Question 175629
Let the tens digit be represented by {{{d[t]}}} and the ones digit be represented by {{{d[o]}}}.  That means you can represent any two digit positive number by {{{10d[t] + d[o]}}} if you restrict {{{d[t]}}} and {{{d[o]}}} to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.


If our original number is {{{10d[t] + d[o]}}} then the number with reversed digits is {{{10d[o] + d[t]}}}.  Then the conditions of the problem give us:


{{{2(10d[t] + d[o]) - 6 = 10d[o] + d[t]}}}


Simplify and solve for {{{d[t]}}}


{{{20d[t] + 2d[o] - 6 = 10d[o] + d[t]}}}


{{{19d[t] = 8d[o] + 6}}}


{{{d[t] = (8d[o] + 6)/19}}}


Now remember that both {{{d[t]}}} and {{{d[o]}}} are restricted to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.  That means that we have to find which of the given set of numbers, when substituted for {{{d[o]}}}, yield {{{d[t]}}} also an element of the given set.  We know that since the denominator is 19 and there are only ten consecutive integer possibilities, we have at most 1 solution.


We can also see that {{{8d[o] + 6}}} must be an integer multiple of 19.


The first five integer multiples of 19 are 19, 38, 57, 76, and 95.  So:


{{{8d[o] + 6 = 19}}} → {{{8d[o] = 12}}} → {{{d[o]}}} not an integer.


{{{8d[o] + 6 = 38}}} → {{{8d[o] = 32}}} → {{{red(d[o] = 4)}}} IS an integer.  This is our solution, but let's continue just to make sure.


{{{8d[o] + 6 = 57}}} → {{{8d[o] = 51}}} → {{{d[o]}}} not an integer.


{{{8d[o] + 6 = 76}}} → {{{8d[o] = 70}}} → {{{d[o]}}} not an integer.


{{{8d[o] + 6 = 95}}} → {{{8d[o] = 89}}} → {{{d[o]}}} not an integer AND {{{d[o] > 9}}} so we can stop looking.


Now we know the ones digit is 4, so using {{{d[t] = (8(4) + 6)/19 = 38/19 = 2}}} we know the tens digit is 2.


The only positive two-digit integer that satisfies the given conditions is 24.


Check the answer:


2 times 24 is 48 which is 6 more than 42 which is 24 with the digits reversed.  Checks.


By the way, the word is integer, not inte<big><b>r</big></b>ger.