Question 312732
<pre><b>
The sum of consecutive integers is an arithmetic series
with d=1 and first term {{{a[1]}}}
 
{{{S[n]=(2a[1] + (n-1)*d)(n/2)}}}
 
Substituting d=1 and {{{S[n]}}}=33
 
{{{33=(2a[1] + (n-1)*1)(n/2)}}}
 
Multiplying through by 2
 
{{{66=(2a[1]+n-1)n}}}
 
{{{66=2a[1]n+n^2-n}}}
 
{{{0=n^2+(2a[1]-1)n-66}}}
 
{{{n^2+(2a[1]-1)n-66=0}}}
 
There are only 4 possible factorizations possible for that
so that the middle term's coefficient will be positive: 
 
1.  (n+66)(n-1)=0 which has middle term 65n
 
So n=1.  We can't have just one "consecutive" integer, for
there has to be at least two for each to be "consecutive" to.
 
2.  (n+33)(n-2)=0 which has middle term 31n

So {{{2a[1]-1=31}}}
   {{{2a[1]=32}}}
   {{{a[1]=16}}}
 
and n=2, and the 2 consecutive integers are 16+17=33

3.  (n+22)(n-3)=0 which has middle term 20n

So {{{2a[1]-1=19}}}
   {{{2a[1]=20}}}
   {{{a[1]=10}}}

and n=3, and the 3 consecutive integers are 10+11+12=33


4.  (n+11)(n-6)=0 which has middle term 5n

So {{{2a[1]-1=5}}}
   {{{2a[1]=6}}}
   {{{a[1]=3}}}

and n=6, and the 6 consecutive integers are 3+4+5+6+7+8=33

So there are three possibilities for n, 2, 3, and 6.

That's all three I, II and III which is choice (E)

Edwin</pre>