Question 915250
The units digit of a two digit number is 5 more than the tens digit. If the digits are reversed and the new number is divided by the original, the qoutient is 2 and the remainder is 7. What is the original number?
***
let u=units digit
let t=tens digit
u=t+5
original number=10t+u
new number=10u+t
{{{(10u+t)/(10t+u)=2+(7/(10t+u))}}}
{{{(10(t+5)+t)/(10t+t+5)=2+(7/(10t+t+5))}}}
{{{(10t+50+t)/(10t+t+5)=2+(7/(10t+t+5))}}}
{{{(11t+50)/(11t+5)=2+(7/(11t+5))}}}
{{{(11t+50)/(11t+5)=(22t+17)/(11t+5))}}}
(11t+5)(22t+17)=(11t+50)(11t+5)
242t^2+297t+85=121t^2+605t+250
121t^2-308t-165=0
solve for t by quadratic formula:
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=121, b=-308, c=165
ans:
t=-55/121≈-0.4545(reject)
or
t=3
u=t+5=8
Check:
original number=38
new number=83
new number/original number=83/38=2+Remainder=7