Question 567957
Call the integers {{{ n }}}, {{{ n+1 }}}, and {{{ n+2 }}}
given:
{{{ 11*( n+2 ) = n*( n+1 ) + 46 }}}
{{{ 11n + 22 = n^2 + n + 46 }}}
{{{ n^2 - 10n + 24 = 0 }}}
complete the square:
{{{ n^2 - 10n + (-10/2)^2 = -24 + (-10/2)^2 }}}
{{{ ( n - 5 )^2 = -24 + 25 }}}
{{{ n - 5 = 1 }}}
{{{ n = 6 }}}
also,
{{{ n - 5 = -1 }}}
{{{ n = 4 }}}
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If the numbers are:
{{{ n = 6 }}}
{{{ n +1 = 7 }}}
{{{ n + 2 = 8 }}}
{{{ 11*( n+2 ) = n*( n+1 ) + 46 }}}
{{{ 11*8 = 6*7 + 46 }}}
{{{ 88 = 42 + 46 }}}
{{{ 88 = 88 }}}
OK
If the numbers are:
{{{ n = 4 }}}
{{{ n + 1 = 5 }}}
{{{ n + 2 = 6 }}}
{{{ 11*( n+2 ) = n*( n+1 ) + 46 }}}
{{{ 11*6 = 4*5 + 46 }}}
{{{ 66 = 20 + 46 }}}
{{{ 66 = 66 }}}
OK
The consecutive integers can be:
6,7, and 8
or
4,5, and 6