<PRE><B>Question 1032955</B>
&lt;font face = &quot;arial&quot;&gt;
The first solution above is correct, however it is too 
complicated and also involves the use of calulator 
to do lots of trial and error.  The second only narrows
down the potential solutions to 14,15 and 16.  He states 
that it is either 14 or 15, but doesn&#39;t rule out 16.
Here is the best complete solution:&lt;br&gt;

&lt;b&gt;
Let the answer be the counting number n.
The sum of the first n counting numbers is {{{n(n+1)/2}}}.&lt;br&gt;

The 92 that Alex got was between&lt;br&gt; 

the sum of the counting numbers minus the sum of the two 
smallest possible counting numbers 1 and 2, which is 3&lt;/br&gt; 

and&lt;br&gt; 

the sum of the counting numbers minus the sum of the largest two
possible counting numbers n and n-1, which is 2n-1.&lt;br&gt;

{{{n(n+1)/2-3&lt;=92&lt;=n(n+1)/2-(2n-1)}}}

I went through, but won&#39;t go through here, the solution to that 
inequality but its solution is&lt;br&gt;

{{{(1+sqrt(365))/2&lt;=n&lt;=(sqrt(761)-1)/2}}}&lt;br&gt;

or approximately:&lt;br&gt;

{{{13.2931&lt;=n&lt;=15.0739}}}&lt;br&gt;

So n is either 14 or 15.&lt;br&gt;

The sum of the first 14 counting numbers is&lt;br&gt; 

{{{14*15/2=105}}} which is odd.&lt;br&gt;

The sum of the first 15 counting numbers is 
{{{15*16/2=120}}} which is even.  Since Beau 
double-counted two counting numbers he in effect 
added an even number to the sum of the first n 
counting numbers. Therefore since he got 145, an odd 
number, he could only have in effect added an 
even number to an odd sum of the first n 
counting numbers.  Hence n=14 is the ONLY solution.&lt;br&gt;

Phil&lt;/b&gt;&lt;/font&gt;
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