Question 700129
Let the consecutive even integers  = 
{{{ n }}}, {{{ n + 2 }}}, and {{{ n + 4 }}}
given:
{{{ 2*( n + n + 4 ) = n + 2 + 21 }}}
{{{ 2*( 2n + 4 ) = n + 23 }}}
{{{ 4n + 8 = n + 23 }}}
{{{ 3n = 23 - 8 }}}
{{{ 3n = 15 }}}
{{{ n = 5 }}}
This didn't work because I', going to get
5,7, and 9 which are odd numbers.
I need to call the smallest number 
{{{ 2n }}} to make sure it will be even
My number are {{{ 2n }}}, {{{ 2n + 2 }}}, {{{ 2n + 4 }}}
given:
{{{ 2*( 2n + 2n + 4 ) = 2n + 2 + 21 }}}
{{{ 2*( 4n + 4 ) = 2n + 23 }}}
{{{ 8n + 8  = 2n + 23 }}}
{{{ 6n = 15 }}}
{{{ n = 15/6 }}}
{{{ 2n = 5 }}}
This didn't work either.
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Now that I think about it, the problem has no solution.
It is saying that twice the sum of 2 numbers ( which is even ) is
equal to 21 plus an even number ( which is odd ), and
even cannot equal odd.