How many positive four-digit integers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one-ninth of $N$?

Question

How many positive four-digit integers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one-ninth of $N$?

I know that if we let the number's digits be $a$, $b$, $c$, and $d$, the number will be $abcd$ (not multiplication). That means that N/9 = $bcd$, so $N = 9 * bcd$. How should I proceed from here?

Answer 1

If $N$ has the form "abcd", and $N/9$ has the form "bcd", then by subtracting $N/9$ from $N$, you can see that $8N/9$ has the form "a000" - in other words (back to actual multiplication!), $8N/9 = 1000a$. Can you use this to get an equation for "bcd", i.e. $N/9$, in terms of $a$? What possibilities does this leave?

Answer 2

It is $(abcd)=9(bcd)$.

The units digit of $9d$ must be $d$, so $d=0;5$. We have $A) (abc0)=9(bc0)$ or $B) (abc5)=9(bc5)$.

$A)$ The units digit of $9c$ must be $c$, so $c=0;5$. We have $1) \ (ab00)=9(b00)$ or $2) \ (ab50)=9(b50)$.

$1)$ The units digit of $9b=b$, so $b=5$. We have $\color{red}{(4500)=9(500)}$.

$2)$ The units digit of $9b+4$ must be $b$, so $b=2;7$. We have $\color{red}{(2250)=9(250)}$ and $\color{red}{(6750)=9(750)}$.

$B)$ The units digit of $9c+4$ must be $c$, so $c=2;7$. We have $1) \ (ab25)=9(b25)$ or $2) \ (ab75)=9(b75)$.

$1)$ The units digit of $9b+2$ must be $b$, so $b=1;6$. We have $\color{red}{(1125)=9(125)}$ and $\color{red}{(5675)=9(675)}$.

$2)$ The units digit of $9b+6$ must be $b$, so $b=3;8$. We have $\color{red}{(3375)=9(375)}$ and $\color{red}{(7875)=9(875)}$.