Question 1045027
{{{red(EDITED)}}} {{{red(SOLUTION)}}} (Sorry about the typo):
We are told that
{{{number[1]=divisor*quotient[1]+11}}} , and
{{{number[2]=divisor*quotient[2]+21}}} .
So, adding up those two equations, we get
{{{number[1]+number[2]=divisor*quotient[1]+divisor*quotient[2]+11+21}}}
{{{number[1]+number[2]=divisor*(quotient[1]+quotient[2])+32}}}

Since we are also told that
{{{number[1]+number[2]=divisor*quotient[3]+4}}} ,
subtracting from the equation above from
{{{number[1]+number[2]=divisor*(quotient[1]+quotient[2])+32}}}
we get
{{{0=divisor*(quotient[3]-quotient[1]-quotient[2])+4-32}}}
{{{0=divisor*(quotient[3]-quotient[1]-quotient[2])-28}}}
{{{28=divisor*(quotient[3]-quotient[1]-quotient[2])}}}
Since all the numbers involved in that equation are integers,
that tells us that {{{divisor}}} is a factor of {{{28}}} ,
and it has to be a factor larger than {{{21}}} ,
because {{{red(cross(28))}}}{{{21}}} was a remainder,
and remainders are always smaller than their divisor.
The only factor of {{{28}}} that is greater than {{{21}}} is {{{28}}} ,
so {{{highlight(28)}}} is the divisor.