document.write( "Question 3461: The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x =\r
\n" ); document.write( "\n" ); document.write( " (A) 4 (B) 5 (C) 6 (D) 7 (E) 8
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Algebra.Com's Answer #1533 by drglass(89)\"\" \"About 
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To do this problem it is helpful to know that the sum of the numbers from 1 to x is \"%28n%28n+%2B+1%29%29%2F2\". This means the sum of the page numbers from 1 to x is:

\n" ); document.write( "\"%28x%28x+%2B+1%29%29%2F2\".

\n" ); document.write( "To find the remaining sum, think about it this way \"5+%2B+4+=+1+%2B+2+%2B+3+%2B+4+%2B+5+-+%281+%2B+2+%2B+3%29\". In other words the sum from (x+1) to 10 is the sum of 1 to 10 minus the sum of 1 to x. This allows us to write a relatively simple expression for the sum from (x+1) to 10

\n" ); document.write( "\"%2810%2810+%2B+1%29%29%2F2+-+%28x%28x+%2B+1%29%29%2F2\"

\n" ); document.write( "Since the sum from one to x is 1 plus the sum of x+1 to 10, we get the following equation

\n" ); document.write( " or

\n" ); document.write( "\"%28x%28x+%2B+1%29%29%2F2+=+56++-+%28x%28x+%2B+1%29%29%2F2\"

\n" ); document.write( "If we add \"%28x%28x+%2B+1%29%29%2F2\" to both sides we get

\n" ); document.write( "\"x%28x+%2B+1%29+=+56\"

\n" ); document.write( "The equation becomes \"x%5E2+%2B+x+-+56+=+0\" which factors nicely to \"%28x+-+7%29%28x+%2B+8%29+=+0\". So the solution to the equation is x = 7 and x = -8. Well, you cannot have negative pages, so the answer to the question must be 7. Let's check it.
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\n" ); document.write( "the sum from 1 to 7 is \"%287%287%2B1%29%29%2F2+=+28\" and \"8+%2B+9+%2B+10+=+27\", which is one more than the sum from 1 to 7. So the answer is indeed 7.
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