Question 1080520
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Let {{{a}}} be the first digit, and let {{{b}}} be the second digit.
The value of the number is {{{10a+b}}} .
The value of the number obtained by reversing its digits is {{{10b+a}}} .
Their sum is
{{{10a+b+10b+a=11a+11b=11(a+b)}}} .
For that sum to be the square of an integer, it must be true that
{{{a+b=11}}} .
There are {{{4}}} possible pastors of digits adding to {{{11}}} :
{{{2+9=11}}} ,
{{{3+8=11}}} ,
{{{4+7=11}}} , and
{{{5+6=11}}} .s
Since we can make two 2-digit number out of each paiR
by changing the other of the digits,
there are {{{2*4=8}}} numbers that could be
the original 2-digit number in the problem.