Question 150606
Let {{{u}}} = the units digit
Let {{{t}}} = the tens digit
{{{u = t - 3}}}
{{{t - u = 3}}}
The sum of the digits is {{{t + u}}}
The actual number is {{{10t + u}}}
{{{(10t + u) / (t + u) = 6 + 8 / (t + u)}}}
If the way I wrote the remainder bothers you, think about
{{{19 / 11 = 1 + (8/11)}}} It's really the same thing
continuing,
Multiply each side by {{{t + u}}}
{{{10t + u = 6*(t + u) + 8}}}
{{{10t + u = 6t + 6u + 8}}}
(1) {{{4t - 5u = 8}}}
And, from above,
{{{t - u = 3}}}
Multiply times {{{4}}}
(2) {{{4t - 4u = 12}}}
Subtract (1) from (2)
{{{u = 4}}}
{{{t - u = 3}}}
{{{t - 4 = 3}}}
{{{t = 7}}}
The number is 74
check:
The units digit is 3 less than the tens digit OK
The number divided by the sum of it's digts is
{{{74 / 11 = 6 + (8/11)}}} OK