Question 1032955
<pre><b>
The sum of the first n integers is {{{expr(n/2)*(1+1(n-1)1)}}} or {{{(n^2+n)/2}}}

Let the sum of the two integers that Alec skipped be s<sub>1</sub>.
Let the sum of the two integers that Beau double counted be s<sub>2</sub>.

Then we have these two equations:

{{{system((n^2+n)/2-s[1]=92,(n^2+n)/2+2s[2]=145)}}}

Subtracting the first equation from the 2nd:

{{{2s[2]+s[1]=53}}} or {{{s[1]=53-2s[2]}}}

Adding the two equations:

{{{n^2+n-s[1]+2s[2]=237}}}

Substituting {{{s[1]=53-2s[2]}}}

{{{n^2+n-(53-2s[2])+2s[2]=237}}}

which simplifies to

{{{n^2+n+4s[2]-290=0}}}

The discriminant must be a perfect square.

discriminant = {{{1-4(4s[2]-290)}}}

which simplifies to

{{{1161-16s2}}}

Now we use our TI-83 or TI-84 calculator to
find feasible value(s) of s<sub>2</sub> that
will cause {{{1161-16s2}}} to be a perfect square.

Press the Y= key and put {{{sqrt(1161-16X)}}} after \Y<sub>1</sub>

Press 2ND WINDOW (TBLSET) and set TblStart=0 and &#916;Tbl=1,
highlight the two Auto's

Press 2ND GRAPH (TABLE)

Use the down arrow key to scroll down until you see
an integer in the Y<sub>1</sub> column.

The first place we find one is when X=20 and Y<sub>1</sub>=29

[ We find one again when X=27 and Y<sub>1</sub>=27, also
when X=45 and Y<sub>1</sub>=21, X=50 and Y<sub>1</sub>=19, etc.,
but we will show that these are not possible.]

For now, let's go with the X=20 and Y<sub>1</sub>=29.

The X on the calculator is our s<sub>2</sub>.

So if we take s<sub>2</sub> = 20

Substituting in {{{s[1]=53-2s[2]}}}, we get {{{s[1]=13}}}.
Substituting in {{{n^2+n-s[1]+2s[2]=237}}}, we get n<sup>2</sup>+n-210=0
or (n-14)(n+15)=0 and the only solution n=14

So they raced up 14 term:  

1+2+3+4+5+6+7+8+9+10+11+12+13+14 = {{{(14^2+14)/2}}} = 105

They would have gotten 105 if they had simply added them all
once each.

Since Alec skipped two integers with sum 13, he only got 105-13 = 92

Since Beau doubled two integers with sum 20, he only got 105+40 = 145.

It would not matter what two integers they skipped or double-counted,
as long as Alec skipped two that had sum 13 and Beau double counted
two that had sum 20.

[We can rule out the other numbers we found on the calculator
since {{{s[1]=53-2s[2]}}}, {{{s[1]>0}}} implies {{{s[2]<26.5}}}]

Edwin</pre></b>