There are two distinct 2-digit numbers which have the same units digit but different tens digits. The quotient when one of them is divided by 9 is equal to the remainder when the other is divided by 9, and vice versa. What is the common units digit?
I got two equations for above question :
((10y+x)/9)=((10z+x) mod 9) ---> equation 1
&
((10z+x)/9)=((10y+x) mod 9) ---> equation 2
But, I am stuck at how to continue solving these two equations, Please advise.
You have $10y+x = 9q+r$ and $10z+x = 9r+q$ for some integers $r$ and $q$. Multiply the second equation by $9$ to get
$$90z+9x=81r+9q.$$
Solve the first equation for $9q$ and plug it in the above:
$$90z+9x = 81r+10y+x-r$$
and simplify
$$45z +4x = 40r + 5y.$$
Solve for $4x$, and we see that $4x$ has to be a multiple of $5$. So $x=5$ or $0$.
If $x=0$, then $10y \pmod{9} = 10z \pmod 9$ so both numbers have the same remainder, and therefore the same quotient, upon division by $9$, which means they're the same number. But the numbers are supposed to be distinct, so we much have $c=5$.