Question 1204310
You want {{{n }}}and {{{r }}}such that

{{{105=n+(n+1)+(n+2)}}}+.....+{{{(n+r)}}}

={{{(r+1)n}}}+({{{1+2}}}+.....+{{{r}}})

={{{(r+1)n+r(r+1)/2=((r+1)(2n+r))/2}}},


which is equivalent to {{{(r+1)(2n+r)=210=2*3*5*7}}}
.

Notice that, obviously, {{{r+1< 2n+r}}}, therefore {{{(r+1)^2 <(r+1)(2n+r)=105}}}, whence {{{r+1<= abs(sqrt(210))=14}}}, so 

{{{r+1}}}∈{ {{{2}}},{{{3}}},{{{5}}},{{{2*3}}},{{{7}}},{{{2*5}}},{{{2*7}}} }...(i.e. all the divisors of {{{105}}} that are ≤{{{14}}}- a total of {{{highlight(7) }}}cases).


here they are:


1. If {{{r+1=2}}} then {{{2n+r=105}}}; it follows that {{{r=1}}} and {{{n=52}}}, so you have {{{105=52+53}}}
.2  If {{{r+1=3}}} then {{{2n+r=70}}}; it follows that {{{r=2}}} and {{{n=34}}}, so you have {{{105=34+35+36}}}
.
3. If {{{r+1=5}}} then 2{{{n+r=42}}}; it follows that {{{r=4}}} and{{{ n=19}}}, so you have {{{105=19+20+21+22+23}}}
.
4. If {{{r+1=6}}} then {{{2n+r=35}}}; it follows that{{{ r=5}}} and {{{n=15}}}, so you have {{{105=15+16+17+18+19+20}}}
.
5 If {{{r+1=7 }}}then {{{2n+r=30}}}; it follows that {{{r=6}}} and {{{n=12}}}, so you have
{{{105=12+13+14+15+16+17+18}}}
.
6. If {{{r+1=10 }}}then {{{2n+r=21}}}; it follows that {{{r=9}}} and{{{ n=6}}}, so you have {{{105=6+7+8+9+10+11+12+13+14+15}}}
.
7. If {{{r+1=14}}} then {{{2n+r=15}}}; it follows that {{{r=13}}} and {{{n=1}}}, so you have {{{105=1+2+3+4+5+6+7+8+9+10+11+12+13+14}}}
.