Logophobic
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Everything posted by Logophobic
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EAGLE
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PIQUE
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MOUSE FLOCK THIRD
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E P O C H QUOTE - 1 EXTRA - 1 POINT - 0 LINES - 0 BUTTE - 0 EXIST - 1 SNARE - 0 AXMEN - 0 POISE - 0 maurice +5 EPOCH - 5 araver +25?
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_ _ _ _ _ QUOTE - 1 EXTRA - 1 POINT - 0 LINES - 0 BUTTE - 0 EXIST - 1 SNARE - 0 AXMEN - 0
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_ _ _ _ _ QUOTE - 1 EXTRA - 1 POINT - 0 LINES - 0 BUTTE - 0 EXIST - 1 SNARE - 0
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_ _ _ _ _ QUOTE - 1 EXTRA - 1 POINT - 0 LINES - 0
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COXAL
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CODEX
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COMBO
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COINS - If 2, proves C because JOINT - 1 and WORDS - 1 PULSE - If 1, proves L because PRINT - 0, EXIST - 0, and MERGE - 0. If both C and L are disproved, then _O_O_ is proven because COLOR - 2 and COLON - 2
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The way I understood the question, we would be counting the sum of payments made rather than the net difference. In that case, any complete circuit route of 49 moves does require that at least one of them pays a total of at least $25.20. It is possible, in 49 moves, that each pays a total of $25.20 for a net of zero with neither exceeding a net of $12.60.
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There are multiple solutions:
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3 x 3 table with constrains (continued)
Logophobic replied to jasen's question in New Logic/Math Puzzles
The are essentially the same solution. 3 6 4 -3 -6 -48 9 2 x -1 = -8 -9 -21 5 7 -1 -5 -7 -3 -6 -4 10 10 10 7 4 6-8 -9 -2 + 10 10 10 = 2 1 8-1 -5 -7 10 10 10 9 5 3 -
I concur: There are no solutions.
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"the middle-middle square is adjacent to all other square" This implies that the restriction applies to diagonally adjacent squares. There are only two solutions, along with their vertical reflections, that qualify in this case.
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And that is the solution for the fewest number of sides, as described by Buddyboy3000
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A note on the distribution of factors and the number of distinct solutions:
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You forgot reflections. You're solution 4 is a reflection of solution 1. Likewise 3 of 2, 6 of 5, and 8 of 7.
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I see, and the best solution is
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What are you getting at? If what I've posted is 'all true', then there is no other solution.
