document.write( "Question 1039608: Prove that one of every three consecutive positive numbers is divisible by 3
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Algebra.Com's Answer #654372 by addingup(3677) $\"\"$ $\"About$

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Let 3 consecutive positive integers be n, n+1 and n+2
\n" ); document.write( "Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.
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\n" ); document.write( "Therefore:
\n" ); document.write( "n = 3p or 3p+1 or 3p+2, where p is some integer
\n" ); document.write( "If n = 3p, then n is divisible by 3
\n" ); document.write( "If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3
\n" ); document.write( "If n = 3p+2, then n+1 = 3p+2+1 = 3p+3 = 3(p+1) is divisible by 3
\n" ); document.write( "Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3 \n" ); document.write( "

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