Question 758234
It looks like {{{ n = 97 }}} is as high as you can go
because if {{{ n = 98 }}}, then {{{ n + 2 = 100 }}},
which is 3 digits
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I could also state a rule that for any pair of 2-digit numbers,
{{{ n }}} and {{{ n+ 2 }}}, if the 10s digit does not change when
you go from one to the other, then the sum of the digits
of {{{ n + 2 }}} will always be greater than the sum of the
digits of {{{ n }}}.
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That means that if the digits of {{{ n }}} are {{{ t }}} ( for tens )
and {{{ u }}} ( for units ), then only in the cases where
{{{ u = 8 }}} and {{{ u = 9 }}} will he digit sums of {{{ n + 2 }}}
be less than the digit sums of {{{ n }}}.
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So, the required number pairs are:
18, 20
19, 21
28, 30
29, 31
38, 40
39, 41
48, 50
49, 51
58, 60
59, 61
. . . . . 
88, 90
89, 91
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