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Modificato da The Cellar Door:
3/27/2016 4:40:03 AM
13
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It's 0 n=0
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I'm part of the altisum
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Lmao you forgot the most important part.
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42! [spoiler]The answer to everything.[/spoiler]
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That's pythagoras' theorem. A^2 + B^2 = C^2
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Is 5 tooooooooooo eashy
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um ABCDEFGHIJKLMNOPQRSTUVWXYZ 1234567891011121314151617181920212223242526
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Assuming A, B, and C must all be whole numbers, there is no solution. [b]However[/b], because it is not stated within the problem, we can use irrational numbers to solve. One solution would be as such: A=∛3 B=∛3 C=∛6 n=3 Since we are cubing each cube root, the equation becomes 3+3=6, which is correct. Another solution could be found by using 0 to make the equation simplify to 0+0=0, since 0 to the power of any non-zero number is 0. A=0 B=0 C=0 n=3 If you want to modify the equation so it does not have a solution, you would have to include the rule that A, B, and C must be whole numbers. Otherwise, irrational numbers and zero allow for solutions.
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[quote]Assuming A, B, and C must all be whole numbers, there is no solution. [b]However[/b], because it is not stated within the problem, we can use irrational numbers to solve. One solution would be as such: A=∛3 B=∛3 C=∛6 n=3 Since we are cubing each cube root, the equation becomes 3+3=6, which is correct. Another solution could be found by using 0 to make the equation simplify to 0+0=0, since 0 to the power of any non-zero number is 0. A=0 B=0 C=0 n=3 If you want to modify the equation so it does not have a solution, you would have to include the rule that A, B, and C must be whole numbers. Otherwise, irrational numbers and zero allow for solutions.[/quote]Uh..what he said!
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[quote]Assuming A, B, and C must all be whole numbers, there is no solution. [b]However[/b], because it is not stated within the problem, we can use irrational numbers to solve. One solution would be as such: A=∛3 B=∛3 C=∛6 n=3 Since we are cubing each cube root, the equation becomes 3+3=6, which is correct. Another solution could be found by using 0 to make the equation simplify to 0+0=0, since 0 to the power of any non-zero number is 0. A=0 B=0 C=0 n=3 If you want to modify the equation so it does not have a solution, you would have to include the rule that A, B, and C must be whole numbers. Otherwise, irrational numbers and zero allow for solutions.[/quote] THE NUMBERS MASON! WHAT DO THEY MEAN?! *shakes violently*
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[quote]A^n+B^n=C^n n>2 You will not succeed.[/quote] That looks tough.
