Question 696128
{{{a}}} = first (tens) digit
{{{b}}} = second (ones) digit
The number looks like ab,
but its value is {{{10a+b}}}.
 
The number formed when the digits are reversed looks like ba,
and its value is {{{10b+a}}} , which is larger.
The reversal of the digits increases the value by 1/5 to
{{{1+1/5=6/5}}} of the original value,
making the reverse number equal to {{{6/5}}} times the original number.
So {{{10b+a=(6/5)(10a+b)}}}
 
First we simplify:
{{{10b+a=(6/5)(10a+b)}}} --> {{{5(10b+a)=5(6/5)(10a+b)}}} --> {{{5(10b+a)=6(10a+b)}}} --> {{{50b+5a=60a+6b}}}
 
Next we solve:
{{{50b+5a=60a+6b}}} --> {{{50b+5a-5a-6b=60a+6b-54-6b}}} --> {{{44b=55a}}} --> {{{44b/11=55a/11}}} --> {{{highlight(4b=5a)}}}
 
Since {{{a}}} and {{{b}}} are integers,
for {{{4b}}} to be the multiple of {{{5}}} indicated by {{{5b}}},
{{{b}}} must be a multiple of {{{5}}}.
Since {{{b}}} is a digit and a multiple of {{{5}}}, it can only be {{{highlight(b=5)}}}.
So {{{4*5=5a}}} --> {{{4*5/5=5a/5}}} --> {{{highlight(a=4)}}},
and the original two-digit number is {{{highlight(45)}}}.