Question 677032
If the units digit is {{{ a }}} and
the tens digit is {{{ b }}}, the the
value of the number is {{{ a + 10b }}}
given:
{{{ a + 10b = b + 10a + 11 }}}
{{{ ( 10b - b ) - ( 10a - a ) = 11 }}}
{{{ ( 10 - 1 )*b - ( 10 - 1 )*a = 11 }}}
{{{ 9b - 9a = 11 }}}
{{{ 9*( b - a ) = 11 }}}
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I don't think it's true for any 2-digit numbers
{{{ b - a }}} must be a whole number and not
a fraction, so
{{{ b - a = ( 9k ) / 9 }}} where {{{ k }}} can
be 0 to 9 . Note that if {{{ k = 10 }}}, then
{{{ ( 9*10 ) / 9 = 10 }}}, but the difference of 
2 digits can't be {{{ 10 }}}
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and also, {{{ 9k }}} can never be {{{ 11 }}}