Question 990357
It is not always so. It works half of the time. It all depends on the digits.


If you start with 67.89, it does not work.
The first swap gives you 89.67;
the first sum is 157.56, which reversed is 156.57,
and the second sum is 314.13.


If you start with 34.25, it does not work either.
The first swap gives you 25.34;
the first sum is 59.59, which reversed is also 59.59,
and the second sum is 119.18.


It worked for your example.


If you start with 14.26, it also works.
The first swap gives you 26.14;
the first sum is 40.40, which reversed is also 40.40,
and the second sum is 80.80.


{{{p}}}= number of whole pounds, with {{{p<=99}}}
{{{c}}}= number of cents, with {{{c<=99}}}
The amount value, in cents, is {{{100p+c}}} .
Swapping the numbers of pounds and cents, we have a value of {{{100c+p}}} .
Adding both values, we get {{{100p+c+100c+p=100(p+c)+(p+c)}}} .
From then on, it all depends on the value of {{{p+c}}} .


If {{{p+c<=49<99}}}, it works.
For the first sum,
{{{p+c}}} is the two-digit number that makes up the tens and ones, and
{{{p+c}}} is also the two-digit number that makes up the cents (the tenths and hundredths).
Then, swapping the tens and ones with tenths and hundredths gives you the same number.
Adding swapped and unswapped, you get
{{{100(p+c)+(p+c)+100(p+c)+(p+c)=100(2p+2c)+(2p+2c)}}} .
Since {{{p+c<=49}}} , {{{2p+2c<=98}}} ,
{{{2p+2c}}} is a two-digit number that makes up the tens and ones, and
{{{2p+2c}}} is also the two-digit number that makes up the cents (the tenths and hundredths).


If {{{50<=p+c<=99}}}, as in the case of 34.25, it does not work.
For the first sum,
{{{p+c}}} is a two-digit number that makes up the tens and ones, and
{{{p+c}}} is also the two-digit number that makes up the cents (the tenths and hundredths).
Then, swapping the tens and ones with tenths and hundredths gives you the same number.
Adding swapped and unswapped, you get
{{{100(p+c)+(p+c)+100(p+c)+(p+c)=100(2p+2c)+(2p+2c)}}} .
Since {{{50<=p+c<=99}}} , {{{100<=2p+2c<=198}}} ,
{{{2p+2c=100+(2p+2c-100)}}} is a three-digit number,
but {{{(2p+2c-100)<=98}}} and is a two-digit number.
The second sum is
{{{100(2p+2c)+(2p+2c)=100(100+(2p+2c-100))+100+(2p+2c-100)=10000+100(2p+2c-100)+100+(2p+2c-100)=10000+100(2p+2c-100+1)+(2p+2c-100)}}} .
{{{(2p+2c-100)}}} is a two-digit number that represents the cents,
{{{(2p+2c-100+1) is the two-digit number that makes up the tens and ones,
and the {{{10000}}} cents represents the {{{1}}} in the hundreds place.


If {{{100<=p+c<=149}}}, as in the case of 83.64, it works.
The first sum is
{{{100(p+c)+(p+c)=100(100+(p+c-100))+100+(p+c-100)=10000+100(p+c-100)+100+(p+c-100)=10000+100(p+c-100+1)+(p+c-100)}}} , where
{{{p+c-100}}} is the two-digit number that makes up the cents (the tenths and hundredths),
{{{p+c-100+1}}} is the two-digit number that makes up the tens and ones,
and the {{{10000}}} cents represents the {{{1}}} in the hundreds place.
Swapping the tens and ones with tenths and hundredths yields
{{{10000+100(p+c-100)+(p+c-100+1)}}} cents.
Adding swapped and unswapped we get (in cents)
{{{10000+100(p+c-100)+(p+c-100+1)+10000+100(p+c-100+1)+(p+c-100) =20000+100(2p+2c-200+1)+(2p+2c-200+1)}}} .
Since {{{100<=p+c<=149}}} , {{{200<=2p+2c<=298}}} ,
and {{{1<=2p+2c-200+1<=99}}} is the two digit number that
makes up the tens and ones, and also makes up the tenths and hundredths.


If {{{150<=p+c<=198}}} , as in the case of 67.89, it does not work.
The first sum is
{{{10000+100(p+c-100+1)+(p+c-100)}}} , where
{{{p+c-100}}} is the two-digit number that makes up the cents (the tenths and hundredths),
{{{p+c-100+1}}} is the two-digit number that makes up the tens and ones,
and the {{{10000}}} cents represents the {{{1}}} in the hundreds place.
Swapping the tens and ones with tenths and hundredths yields
{{{10000+100(p+c-100)+(p+c-100+1)}}} cents.
Adding swapped and unswapped we get (in cents)
{{{20000+100(2p+2c-200+1)+(2p+2c-200+1)}}} .
However, since {{{150<=p+c<=198}}},
{{{300<=2p+2c<=396}}} , and {{{101<=2p+2c-200+1<=197}}} ,
so the second sum is
{{{20000+100(2p+2c-200+1)+(2p+2c-200+1)=
20000+100(100+(2p+2c-300+1))+100+(2p+2c-300+1)=
20000+10000+100(2p+2c-300+1)+100+(2p+2c-300+1)}}}
={{{30000+100(2p+2c-300+2)+(2p+2c-300+1)}}} , where
{{{(2p+2c-300+1)}}} is the two-digit number that makes up the cents (the tenths and hundredths),
{{{(2p+2c-300+1)}}} is the two-digit number that makes up the tens and ones,
and the {{{30000}}} cents represents the {{{3}}} in the hundreds place.