journal:fall2019:jgleas15:week12

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journal:fall2019:jgleas15:week12 [2019/11/08 13:21] jgleas15 |
journal:fall2019:jgleas15:week12 [2019/11/11 15:59] (current) jgleas15 |
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+ | First thing we can see clearly right away is that E must = 0 For example we see D - E = D where D cannot be borrowing so E must = 0 for this to be true | ||

+ | Next we can find the value of C if we take a look at C - E = E since we know that E = 0 and C cannot = E so C must be being borrowed from meaning C would = 1 | ||

+ | |||

+ | Next we can look at the bottom at E - Z = Z which is next to G - G = E so E must be borrowing but not being borrowed from so it is really (E + 10) - Z = Z so the only value that would work for Z would be for Z to = 6 | ||

+ | |||

+ | Now we can look at the middle at G - C = Z which is next to E - E = E so G is not borrowing or being borrowed from so the equation is G - 1 = 6 so G = 7 | ||

+ | |||

+ | Next we can narrow down some values, so now taking a look at the bottom we see Q - G = S which is next to G - G = E so Q is not borrowing meaning S << Q and G << Q now we can look at G - M = Q and since G << Q then G must be borrowing meaning G << M | ||

+ | |||

+ | Next we will look at D - O = G where we know that D in not borrowing but D is being borrowed from so we know that D is >> G and since O has to be greater than 1 since we found the values of 1 and 0 and D is being borrowed from so D is at east 3 more than G which = 7 so D is either A or B in base 12 while O = 2 or 3 | ||

+ | |||

+ | Now we will look at Z - S = C where Z is being borrowed from but not borrowing so the equation is really (6 - 1) - S = 1 so S = 4 | ||

+ | |||

+ | Next we will look at Q - G = S which is not being borrowed from or borrowing so the equation is Q - 7 = 4 so Q = B meaning D = A and O = 2 | ||

+ | |||

+ | Now we can look at D - J = C where there is no borrowing taking place so the equation in base 12 is A - J = 1 so J = 9 and since M >> G, the only value left for M to = is 8 | ||

+ | |||

+ | Finally we can look at T - I = C where T is being borrowed from so the equation is (T - 1) - I = 1 or T - I = 2 and since 5 and 3 are the only values left T = 5 and I = 3 | ||

+ | |||

+ | so we have | ||

+ | |||

+ | 0 = E | ||

+ | |||

+ | 1 = C | ||

+ | |||

+ | 2 = O | ||

+ | |||

+ | 3 = I | ||

+ | |||

+ | 4 = S | ||

+ | |||

+ | 5 = T | ||

+ | |||

+ | 6 = Z | ||

+ | |||

+ | 7 = G | ||

+ | |||

+ | 8 = M | ||

+ | |||

+ | 9 = J | ||

+ | |||

+ | A = D | ||

+ | |||

+ | B = Q | ||

+ | |||

+ | For verification we put the values back in to the puzzle and solve in base 12 | ||

+ | |||

+ | |||

+ | 285 | ||

+ | _________ | ||

+ | 1A6164 | 50AA167B | ||

+ | -390308 | ||

+ | ====== | ||

+ | 13A70A7 | ||

+ | -1301028 | ||

+ | ======= | ||

+ | A607BB | ||

+ | -946778 | ||

+ | ====== | ||

+ | 116043 | ||

+ | |||

+ | Through plugging in the values for verification the values match | ||

journal/fall2019/jgleas15/week12.1573237317.txt.gz · Last modified: 2019/11/08 13:21 by jgleas15

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