Question 929114


Let {{{x}}} = one's digit in original number
Let {{{y}}} = ten's digit in original number
value of original number = {{{10y + x}}}

In new number (with reversed digits):
{{{y}}} = one's digit
{{{x}}} = ten's digit
value of new number = {{{10x + y}}}

We can now form these equations:
"Sum of digits of two-digit no. is {{{10}}}" (original number)

{{{y + x = 10}}}...eq.1

"Reverse the digits, the no. is increased by {{{36}}}."

{{{10y + x + 36 = 10x + y}}}

{{{10y - y + x + 36 = 10x + y - y}}}

{{{9y + x + 36 = 10x}}}

{{{9y + x - x + 36 = 10x - x}}}

{{{9y + 36 = 9x}}}

{{{9y/9 + 36/9 = 9x/9}}}

{{{y + 4= x}}}....eq.2

Now compare the two equations:

{{{y + x = 10}}}
{{{y + 4 = x}}}

We can substitute {{{(y + 4)}}} for {{{x}}} in the first equation to solve:

{{{y + (y + 4) = 10}}}

{{{2y + 4= 10}}}

{{{2y + 4 - 4 = 10 - 4}}}

{{{2y = 6}}}

{{{2y/2 = 6/2}}}

{{{highlight(y = 3)}}}

Now substitute this value for {{{y}}} into one of the equations to find {{{x}}}:

{{{y + 4 = x}}}

{{{3 + 4 = x}}}

{{{highlight(x = 7)}}}

Our original number is {{{highlight(37)}}}.