Question 778125

let {{{3}}} consecutive integers be {{{x}}},{{{x+1}}},and {{{x+2}}}

 if their sum is between {{{95}}} and {{{115}}}, then we have

{{{95<x+(x+1)+(x+2)<115}}}......solve for {{{x}}}

{{{95<x+x+1+x+2<115}}}

{{{95<3x +3<115}}}

{{{95-3<3x +3-3<115-3}}}

{{{92<3x<112}}}

{{{92/3<3x/3<112/3}}}

{{{30.67<x<37.33}}}

so, {{{x}}} could be any integer between {{{31}}} and {{{36}}}

if {{{x=31}}}, then {{{x+1=32}}},and {{{x+2=33}}}

or if {{{x=34}}}, then {{{x+1=35}}},and {{{x+2=36}}}