Question 1086342
{{{(n(n+1))/2>1000000}}}
{{{n(n+1)>2000000}}}
{{{n^2+n>2000000}}}
{{{n^2+n-2000000>0}}}
We can solve for when it equals zero and then take the next largest integer,
{{{n^2+n-2000000=0}}}
{{{(n^2+n+1/4)=2000000+1/4}}}
{{{(n+1/2)^2=8000001/4}}}
{{{n+1/2=0 +- sqrt(8000001)/2}}}
{{{n=1/2 +- (3sqrt(888889))/2}}}
Only positive n makes sense in this problem,
{{{n=(1+3sqrt(888889))/2}}}
or approximately,
{{{n=1413.71}}}
So the next largest integer is,
{{{N=1414}}}
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Assume the sum does equal 100000,
{{{(n(n+1))/2=10000}}}
{{{n^2+n=200000}}}
{{{n^2+n-200000=0}}}
Similarly,
{{{(n+1/2)^2=200000+1/4}}}
{{{(n+1/2)^2=800001/4}}}
Since the {{{sqrt(800001)}}} is not an integer, there is no hope that the required sum would be an integer. 
So it would never equal 100000.