document.write( "Question 1032955: Alec and Beau race to compute 1 + 2 + 3 + ХХХ + n. Alec skips two numbers and gets a sum of 92. Beau double-counts two numbers and gets a sum of 145. What is the value of n? \n" ); document.write( "

Algebra.Com's Answer #647605 by AnlytcPhil(1807) $\"\"$ $\"About$

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\n" ); document.write( "The first solution above is correct, however it is too
\n" ); document.write( "complicated and also involves the use of calulator
\n" ); document.write( "to do lots of trial and error. The second only narrows
\n" ); document.write( "down the potential solutions to 14,15 and 16. He states
\n" ); document.write( "that it is either 14 or 15, but doesn't rule out 16.
\n" ); document.write( "Here is the best complete solution:
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\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "Let the answer be the counting number n.
\n" ); document.write( "The sum of the first n counting numbers is $\"n%28n%2B1%29%2F2\"$ .
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\n" ); document.write( "\n" ); document.write( "The 92 that Alex got was between
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\n" ); document.write( "\n" ); document.write( "the sum of the counting numbers minus the sum of the two
\n" ); document.write( "smallest possible counting numbers 1 and 2, which is 3
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\n" ); document.write( "\n" ); document.write( "and
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\n" ); document.write( "\n" ); document.write( "the sum of the counting numbers minus the sum of the largest two
\n" ); document.write( "possible counting numbers n and n-1, which is 2n-1.
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\n" ); document.write( "\n" ); document.write( " $\"n%28n%2B1%29%2F2-3%3C=92%3C=n%28n%2B1%29%2F2-%282n-1%29\"$ \r
\n" ); document.write( "\n" ); document.write( "I went through, but won't go through here, the solution to that
\n" ); document.write( "inequality but its solution is
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\n" ); document.write( "\n" ); document.write( " $\"%281%2Bsqrt%28365%29%29%2F2%3C=n%3C=%28sqrt%28761%29-1%29%2F2\"$
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\n" ); document.write( "\n" ); document.write( "or approximately:
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\n" ); document.write( "\n" ); document.write( " $\"13.2931%3C=n%3C=15.0739\"$
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\n" ); document.write( "\n" ); document.write( "So n is either 14 or 15.
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\n" ); document.write( "\n" ); document.write( "The sum of the first 14 counting numbers is
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\n" ); document.write( "\n" ); document.write( " $\"14%2A15%2F2=105\"$ which is odd.
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\n" ); document.write( "\n" ); document.write( "The sum of the first 15 counting numbers is
\n" ); document.write( " $\"15%2A16%2F2=120\"$ which is even. Since Beau
\n" ); document.write( "double-counted two counting numbers he in effect
\n" ); document.write( "added an even number to the sum of the first n
\n" ); document.write( "counting numbers. Therefore since he got 145, an odd
\n" ); document.write( "number, he could only have in effect added an
\n" ); document.write( "even number to an odd sum of the first n
\n" ); document.write( "counting numbers. Hence n=14 is the ONLY solution.
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\n" ); document.write( "\n" ); document.write( "Phil
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