document.write( "Question 699138: I must find, by a method other than induction, that 1+2^(6K+1)is divisible by 3. \n" ); document.write( "

Algebra.Com's Answer #431387 by KMST(5328) $\"\"$ $\"About$

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I can prove that even more similarly defined numbers are divisible by 3.
\n" ); document.write( "If you (or the teacher do not like a generalization, you can always adapt the proof to just the subset of $\"1%2B2%5E%286K%2B1%29\"$ numbers.
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\n" ); document.write( "All numbers of the form $\"1%2B2%5E%282n%2B1%29\"$ are divisible by 3.
\n" ); document.write( "Some of them (but not all) are of the form $\"1%2B2%5E%286K%2B1%29\"$ , if $\"n=3K\"$ ,
\n" ); document.write( "as for $\"1%2B2%5E%282%2A3%2B1%29=1%2B2%5E7=1%2B2%5E%286%2A1%2B1%29\"$ (with $\"n=3\"$ , or $\"K=1\"$ )
\n" ); document.write( " $\"1%2B2%5E7=1%2B128=129=3%2A43\"$ .
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\n" ); document.write( " $\"2%5E2=4=3%2B1\"$ so

.
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+ ... + $\"%28n%28n-1%29%2F2%29%2A3%5E2%2Bn%2A3%2B1%5En=3\"$ [ $\"3%5E%28n-1%29%2Bn%2A3%5E%28n-2%29%2B%28n%28n-1%29%2F2%29%2A3%5E%28n-3%29\"$ + ... + $\"%28n%28n-1%29%2F2%29%2A3%2Bn\"$ ]+ $\"1=3M%2B1\"$ , calling the messy bracket $\"M\"$ .
\n" ); document.write( "The conclusion so far is (in words) the even powers of 2, when divided by 3, have 1 as the remainder.
\n" ); document.write( "In formulas: $\"2%5E%282n%29=3%2AM%2B1\"$ , where $\"M\"$ is a whole number.
\n" ); document.write( " $\"2%5E%282n%2B1%29=+2%5E%282n%29%2A2%5E1=2%5E%282n%29%2A2=%283M%2B1%29%2A2=6M%2B2\"$
\n" ); document.write( " $\"1%2B2%5E%282n%2B1%29=1%2B6M%2B2=6M%2B3=3%282M%2B1%29\"$
\n" ); document.write( "Since $\"1%2B2%5E%282n%2B1%29\"$ is $\"3\"$ times the whole number $\"%282M%2B1%29\"$ ,
\n" ); document.write( " $\"1%2B2%5E%282n%2B1%29\"$ is divisible by $\"3\"$ . \n" ); document.write( "

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