Question 1171228




There are three children in a family. Each of their ages is a prime number. 

let their ages be {{{a}}}, {{{b}}}, and {{{c}}}

if the sum of their ages is {{{41}}}, we have
{{{a+b+c=41}}}....eq.1

 and if at least two of the children have ages that differ by {{{16}}}, we have
{{{b-c=16}}}

listing all the primes of {{{0}}} to {{{41}}} we get:

{{{2}}},{{{3}}},{{{5}}},{{{7}}},{{{11}}},{{{13}}},{{{17}}},{{{19}}},{{{23}}},{{{29}}},{{{31}}},{{{37}}},{{{41}}}

let’s separate primes who differ by {{{16}}}

{{{3}}},{{{19}}}
{{{7}}},{{{23}}}

let’s consider this set as {{{b}}},{{{c}}}:

{{{a=41-(b+c)}}} ......substitute {{{19}}},{{{3}}} as {{{b}}},{{{c}}}
{{{a=41-(19+3)}}} 
{{{a=41-22}}}
{{{a=19}}}

hence, {{{a=19}}}, {{{b=19}}},{{{c=3}}}

{{{a=41-(b+c)}}} ......substitute {{{23}}},{{{7}}} as {{{b}}},{{{c}}}
{{{a=41-(23+7)}}}
{{{a=41-30}}}
{{{a=11}}}

hence, {{{a=11}}}, {{{b=23}}}, {{{c=7}}}