invalid words were declared by either contestant in Episode M73.

R
Problem: On one occasion during the show's history, it was possible to reach the target in a numbers game using exactly Q mathematical operators (an operator being +, -, * or /). Let R equal this target.
Answer: R = 100
Source: Possibly none at all; possibly http://wiki.apterous.org/CECIL; very unlikely http://wiki.apterous.org/Episode_1504
Explanation: An older version of CECIL was able to generate 100 as a target, doing so on four occasions. This is the only target to have ever been the same as one of the possible starting numbers, and on one occasion (Episode 1504), 100 came up both as a starting number and as the target. Thus, in this instance, zero operators were needed to achieve the target.

S
Problem: Let S equal the quantity of wins achieved by the player whom the Rth 15-round octochamp in Countdown history beat on his/her debut.
Answer: S = 6
Source: http://wiki.apterous.org/Jack_Worsley, http://wiki.apterous.org/Phyllis_Styles and http://wiki.apterous.org/Octochamp
Explanation: Text on Jack Worsley's wiki page states that he is "the 100th 15-round octochamp in Countdown history." On his debut, he beat six-time winner Phyllis Styles.

T
Problem: Series [S squared] contained an even quantity of preliminary games (heats). Let T equal the sum of the winners' (winner's?) scores from the two chronologically centremost heats of the series.
Answer: T = 117
Source: http://wiki.apterous.org/Series_36 and more conveniently http://www.thecountdownpage.com/series36.htm
Explanation: The square of 6 is 36; Series 36 contained 58 heats, as is more explicitly mentioned on The Countdown Page, making the 29th and 30th heats the middlemost. Tony Baylis won both of these heats, achieving scores of 56 against Will Paine and 61 against Rodders Grafton for a total of 117.

U
Problem: Let U equal the highest integer such that the product of U and T is a valid target in the Preposterous variant, as described on the forum in the CobliviLon 2012 thread.
Answer: U = 854
Source: viewtopic.php?f=10&t=9220 and a bit of common sense
Explanation: The Preposterous variant, used for several puzzles at CobliviLon 2012, is described in this puzzle thread as featuring five-digit targets, naturally making the highest possible target therein 99999. Dividing this by 117 produces 854.6923, and thus the highest integer is clearly 854.

V
Problem: Let V equal the seed number achieved by the winner of the most recent episode whose episode number ended in U.
Answer: V = 7
Source: http://wiki.apterous.org/Episode_5854 and http://wiki.apterous.org/Series_70
Explanation: At the time of writing, Episode 5854 was the last episode whose number ended in 854. This episode was won by Priscilla Munday, who became #7 seed of Series 70.

W
Problem: Let W equal the sum of all the losing scores achieved during Championship of Champions V (not the Roman numeral V, the variable V!).
Answer: W = 290
Source: http://wiki.apterous.org/Championship_of_Champions_VII
Explanation: The sum of the defeated players' scores in the seven matches of the seventh Championship of Champions was 45 + 46 + 38 + 36 + 35 + 17 + 73 = 290.

X
Problem: Let X equal the sum of all "large numbers" that the winner of Jeff Stelling and Rachel Riley's Wth non-special episode used in his/her solutions during this episode.
Answer: X = 150
Source: http://wiki.apterous.org/Episode_5022 and possibly http://wiki.apterous.org/Jeff_Stelling to find out when the first episode was
Explanation: Jeff Stelling and Rachel Riley took over as Countdown presenters in Episode 4733, making the number of their 290th episode 4733 + 289 = 5022. Nicki Sellars won Episode 5022, and therein used the large numbers 25, 75 and 50 in her first, second, and third numbers game solutions. The sum of these is, naturally, 150.