Question 235861
<pre>
To have a product 0, one of the 15 consecutive whole numbers 
has to be 0. And since the amallest whole number is 0, the 15 
consecutive whole numbers are 0,1,2,3,...,14 and

{{{0+1+2+3+4+5+6+7+8+9+10+11+12+13+14 = 105}}}.

So that's not just the GREATEST possible sum, it's also the
LEAST and the ONLY possible sum of 15 consecutive whole numbers
whose product is 0.

Maybe your teacher wanted you to find that sum by formula and
not by adding directly.  If so the formula is

{{{S[n]=(n(2a[1] + (n-1)d))/2}}}

with {{{a[1]=0}}}, {{{d=1}}}, {{{n=15}}}

{{{S[15]=(15(2*0+(15-1)*1))/2}}}

{{{S[15]=(15(0+14*1))/2}}}

{{{S[15]=(15(0+14))/2}}}

{{{S[15]=(15(14))/2}}}

         {{{7}}}
{{{S[15]=(15(cross(14)))/cross(2)}}}

{{{S[15]=15*7}}}

{{{S[15]=105}}}

Edwin</pre>